# What Is a Matrix Norm?

A matrix norm is a function $\|\cdot\| : \mathbb{C}^{m\times n} \to \mathbb{R}$ satisfying

• $\|A\| \ge 0$ with equality if and only if $A=0$ (nonnegativity),
• $\|\alpha A\| =|\alpha| \|A\|$ for all $\alpha\in\mathbb{C}$, $A\in\mathbb{C}^{m\times n}$ (homogeneity),
• $\|A+B\| \le \|A\|+\|B\|$ for all $A,B\in\mathbb{C}^{m\times n}$ (the triangle inequality).

These are analogues of the defining properties of a vector norm.

An important class of matrix norms is the subordinate matrix norms. Given a vector norm on $\mathbb{C}^n$ the corresponding subordinate matrix norm on $\mathbb{C}^{m\times n}$ is defined by

$\notag \|A\| = \displaystyle\max_{x\ne0} \frac{ \|Ax\| }{ \|x\| },$

or, equivalently,

$\|A\| = \displaystyle\max_{\|x\| = 1} \|Ax\|.$

For the $1$-, $2$-, and $\infty$-vector norms it can be shown that

\notag \begin{alignedat}{2} \|A\|_1 &= \max_{1\le j\le n}\sum_{i=1}^m |a_{ij}|, &&\qquad \mbox{max column sum'',} \\ \|A\|_{\infty} &= \max_{1\le i\le m}\sum_{j=1}^n |a_{ij}|, &&\qquad \mbox{max row sum'',} \\ \|A\|_2 &= \sigma_{\max}(A), &&\qquad \mbox{spectral norm,} \end{alignedat}

where $\sigma_{\max}(A)$ denotes the largest singular value of $A$. To remember the formulas for the $1$– and $\infty$-norms, note that 1 is a vertical symbol (for columns) and $\infty$ is a horizontal symbol (for rows). For the general Hölder $p$-norm no explicit formula is known for $\|A\|_p$ for $p\ne 1,2,\infty$.

An immediate implication of the definition of subordinate matrix norm is that

$\notag \|AB\| \le \|A\| \|B\|$

whenever the product is defined. A norm satisfying this condition is called consistent or submultiplicative.

Another commonly used norm is the Frobenius norm,

$\notag \|A\|_F = \biggl( \displaystyle\sum_{i=1}^m \sum_{j=1}^n |a_{ij}|^2 \biggr)^{1/2} = \bigl( \mathrm{trace}(A^*\!A) \bigr)^{1/2}.$

The Frobenius norm is not subordinate to any vector norm (since $\|I_n\|_F = n^{1/2}$, whereas $\|I_n\| = 1$ for any subordinate norm), but it is consistent.

The $2$-norm and the Frobenius norm are unitarily invariant: they satisfy $\|UAV\| = \|A\|$ for any unitary matrices $U$ and $V$. For the Frobenius norm the invariance follows easily from the trace formula.

As for vector norms, all matrix norms are equivalent in that any two norms differ from each other by at most a constant. This table shows the optimal constants $\alpha_{pq}$ such that $\|A\|_p \le \alpha_{pq} \|A\|_q$ for $A\in\mathbb{C}^{m\times n}$ and the norms mentioned above. Each inequality in the table is attained for some $A$.

For any $A\in\mathbb{C}^{n\times n}$ and any consistent matrix norm,

$\notag \rho(A) \le \|A\|, \qquad\qquad (1)$

where $\rho$ is the spectral radius (the largest absolute value of any eigenvalue). To prove this inequality, let $\lambda$ be an eigenvalue of $A$ and $x$ the corresponding eigenvector (necessarily nonzero), and form the matrix $X=[x,x,\dots,x] \in \mathbb{C}^{n\times n}$. Then $AX = \lambda X$, so $|\lambda| \|X\| = \|AX\| \le \|A\| \, \|X\|$, giving $|\lambda|\le\|A\|$ since $X \ne 0$. For a subordinate norm it suffices to take norms in the equation $Ax = \lambda x$.

A useful relation is

$\notag \|A\|_2 \le \bigl( \|A\|_1 \|A\|_{\infty}\bigr)^{1/2}, \qquad\qquad (2)$

which follows from $\|A\|_2^2 = \|A^*A\|_2 = \rho(A^*A) \le \| A^*A \|_{\infty} \le \|A^*\|_{\infty} \|A\|_{\infty} = \|A\|_1 \|A\|_{\infty}$, where the first two equalities are obtained from the singular value decomposition and we have used (1).

## Mixed Subordinate Matrix Norms

A more general subordinate matrix norm can be defined by taking different vector norms in the numerator and denominator:

$\notag \|A\|_{\alpha,\beta} = \displaystyle\max_{x\ne 0} \frac{ \|Ax\|_{\beta} }{ \|x\|_{\alpha} }.$

Some authors denote this norm by $\|A\|_{\alpha\to\beta}$.

A useful characterization of $\|A\|_{\alpha,\beta}$ is given in the next result. Recall that $\|\cdot\|^D$ denotes the dual of the vector norm $\|\cdot\|$.

Theorem 1. For $A\in\mathbb{C}^{m\times n}$,

$\notag \|A\|_{\alpha,\beta} = \displaystyle\max_{x\in\mathbb{C}^n \atop y \in\mathbb{C}^m} \frac{\mathrm{Re}\, y^*Ax}{\|y\|_\beta^D \|x\|_\alpha}.$

Proof. We have

\notag \begin{aligned} \displaystyle \max_{x\in\mathbb{C}^n \atop y \in\mathbb{C}^m} \frac{\mathrm{Re}\, y^*Ax}{\|y\|_\beta^D \|x\|_\alpha} &= \max_{x\in\mathbb{C}^n} \frac{1}{\|x\|_\alpha} \max_{y\in\mathbb{C}^m} \frac{\mathrm{Re}\, y^*Ax}{\|y\|_\beta^D} \\ &= \max_{x\in\mathbb{C}^n} \frac{\| Ax \|_\beta }{\|x\|_\alpha} = \|A\|_{\alpha,\beta}, \end{aligned}

where the second equality follows from the definition of dual vector norm and the fact that the dual of the dual norm is the original norm.

We can now obtain a connection between the norms of $A$ and $A^*$. Here, $\|A^*\|_{\beta^D,\alpha^D}$ denotes $\max_{x\ne 0} \|Ax\|_\alpha^D / \|x\|_\beta^D$.

Theorem 2. If $A\in\mathbb{C}^{m\times n}$ then $\|A\|_{\alpha,\beta} = \|A^*\|_{\beta^D,\alpha^D}$.

Proof. Using Theorem 1, we have

$\notag \|A^*\|_{\alpha,\beta} = \displaystyle\max_{x\in\mathbb{C}^n \atop y \in\mathbb{C}^m} \frac{\mathrm{Re}\, y^*Ax}{\|y\|_\beta^D \|x\|_\alpha} = \displaystyle\max_{x\in\mathbb{C}^n \atop y \in\mathbb{C}^m} \frac{\mathrm{Re}\, x^*(A^*y)}{\|x\|_\alpha \|y\|_{\beta^D}} = \|A^*\|_{\beta^D,\alpha^D}. \quad\square$

If we take the $\alpha$– and $\beta$-norms to be the same $p$-norm then we have $\|A\|_p = \|A^*\|_q$, where $p^{-1} + q^{-1} = 1$ (giving, in particular, $\|A\|_2 = \|A^*\|_2$ and $\|A\|_1 = \|A^*\|_\infty$, which are easily obtained directly).

Now we give explicit formulas for the $(\alpha,\beta)$ norm when $\alpha$ or $\beta$ is $1$ or $\infty$ and the other is a general norm.

Theorem 3. For $A\in\mathbb{C}^{m\times n}$,

\notag \begin{alignedat}{2} \|A\|_{1,\beta} &= \max_j \| A(:,j) \|_{\beta}, &\qquad\qquad& (3) \\ \|A\|_{\alpha,\infty} &= \max_i \|A(i,:)^*\|_{\alpha}^D, &\qquad\qquad& (4) \\ \end{alignedat}

For $A\in\mathbb{R}^{m\times n}$,

$\notag \|A\|_{\infty,\beta} = \displaystyle\max_{x\in Z} \|Az\|_{\beta}, \qquad\qquad (5)$

where

$Z = \{ z\in\mathbb{R}^n: z_i = \pm 1~\mathrm{for~all}~i \,\},$

and if $A$ is symmetric positive semidefinite then

$\notag \|A\|_{\infty,1} = \displaystyle\max_{x\in Z} z^T\!Az.$

Proof. For (3),

$\notag \|Ax\|_{\beta} = \Big\| \sum_j x_j A(\colon,j) \Bigr\|_{\beta} \le \displaystyle\max_j \| A(\colon,j) \|_{\beta} \sum_j |x_j|,$

with equality for $x=e_k$, where the maximum is attained for $j=k$. For (4), using the Hölder inequality,

$\|Ax\|_{\infty} = \displaystyle\max_i | A(i,\colon) x | \le \max_i \bigl( \|A(i,\colon)^*\|_{\alpha}^D \|x\|_{\alpha} \bigr) = \|x\|_{\alpha} \max_i \|A(i,\colon)^*\|_{\alpha}^D.$

Equality is attained for an $x$ that gives equality in the Hölder inequality involving the $k$th row of $A$, where the maximum is attained for $i=k$.

Turning to (5), we have $\|A\|_{\infty,\beta} = \max_{ \|x\|_\infty =1} \|Ax\|_\beta$. The unit cube $\{\, x\in\mathbb{R}^n: -e \le x \le e\,\}$, where $e = [1,1,\dots,1]^T$, is a convex polyhedron, so any point within it is a convex combination of the vertices, which are the elements of $Z$. Hence $\|x\|_\infty = 1$ implies

$\notag x = \displaystyle\sum_{z\in Z} \lambda_z Z, \quad \mathrm{where} \quad \lambda_z \ge 0, \quad \sum_{z\in Z} \lambda_z = 1$

and then

$\notag \|Ax\|_\beta = \biggl\| \displaystyle\sum_{z\in Z} \lambda_z Az \biggr\|_\beta \le \max_{z\in Z} \|Az\|_\beta.$

Hence $\max_{\|x\|_\infty = 1} \|Ax\|_\beta \le \max_{z\in Z} \|Az\|_\beta$, but trivially $\max_{z\in Z} \|Az\|_\beta \le \max_{\|x\|_\infty = 1} \|Ax\|_\beta$ and (5) follows.

Finally, if $A$ is symmetric positive semidefinite let $\xi_z = \mathrm{sign}(Az) \in Z$. Then, using a Cholesky factorization $A = R^T\!R$ (which exists even if $A$ is singular) and the Cauchy–Schwarz inequality,

\notag \begin{aligned} \max_{z\in Z} \|Az\|_1 &= \max_{z\in Z} \xi_z^T Az = \max_{z\in Z} \xi_z^T R^T\!Rz\\ &= \max_{z\in Z} (R\xi_z)^T Rz \le \max_{z\in Z} \|R\xi_z\|_2 \|Rz\|_2\\ &\le \max_{z\in Z} \|Rz\|_2^2\\ &= \max_{z\in Z} z^T\!Az. \end{aligned}

Conversely, for $z\in Z$ we have

$\notag z^T\!Az = |z^T\!Az| \le |z|^T|Az| = e^T |Az| = \|Az\|_1,$

so $\max_{z\in Z} z^T\!Az \le \max_{z\in Z} \|Az\|_1$. Hence $\max_{z\in Z} z^T\!Az = \max_{z\in Z} \|Az\|_1 = \|A\|_{\infty,1}$, using (5). $\square$

As special cases of (3) and (4) we have

\notag \begin{aligned} \|A\|_{1,\infty} &= \max_{i,j} |a_{ij}|, \qquad\qquad\qquad (6) \\ \|A\|_{2,\infty} &= \max_i \|A(i,:)^*\|_2, \\ \|A\|_{1,2} &= \max_J \|A(:,j)\|_2. \end{aligned}

We also obtain by using Theorem 2 and (5), for $A\in\mathbb{R}^{m\times n}$,

$\notag \|A\|_{2,1} = \displaystyle\max_{z\in Z} \|A^Tz\|_2.$

The $(2,\infty)$-norm has recently found use in statistics (Cape, Tang, and Priebe, 2019), the motivation being that because it satisfies

$\notag \|A\|_{2,\infty} \le \|A\|_2 \le m^{1/2} \|A\|_{2,\infty},$

the $(2,\infty)$-norm can be much smaller than the $2$-norm when $m \gg n$ and so can be a better norm to use in bounds. The $(2,\infty)$– and $(\infty,2)$-norms are used by Rebrova and Vershynin (2018) in bounding the $2$-norm of a random matrix after zeroing a submatrix. They note that the $2$-norm of a random $n\times n$ matrix involves maximizing over infinitely many random variables, while the $(\infty,2)$-norm and $(2,\infty)$-norm involve only $2^n$ and $n$ random variables, respectively.

The ($\alpha,\beta$) norm is not consistent, but for any vector norm $\gamma$, we have

\notag \begin{aligned} \|AB\|_{\alpha,\beta} &= \max_{x\ne0} \frac{ \|ABx\|_\beta }{ \|x\|_\alpha} = \max_{x\ne0} \frac{ \|A(Bx)\|_\beta }{ \|Bx\|_\gamma} \frac{ \|Bx\|_\gamma}{ \|x\|_\alpha} \le \max_{x\ne0} \frac{ \|A(Bx)\|_\beta }{ \|Bx\|_\gamma} \max_{x\ne0} \frac{ \|Bx\|_\gamma}{ \|x\|_\alpha}\\ &\le \|A\|_{\gamma,\beta} \|B\|_{\alpha,\gamma}. \end{aligned}

## Matrices With Constant $p$-Norms

Let $A$ be a nonnegative square matrix whose row and column sums are all equal to $\mu$. This class of matrices includes magic squares and doubly stochastic matrices. We have $\|A\|_1 = \|A\|_{\infty} = \mu$, so $\|A\|_2 \le \mu$ by (2). But $Ae = \mu e$ for the vector $e$ of $1$s, so $\mu$ is an eigenvalue of $A$ and hence $\|A\|_2 \ge \mu$ by (1). Hence $\|A\|_1 = \|A\|_2 = \|A\|_{\infty} = \mu$. In fact, $\|A\|_p = \mu$ for all $p\in[1,\infty]$ (see Higham (2002, Prob. 6.4) and Stoer and Witzgall (1962)).

## Computing Norms

The problem of computing or estimating $\|A\|_{\alpha,\beta}$ may be difficult for two reasons: we may not have an explicit formula for the norm, or $A$ may be given implicitly, as $A = f(B)$, $A = B^{-1}$, or $A = BC$, for example, and we may wish to compute the norm without forming $A$. Both difficulties are overcome by a generalization of the power method proposed by Tao in 1975 for arbitrary norms and independently Boyd in 1984 for $\alpha$ and $\beta$ both $p$-norms (see Higham, 2002, Sec.$\,$15.2). The power method accesses $A$ only through matrix–vector products with $A$ and $A^*$, so $A$ does not need to be explicitly available.

Let us focus on the case where $\alpha$ and $\beta$ are the same $p$-norm. The power method is implemented in the Matrix Computation Toolbox as pnorm and we use it here to estimate the $\pi$-norm and the $99$-norm, which we compare with the $1$-, $2$-, and $\infty$-norms.

>> A = gallery('frank',4)
A =
4     3     2     1
3     3     2     1
0     2     2     1
0     0     1     1
>> [norm(A,1) norm(A,2) pnorm(A,pi), pnorm(A,99), norm(A,inf)]
ans =
8.0000e+00   7.6237e+00   8.0714e+00   9.8716e+00   1.0000e+01


The plot in the top half of the following figure shows the estimated $p$-norm for the matrix gallery('binomial',8) for $p \in[1,25]$. This shape of curve is typical, because, as the plot in the lower half indicates, $\log(\|A\|_p)$ is a convex function of $1/p$ for $p\ge 1$ (see Higham, 2002, Sec.$\,$6.3).

The power method for the $(1,\infty)$ norm, which is the max norm in (6), is investigated by Higham and Relton (2016) and extended to estimate the $k$ largest values $a_{ij}$ or $|a_{ij}|$.

## A Norm Related to Grothendieck’s Inequality

A norm on $\mathbb{C}^{n\times n}$ can be defined by

\notag \begin{aligned} \|A\|^{(t)} = \max \biggl\{ \bigg|\sum_{i,j=1}^n a_{ij} y^*_jx_i \bigg| : x_i,y_j \in\mathbb{C}^t, \; \|x_i\|_2 \le 1, \; \|y_j\|_2 \le 1, \; i,j=1\colon t \, \biggr\}. \end{aligned}

Note that

$\notag \|A\|^{(1)} = \displaystyle\max_{0\ne x,y\in\mathbb{C}^n} \frac{|y^*Ax|}{\|y\|_\infty \|x\|_\infty} = \|A\|_{\infty,1}.$

A famous inequality of Grothendieck states that $\|A\|^{(n)} \le c \|A\|^{(1)}$ for all $A$, for some constant $c$ independent of $n$. Much work has been done to estimate the smallest possible constant $c$, which is known to be in the interval $[1.33,1.41]$, or $[1.67,1.79]$ for the analogous inequality for real data. See Friedland, Lim, and Zhang (2018) and the references therein.

## Notes

The formula (5) with $\beta = 1$ is due to Rohn (2000), who shows that evaluating it is NP-hard. For general $\beta$, the formula is given by Lewis (2010).

Computing mixed subordinate matrix norms based on $p$-norms is generally NP-hard (Hendrickx and Olshevsky, 2010).

## References

This is a minimal set of references, which contain further useful references within.

# What Is a Vector Norm?

A vector norm measures the size, or length, of a vector. For complex $n$-vectors, a vector norm is a function $\|\cdot\| : \mathbb{C}^n \to \mathbb{R}$ satisfying

• $\|x\| \ge 0$ with equality if and only if $x=0$ (nonnegativity),
• $\|\alpha x\| =|\alpha| \|x\|$ for all $\alpha\in\mathbb{C}$ $x\in\mathbb{C}^n$ (homogeneity),
• $\|x+y\| \le \|x\|+\|y\|$ for all $x,y\in\mathbb{C}^n$ (the triangle inequality).

An important class of norms is the Hölder $p$-norms

$\|x\|_p = \biggl( \displaystyle\sum_{i=1}^n |x_i|^p \biggr)^{1/p}, \quad p\ge 1. \qquad\qquad (1)$

The $\infty$-norm is interpreted as $\lim_{p\to\infty}\|x\|_p$ and is given by

$\notag \|x\|_{\infty} = \displaystyle\max_{1\le i\le n} |x_i|.$

Other important special cases are

\notag \begin{alignedat}{2} \|x\|_1 &= \sum_{i=1}^n |x_i|, &\quad& \mbox{Manhattan'' or taxi~cab'' norm}, \\ \|x\|_2 &= \biggl( \sum_{i=1}^n |x_i|^2 \biggr)^{1/2} = (x^*x)^{1/2}, &\quad& \mbox{Euclidean length}. \end{alignedat}

A useful concept is that of the dual norm associated with a given vector norm, which is defined by

$\notag \|y\|^D = \displaystyle\max_{x\ne0} \displaystyle\frac{\mathop{\mathrm{Re}}x^* y}{\|x\|}.$

The maximum is attained because the definition is equivalent to $\|y\|^D = \max\{ \, \mathop{\mathrm{Re}} x^*y: \|x\| = 1\,\}$, in which we are maximizing a continuous function over a compact set. Importantly, the dual of the dual norm is the original norm (Horn and Johnson, 2013, Thm.$\,$ 5.5.9(c)).

We can rewrite the definition of dual norm, using the homogeneity of vector norms, as

$\notag \|y\|^D = \displaystyle\max_{|c| = 1} \| cy \|^D = \max_{|c| = 1} \max_{x\ne 0} \frac{\mathop{\mathrm{Re}} x^*(cy) }{\|x\|} = \max_{x\ne 0} \max_{|c| = 1} \frac{\mathop{\mathrm{Re}} c(x^*y) }{\|x\|} = \max_{x\ne 0} \frac{ |x^*y| }{\|x\|}.$

Hence we have the attainable inequality

$\notag |x^*y| \le \|x\| \, \|y\|^D, \qquad\qquad (2)$

which is the generalized Hölder inequality.

The dual of the $p$-norm is the $q$-norm, where $p^{-1} + q^{-1} = 1$, so for the $p$-norms the inequality (2) becomes the (standard) Hölder inequality,

$\notag |x^*y| \le \|x\|_p \, \|y\|_q, \quad \displaystyle\frac{1}{p} + \frac{1}{q} = 1.$

An important special case is the Cauchy–Schwarz inequality,

$\notag |x^*y| \le \|x\|_2 \, \|y\|_2.$

The notation $\|x\|_0$ is used to denote the number of nonzero entries in $x$, even though it is not a vector norm and is not obtained from (1) with $p = 0$. In portfolio optimization, if $x_k$ specifies how much to invest in stock $k$ then the inequality $\|x\|_0 \le k$ says “invest in at most $k$ stocks”.

In numerical linear algebra, vector norms play a crucial role in the definition of a subordinate matrix norm, as we will explain in the next post in this series.

All norms on $\mathbb{C}^n$ are equivalent in the sense that for any two norms $\|\cdot\|_\alpha$ and $\|\cdot\|_\beta$ there exist positive constants $\nu_1$ and $\nu_2$ such that

$\nu_1\|x\|_\alpha \le \|x\|_\beta \le \nu_2 \|x\|_\alpha \quad \mathrm{for~all}~x\in \mathbb{C}^n.$

For example, it is easy to see that

\notag \begin{aligned} \|x\|_2 &\le \|x\|_1 \le \sqrt{n} \|x\|_2,\\ \|x\|_\infty &\le \|x\|_2 \le \sqrt{n} \|x\|_\infty,\\ \|x\|_\infty &\le \|x\|_1 \le n \|x\|_\infty. \end{aligned}

The 2-norm is invariant under unitary transformations: if $Q^*Q = I$, then $\|Qx\|^2 = x^*Q^* Qx = x^*x = \|x\|_2^2$.

Care must be taken to avoid overflow and (damaging) underflow when evaluating a vector $p$-norm in floating-point arithmetic for $p\ne 1,\infty$. One can simply use the formula $\|x\|_p = \| (x/\|x\|_{\infty}) \|_p \|x\|_{\infty}$, but this requires two passes over the data (the first to evaluate $\|x\|_{\infty}$). For more efficient one-pass algorithms for the $2$-norm see Higham (2002, Sec. 21.8) and Harayama et al. (2021).

## References

This is a minimal set of references, which contain further useful references within.

Note: This article was revised on October 12, 2021 to change the definition of dual norm to use $\mathrm{Re}$.

# What Is a Totally Nonnegative Matrix?

The determinant of a square submatrix of a matrix is called a minor. A matrix $A\in\mathbb{R}^{n\times n}$ is totally positive if every minor is positive. It is totally nonnegative if every minor is nonnegative. These definitions require, in particular, that all the matrix elements must be nonnegative or positive, as must $\det(A)$.

An important property is that total nonnegativity is preserved under matrix multiplication and hence under taking positive integer powers.

Theorem 1. If $A,B\in\mathbb{R}^{n\times n}$ are totally nonnegative then so is $AB$.

Theorem 1 is a direct consequence of the Binet–Cauchy theorem on determinants (also known as the Cauchy–Binet theorem). To state it, we need a way of specifying submatrices. We say the vector $\alpha = [\alpha_1,\alpha_2,\dots,\alpha_k]$ is an index vector of order $k$ if its components are integers from the set $\{1,2,\dots,n\}$ satisfying $\alpha_1 < \alpha_2 < \cdots < \alpha_k$. If $\alpha$ and $\beta$ are index vectors of order $k$ and $\ell$, respectively, then $A(\alpha, \beta)$ denotes the $k\times \ell$ matrix with ($i,j$) element $a_{\alpha_i,\beta_j}$.

Theorem 2. (Binet–Cauchy) Let $A\in\mathbb{R}^{m\times n}$, $B\in\mathbb{R}^{n\times p}$, and $C = AB$. If $\alpha$ and $\beta$ are index vectors of order $k$ and $1 \le k \le \min(m,n,p)$ then

$\notag \det(C(\alpha,\beta)) = \sum_{\gamma} \det( A(\alpha,\gamma) ) \det( B(\gamma,\beta) ), \qquad (1)$

where the sum is over all index vectors $\gamma$ of order $k$.

Note than when $k = m = n = p$, (1) reduces to the well-known relation $\det(AB) = \det(A)\det(B)$, while when $k = 1$, (1) reduces to the definition of matrix multiplication.

Totally nonnegative matrices have many interesting determinantal properties. For example, they satisfy Fischer’s inequality, first proved for symmetric positive definite matrices.

Theorem 3. (Fischer) If $A\in\mathbb{R}^{n\times n}$ is totally nonnegative then for any index vector $\alpha$,

$\notag \det(A) \le \det(A(\alpha)) \det(A(\alpha^c)), \qquad (2)$

where $\alpha^c$ comprises the indices not in $\alpha$.

By repeatedly applying (2) with $\alpha$ containing just one element, we obtain Hadamard’s inequality for totally nonnegative $A$:

$\notag \det(A) \le a_{11} a_{22} \cdots a_{nn}.$

## Examples

We give some examples of totally positive matrices, showing how they can be generated in MATLAB. We use the Anymatrix toolbox.

A matrix well known to be positive definite, but which is also totally positive, is the Hilbert matrix $H\in\mathbb{R}^{n\times n}$, with $h_{ij} = 1/(i+j-1)$. The Hilbert matrix is a particular case of a Cauchy matrix $C$, with $c_{ij} = 1/(x_i + y_j)$ for given vectors $x,y\in\mathbb{R}^{n\times n}$. A Cauchy matrix is totally positive if $0 < x_1 < x_2 < \cdots < x_n$ and $0 < y_1 < y_2 < \cdots < y_n$, which follows from the formula

$\notag \det(C_n) = \displaystyle\frac{\displaystyle\prod_{1\le i < j \le n} (x_j-x_i) (y_j-y_i) } {\displaystyle\prod_{1\le i,j \le n} (x_i+y_j) }.$

In MATLAB, the Hilbert matrix is hilb(n) and the Cauchy matrix can be generated by gallery('cauchy',x,y) (or anymatrix('gallery/cauchy',x,y)).

$\notag V = V(x_1,x_2,\dots,x_n) = \begin{bmatrix} 1 & 1 & \dots & 1 \\ x_1 & x_2 & \dots & x_n \\ \vdots &\vdots & & \vdots \\ x_1^{n-1} & x_2^{n-1} & \dots & x_n^{n-1} \end{bmatrix} \in \mathbb{C}^{n\times n}$

is totally positive if the points $x_i$ satisfy $0 < x_1 < x_2 < \cdots < x_n$. As a partial check, the general formula

$\notag \det(V) = \displaystyle\prod_{1\le i < j \le n}^n (x_i - x_j)$

shows that every leading principal minor is positive. In MATLAB, a Vandermonde matrix can be generated by anymatrix('core/vand',x).

The Pascal matrix $P_n\in\mathbb{R}^{n\times n}$ is defined by

$p_{ij} = \displaystyle\frac{ (i+j-2)! }{ (i-1)! (j-1)! } = {i+j-2 \choose j-1}.$

For example, in MATLAB:

>> P = pascal(5)
P =
1     1     1     1     1
1     2     3     4     5
1     3     6    10    15
1     4    10    20    35
1     5    15    35    70


The Pascal matrix is totally positive for all $n$ (see the section below on bidiagonal factorizations).

The one-parameter correlation matrix $C_n(\theta)$ with off-diagonal elements given by $\theta$ with $0 \le \theta < 1$, illustrated by

$\notag C_3(\theta) = \begin{bmatrix} 1 & \theta & \theta \\ \theta & 1 & \theta \\ \theta & \theta & 1 \\ \end{bmatrix},$

is not totally positive because while the principal minors are all positive, the submatrix $A([1,2],[2,3]) = \bigl[\begin{smallmatrix} \theta & \theta \\ 1 & \theta \end{smallmatrix}\bigr]$ has nonpositive determinant. However, the Kac–Murdock–Szegö matrix $K_n(\theta) = (\theta^{|i-j|})$, with $0 \le \theta < 1$, illustrated by

$\notag K_3(\theta) = \begin{bmatrix} 1 & \theta & \theta^2 \\ \theta & 1 & \theta \\ \theta^2 & \theta & 1 \\ \end{bmatrix}$

is totally positive thanks to the decay of the elements way from the diagonal. In MATLAB, the Kac–Murdock–Szegö matrix can be generated by gallery('kms',n,rho).

The lower Hessenberg Toeplitz matrix $H_n$ with all elements $1$ on and below the superdiagonal, illustrated for $n = 4$ by

$\notag H_4 = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{bmatrix},$

is totally nonnegative. It has $\lfloor n/2 \rfloor$ zero eigenvalues, which appear in a single Jordan block, and its largest eigenvalue is $2(1+\cos(2\pi/(n+2)))$. In MATLAB, this matrix can be generated by anymatrix('core/hessfull01',n). This and other binary totally nonnegative matrices are studied by Brualdi and Kirkland (2010).

Finally, consider a nonnegative $4\times 4$ bidiagonal matrix factorized into a product of elementary nonnegative bidiagonal matrices (nonnegative means that the elements of the matrix are nonnegative):

\notag \begin{aligned} L = \begin{bmatrix} 1 & 0 & 0 & 0 \\ \ell_{21} & 1 & 0 & 0 \\ 0 & \ell_{32} & 1 & 0 \\ 0 & 9 & \ell_{43} & 1 \\ \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 & 0 \\ \ell_{21} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & \ell_{32} & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & \ell_{43} & 1 \\ \end{bmatrix}\\ &\equiv L_1(\ell_{21}) L_2(\ell_{32}) L_3(\ell_{43}). \end{aligned}

It is easy to see by inspection that $L_1$, $L_2$, and $L_3$ are totally nonnegative, so $L$ is totally nonnegative by Theorem 1. With $D = \mathrm{diag}(1,-1,1,1,-1)$, we have

\notag \begin{aligned} DL^{-1}D &= (DLD)^{-1} = (DL_1D \cdot DL_2D \cdot DL_3D )^{-1}\\ &= (DL_3D)^{-1} (DL_2D)^{-1} (DL_3D)^{-1}\\ &= L_3(\ell_{43}) L_2(\ell_{32}) L_1(\ell_{21}), \qquad (3)\ \end{aligned}

which is a product of totally nonnegative matrices and hence is totally nonnegative by Theorem 1. This example clearly generalizes to show that an $n\times n$ nonnegative bidiagonal matrix is totally nonnegative.

## Inverse

Recall that the inverse of a nonsingular $A\in\mathbb{R}^{n\times n}$ is given by $A^{-1} = \mathrm{adj}(A)/\det(A)$, where

$\mathrm{adj}(A) = \bigl( (-1)^{i+j} \det(A_{ji}) \bigr)$

and $A_{pq}$ denotes the submatrix of $A$ obtained by deleting row $p$ and column $q$. If $A$ is nonsingular and totally nonnegative then it follows that $A^{-1}$ has a checkerboard (alternating) sign pattern. Indeed, we can write $A^{-1} = DBD$, where $D = \mathrm{diag}((-1)^{i+1})$ and $B$ has nonnegative elements, and in fact it can be shown that $B$ is totally nonnegative using Theorem 1, Theorem 6, and (3). For example, here is the inverse of the $4\times 4$ Pascal matrix:

>> inv(sym(pascal(5)))
ans =
[  5, -10,  10,  -5,  1]
[-10,  30, -35,  19, -4]
[ 10, -35,  46, -27,  6]
[ -5,  19, -27,  17, -4]
[  1,  -4,   6,  -4,  1]


## Eigensystem

A totally nonnegative matrix has nonnegative trace and determinant, so the sum and product of its eigenvalues are both nonnegative. In fact, all the eigenvalues are real and nonnegative. Since a Jordan block corresponding to a nonnegative eigenvalue is totally nonnegative any Jordan form with nonnegative eigenvalues is possible. More can be said of $A$ is irreducible. Recall that a matrix $A\in\mathbb{C}^{n\times n}$ is irreducible if there does not exist a permutation matrix $P$ such that

$\notag P^TAP = \begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix}$

where $A_{11}$ and $A_{22}$ are square, nonempty submatrices.

Theorem 4. If $A\in\mathbb{R}^{n\times n}$ is totally nonnegative then its eigenvalues are all real and nonnegative. If $A$ is also irreducible then the positive eigenvalues are distinct.

If $A$ is nonsingular and totally nonnegative and irreducible then by the theorem we can write the eigenvalues as $\lambda_1 > \lambda_2 > \cdots > \lambda_n >0$. It is known that the eigenvector $x_k$ associated with $\lambda_k$ has $k-1$ sign changes, that is, $(x_k)_{i+1}$ and ($x_k)_i$ have opposite signs for $k-1$ values of $i$ (any zero elements are deleted before counting sign changes). Note that for $k=1$, we already know from Perron–Frobenius theory that there is a positive eigenvector $x_1$. This result is illustrated by the Pascal matrix above:

>> A = pascal(5); [V,d] = eig(A,'vector'); [~,k] = sort(d,'descend');
>> evals = d', evecs = V(:,k)
evals =
1.0835e-02   1.8124e-01   1.0000e+00   5.5175e+00   9.2290e+01
evecs =
1.7491e-02   2.4293e-01  -7.6605e-01  -5.7063e-01   1.6803e-01
7.4918e-02   4.8079e-01  -3.8302e-01   5.5872e-01  -5.5168e-01
2.0547e-01   6.1098e-01   1.6415e-01   2.5292e-01   7.0255e-01
4.5154e-01   4.1303e-01   4.3774e-01  -5.1785e-01  -4.0710e-01
8.6486e-01  -4.0736e-01  -2.1887e-01   1.7342e-01   9.0025e-02


Note that the number of sign changes (but not the number of negative elements) increases by $1$ as we go from one column to the next

The class of nonsingular totally nonnegative irreducible matrices is known as the oscillatory matrices, because such matrices arise in the analysis of small oscillations of elastic systems. An equivalent definition (in fact, the usual definition) is that an oscillatory matrix is a totally nonnegative matrix for which $A^q$ is totally positive for some positive integer $q$.

## LU Factorization

The next result shows that a totally nonnegative matrix has an LU factorization with special properties. We will need the following special case of Fischer’s inequality (Theorem 3):

$\notag \det(A) \le \det \bigl( A(1\colon p,1\colon p) \bigr) \det \bigl( A(p+1\colon n,p+1\colon n) \bigr), \quad p=1\colon n-1. \qquad (4)$

Theorem 5. If $A\in\mathbb{R}^{n\times n}$ is nonsingular and totally nonnegative then it has an LU factorization with $L$ and $U$ totally nonnegative and the growth factor $\rho_n = 1$.

Proof. Since $A$ is nonsingular and every minor is nonnegative, (4) shows that $\det(A(1\colon p,1\colon p))>0$ for $p=1\colon n-1$, which guarantees the existence of an LU factorization. That the elements of $L$ and $U$ are nonnegative follows from explicit determinantal formulas for the elements of $L$ and $U$. The total nonnegativity of $L$ and $U$ is proved by Cryer (1976). Gaussian elimination starts with $A^{(1)} = A$ and computes $a_{ij}^{(k+1)} = a_{ij}^{(k)} - m_{ik}a_{kj}^{(k)} = a_{ij}^{(k)} - \ell_{ik} u_{kj} \le a_{ij}^{(k)}$, since $\ell_{ik}, u_{kj} \ge 0$. Thus $a_{ij} = a_{ij}^{(1)} \ge a_{ij}^{(2)} \ge \cdots \ge a_{ij}^{(r)}$, $r = \min (i,j)$. For $i > j$, $a_{ij}^{(r)} \ge a_{ij}^{(r+1)} = \cdots = a_{ij}^{(n)} = 0$; for $j \ge i$, $a_{ij}^{(r)} = \cdots = a_{ij}^{(n)} = u_{ij} \ge 0$. Thus $0 \le a_{ij}^{(k)} \le a_{ij}$ for all $i,j,k$ and hence $\rho_n \le 1$. But $\rho_n\ge1$, so $\rho_n=1$.

Theorem 5 implies that it is safe to compute the LU factorization without pivoting of a nonsingular totally nonnegativity matrix: the factorization does not break down and it is numerically stable. In fact, the computed LU factors have a strong componentwise form of stability. As shown by De Boor and Pinkus (1977), for small enough unit roundoff $u$ the computed factors $\widehat{L}$ and $\widehat{U}$ will have nonnegative elements and so from the standard backward error result for LU factorization,

$\notag \widehat{L}\widehat{U} = A + \Delta A, \quad |\Delta A| \le \gamma_n |\widehat{L}||\widehat{U}| \quad \Bigl(\gamma_n = \displaystyle\frac{nu}{1-nu} \Bigr),$

we have

$\notag |\widehat{L}||\widehat{U}| = |\widehat{L}\widehat{U}| = |A + \Delta A| \le |A| + \gamma_n |\widehat{L}||\widehat{U}|,$

which gives $|\widehat{L}||\widehat{U}| \le (1 - \gamma_n)^{-1}|A|$ and hence

$\notag \widehat{L}\widehat{U} = A + \Delta A, \quad |\Delta A| \le \displaystyle\frac{\gamma_n}{1-\gamma_n} |A|,$

which is about as strong a backward error result as we could hope for. The significance of this result is reduced, however, by the fact that for some important classes of totally nonnegative matrices, including Vandermonde matrices and Cauchy matrices, structure-exploiting linear system solvers exist that are substantially faster, and potentially more accurate, than LU factorization.

## Factorization into a Product of Bidiagonal Matrices

We showed above that any nonnegative bidiagonal matrix is totally nonnegative. The next result shows that any nonsingular totally nonnegative matrix has an LU factorization in which $L$ and $U$ can be factorized into a product of nonnegative bidiagonal matrices.

Theorem 6. (Gasca and Peña, 1996) A nonsingular matrix $A\in\mathbb{R}^{n\times n}$ is totally nonnegative if and only if it it can be factorized as

$\notag A = L_{n-1} L_{n-2} \dots L_1 D U_1 U_2 \dots U_{n-1}, \qquad (5)$

where $D$ is a diagonal matrix with positive diagonal entries and $L_i$ and $U_i$ are unit lower and unit upper bidiagonal matrices, respectively, with the first $i-1$ entries along the subdiagonal of $L_i$ and $U_i^T$ zero and the rest nonnegative.

An analogue of Theorem 6 holds for totally positive matrices, the only difference being that the last $n-i+1$ subdiagonal entries of $L_i$ and $U_i^T$ are positive.

The factorization (5) can be computed by Neville elimination, which is a version of Gaussian elimination in which the eliminations are between adjacent rows, working from the bottom of each column upwards.

This factorization into bidiagonal factors can be used to obtain simple proofs of various properties of totally nonnegative matrices and totally positive matrices (Fallat, 2001). It also provides an efficient way to generates such matrices. If all the parameters in $D$ and the $L_i$ and $U_i$ are set to $1$ then the Pascal matrix is generated.

## Testing for Total Positivity

An $n\times n$ matrix has $\sum_{i=1}^n {n \choose k} = 2^n-1$ principal minors (ones based on submatrices centred on the diagonal) and $\sum_{i=1}^n {n \choose k}^2 = {2n \choose n} -1 \approx 4^n/(n\pi)^{1/2}$ minors in total. However, it is not necessary to check all these minors to test for total positivity.

Theorem 7. (Gasca and Peña, 1996) The matrix $A\in\mathbb{R}^{n\times n}$ is totally positive if and only if $\det(A(\alpha,\beta)) > 0$ for all index vectors $\alpha$ and $\beta$ such that one of $\alpha$ and $\beta$ is $[1,2,\dots,k]$ and the entries of the other are $k$ consecutive integers.

Theorem 7 shows that only $n(n+1)$ minors need to be tested. Gasca and Peña have also show that total nonnegativity can be tested by checking about $2^{n+1} + n^2/2$ minors. A more efficient way to test for total nonnegativity is to compute the factorization in Theorem 6 and check the signs of the entries.

## Notes

The results we have described show that totally nonnegative and totally positive matrices are analogous in many ways to symmetric positive (semi)definite matrices. The analogies go further because totally nonnegative and totally positive matrices also satisfy eigenvalue interlacing inequalities (albeit weaker than for symmetric matrices) and the eigenvalues of an oscillatory matrix majorize the diagonal elements. See Fallat and Johnson (2011) or Fallat (2014) for details.

## References

This is a minimal set of references, which contain further useful references within.

# What Is the Perron–Frobenius Theorem?

A real matrix is nonnegative if all its elements are nonnegative and it is positive if all its elements are positive. Nonnegative matrices arise in a wide variety of applications, for example as matrices of probabilities in Markov processes and as adjacency matrices of graphs. Information about the eigensystem is often essential in these applications.

Perron (1907) proved results about the eigensystem of a positive matrix and Frobenius (1912) extended them to nonnegative matrices.

The following three results of increasing specificity summarize the key spectral properties of nonnegative matrices proved by Perron and Frobenius. Recall that a simple eigenvalue of an $n\times n$ matrix is one with algebraic multiplicity $1$, that is, it occurs only once in the set of $n$ eigenvalues. We denote by $\rho(A)$ the spectral radius of $A$, the largest absolute value of any eigenvalue of $A$.

Theorem 1. (Perron–Frobenius) If $A\in\mathbb{R}^{n\times n}$ is nonnegative then

1. $\rho(A)$ is an eigenvalue of $A$,
2. there is a nonnegative eigenvector $x$ such that $Ax = \rho(A)x$.

A matrix $A\in\mathbb{C}^{n\times n}$ is reducible if there is a permutation matrix $P$ such that

$\notag P^TAP = \begin{bmatrix} A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix}$

where $A_{11}$ and $A_{22}$ are square, nonempty submatrices; it is irreducible if it is not reducible. Examples of reducible matrices are triangular matrices and matrices with a zero row or column. A positive matrix is trivially irreducible.

Theorem 2. (Perron–Frobenius) If $A\in\mathbb{R}^{n\times n}$ is nonnegative and irreducible then

1. $\rho(A)$ is an eigenvalue of $A$,
2. $\rho(A)>0$,
3. there is a positive eigenvector $x$ such that $Ax = \rho(A) x$,
4. $\rho(A)$ is a simple eigenvalue.

Theorem 3. (Perron) If $A\in\mathbb{R}^{n\times n}$ is positive then Theorem 2 holds and, in addition, $|\lambda| < \rho(A)$ for any eigenvalue $\lambda$ with $\lambda \ne \rho(A)$.

For nonnegative, irreducible $A$, the eigenvalue $\rho(A)$ is called the Perron root of $A$ and the corresponding positive eigenvector $x$, normalized so that $\|x\|_1 = 1$, is called the Perron vector.

It is a good exercise to apply the theorems to all binary $2\times 2$ matrices. Here are some interesting cases.

• $A = \bigl[\begin{smallmatrix}0 & 1 \\ 0 & 0 \end{smallmatrix}\bigr]$: Theorem 1 says that $\rho(A) = 0$ is an eigenvalue and and that it has a nonnegative eigenvector. Indeed $[1~0]^T$ is an eigenvector. Note that $A$ is reducible and $0$ is a repeated eigenvalue.
• $A = \bigl[\begin{smallmatrix}0 & 1 \\ 1 & 0 \end{smallmatrix}\bigr]$: $A$ is irreducible and Theorem 2 says that $\rho(A)$ is a simple eigenvalue with positive eigenvector. Indeed the eigenvalues are $\pm 1$ and $[1~1]^T/2$ is the Perron vector for the Perron root $1$. This matrix has two eigenvalues of maximal modulus.
• $A = \bigl[\begin{smallmatrix}1 & 1 \\ 1 & 1 \end{smallmatrix}\bigr]$: Theorem 3 says that $\rho(A) = 2$ is an eigenvalue with positive eigenvector and that the other eigenvalue has modulus less than $2$. Indeed the eigenvalues are the Perron root $2$, with Perron vector $[1~1]^T/2$, and $0$.

For another example, consider the irreducible matrix

$\notag B = \begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix}, \quad \Lambda(B) = \bigl\{ 1, \textstyle\frac{1}{2}( -1 \pm \sqrt{3}\mskip1mu\mathrm{i} ) \bigr\}.$

Note that $B$ is a companion matrix and a permutation matrix. Theorem 2 correctly tells us that $\rho(A) = 1$ is an eigenvalue of $A$, and that it has a corresponding positive eigenvector, the Perron vector $[1~1~1]^T/3$. Two of the eigenvalues are complex, however, and all three eigenvalues have modulus 1, as they must because $B$ is orthogonal.

A stochastic matrix is a nonnegative matrix whose row sums are all equal to $1$. A stochastic matrix satisfies $Ae = e$, where $e = [1,1,\dots,1]^T$, which means that $A$ has an eigenvalue $1$, and so $\rho(A) \ge 1$. Since $\rho(A) \le \|A\|$ for any norm, by taking the $\infty$-norm we conclude that $\rho(A) = 1$. For a stochastic matrix, Theorem 1 does not give any further information. If $A$ is irreducible then Theorem 2 tells us that $\rho(A)$ is a simple eigenvalue, and if $A$ is positive Theorem 3 tells us that every other eigenvalue has modulus less than $\rho(A)$.

The next result is easily proved using Theorem 3 together with the Jordan canonical form. It shows that the powers of a positive matrix behave like multiples of a rank-1 matrix.

Theorem 4. If $A\in\mathbb{R}^{n\times n}$ is positive, $x$ is the Perron vector of $A$, and $y$ is the Perron vector of $A^T$ then

$\notag \displaystyle\lim_{k\to\infty} \left( \displaystyle\frac{A}{\rho(A)} \right)^k = \displaystyle\frac{xy^T}{y^Tx}.$

Note that $y$ in the theorem is a left eigenvector of $A$ corresponding to $\rho(A)$, that is, $y^TA = \rho(A)y^T$ (since $\rho(A^T) = \rho(A)$).

If $A$ is stochastic and positive then Theorem 4 is applicable and $x = n^{-1}e$. If $A$ also has unit column sums, so that it is doubly stochastic, then $y = n^{-1}e$ and Theorem 4 says that $\lim_{k\to\infty}A^k = n^{-1}ee^T$. We illustrate this result in MATLAB using a scaled magic square matrix.

>> n = 4; M = magic(n), A = M/sum(M(1,:)) % Doubly stochastic matrix.
A =
16     2     3    13
5    11    10     8
9     7     6    12
4    14    15     1
A =
4.7059e-01   5.8824e-02   8.8235e-02   3.8235e-01
1.4706e-01   3.2353e-01   2.9412e-01   2.3529e-01
2.6471e-01   2.0588e-01   1.7647e-01   3.5294e-01
1.1765e-01   4.1176e-01   4.4118e-01   2.9412e-02

>> for k = 8:8:32, fprintf('%11.2e',norm(A^k-ones(n)/n,1)), end, disp(' ')
3.21e-05   7.37e-10   1.71e-14   8.05e-16


## References

This is a minimal set of references, which contain further useful references within.

• Roger A. Horn and Charles R. Johnson, Matrix Analysis, second edition, Cambridge University Press, 2013. Chapter 8. My review of the second edition.
• Carl D. Meyer, Matrix Analysis and Applied Linear Algebra, Society for Industrial and Applied Mathematics, Philadelphia, PA, USA, 2000. Chapter 8.
• Helene Shapiro, Linear Algebra and Matrices. Topics for a Second Course, American Mathematical Society, Providence, RI, USA, 2015. Chapter 17.

# What Is the Kac–Murdock–Szegö Matrix?

The Kac–Murdock–Szegö matrix is the symmetric Toeplitz matrix

$\notag A_n(\rho) = \begin{bmatrix} 1 & \rho & \rho^2 & \dots & \rho^{n-1} \\ \rho & 1 & \rho & \dots & \rho^{n-2} \\ \rho^2 & \rho & 1 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \rho \\ \rho^{n-1} & \rho^{n-2} & \dots & \rho & 1 \end{bmatrix} \in\mathbb{R}^{n\times n}. \qquad(1)$

It was considered by Kac, Murdock, and Szegö (1953), who investigated its spectral properties. It arises in the autoregressive AR(1) model in statistics and signal processing.

The matrix is singular for $\rho=1$, as $A_n(1)$ is the rank-$1$ matrix $ee^T$, and it is also rank-$1$ for $\rho = -1$, as in this case every column is a multiple of the vector with alternating elements $\pm 1$. The determinant $\det(A_n(\rho)) = (1-\rho^2)^{n-1}$. For $\rho \ne \pm 1$, $A_n$ is nonsingular and the inverse is the tridiagonal (but not Toeplitz) matrix

$\notag A_n(\rho)^{-1} = \displaystyle\frac{1}{1-\rho^2} \begin{bmatrix} 1 & -\rho & 0 & \dots & \dots & 0 \\ -\rho & 1+\rho^2 & -\rho & \dots & \dots & 0 \\ 0 & -\rho & 1+\rho^2 & \ddots & \dots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \dots & \dots & -\rho & 1+\rho^2 & -\rho \\ 0 & \dots & \dots & 0 & -\rho & 1 \end{bmatrix}. \qquad (2)$

For $-1 < \rho < 1$, $A_n(\rho)$ is positive definite, since every leading principal submatrix has positive determinant, as can also be seen by noting that the inverse is diagonally dominant with positive diagonal, so that $A_n^{-1}$ is positive definite and hence $A_n$ is positive definite.

For $-1 \le \rho \le 1$, $A_n(\rho)$ is positive semidefinite, so it is a correlation matrix for $\rho$ in this range.

For $0 \le \rho \le 1$, $A_n(\rho)$ is totally nonnegative, that is. every submatrix has nonnegative determinant. For $0 < \rho < 1$, we know that $A_n(\rho)$ is nonsingular, and it is clearly irreducible, and together with the total nonnegativity these properties imply that the eigenvalues are distinct and positive (this can also be deduced from the fact that the inverse is tridiagonal with nonzero subdiagonal and superdiagonal entries).

It is straightforward to verify that $A_n$ has a factorization $A_n = LDL^*$ with $L$ the inverse of a unit lower bidiagonal matrix:

$\notag L = \begin{bmatrix} 1 & & & &\\ -\rho & 1 & & &\\ & -\rho & 1 & &\\ & & \ddots & \ddots &\\ & & &-\rho &1 \end{bmatrix}^{-1}, \quad D = \mathrm{diag}(1, 1-\rho^2, 1-\rho^2, \dots, 1-\rho^2). \qquad (3)$

This factorization can be used to prove all the properties stated above.

From (1) and (2) we can derive the formulas

\notag \begin{aligned} \|A_n\|_{1,\infty} &= 2 \left(\displaystyle\frac{1-\rho^{k+1}}{1-\rho}\right) -1 - (2k-n+1) \rho^k, \quad k = \lfloor n/2 \rfloor, \\ \|A_n^{-1}\|_{\infty} &= (1+2\rho+\rho^2)/(1-\rho^2) = (1+\rho)/(1-\rho). \end{aligned}

Hence we have an explicit formula for the condition number $\kappa_p(A_n) = \|A_n\|_{1,\infty} \|A_n^{-1}\|_{1,\infty}$ for $p = 1,\infty$.

We can allow $\rho$ to be complex, in which case the definition (1) is modified to conjugate the elements below the diagonal. The factorization $A = LDL^*$ continues to hold with $D$ in $(2)$ replaced by $\mathrm{diag}(1, 1-|\rho|^2, 1-|\rho|^2, \dots, 1-|\rho|^2)$.

The Kac–Murdock–Szegö matrix (for real or complex $\rho$) can be generated in MATLAB as gallery('kms',n,rho).

## References

This is a minimal set of references, which contain further useful references within.

# What Is the Determinant of a Matrix?

The determinant of an $n\times n$ matrix $A$ is defined by

$\notag \det(A) = \displaystyle\sum_j (-1)^{\mathop{\mathrm{sgn}}j} a_{1,j_1}a_{2,j_2} \dots a_{n,j_n}, \qquad (1)$

where the sum is over all $n!$ permutations $j = (j_1,j_2,\dots,j_n$) of the sequence $(1,2,\dots,n)$ and $\mathop{\mathrm{sgn}}j$ is the number of inversions in $j$, that is, the number of pairs $(j_k,j_\ell)$ with $k j_\ell$. Each term in the sum is a signed product of $n$ entries of $A$ and the product contains one entry taken from each row and one from each column.

The determinant is sometimes written with vertical bars, as $|A|$.

Three fundamental properties are

\notag \begin{aligned} \det(\alpha A) &= \alpha^n \det(A)\; \mathrm{for~any~scalar~}\alpha,\qquad(2)\\ \det(A^T) &= \det(A), \qquad(3)\\ \det(AB) &= \det(A)\det(B) \mathrm{~for~} n\times n~ A \mathrm{~and~} B.\qquad(4) \end{aligned}

The first property is immediate, the second can be proved using properties of permutations, and the third is proved in texts on linear algebra and matrix theory.

An alternative, recursive expression for the determinant is the Laplace expansion

$\notag \det(A) = \displaystyle\sum_{j=1}^n (-1)^{i+j} a_{ij} \det (A_{ij}). \qquad(5)$

for any $i\in\{1,2,\dots,n\}$, where $A_{ij}$ denotes the $(n-1)\times (n-1)$ submatrix of $A$ obtained by deleting row $i$ and column $j$, and $\det(a) = a$ for a scalar $a$. This formula is called the expansion by minors because $\det (A_{ij})$ is a minor of $A$.

For some types of matrices the determinant is easy to evaluate. If $T$ is triangular then $\det(T) = \prod_{i=1}^n t_{ii}$. If $Q$ is unitary then $Q^*Q = I$ implies $|\det(Q)| = 1$ on using (3) and (4). An explicit formula exists for the determinant of a Vandermonde matrix.

The determinant of $A$ is connected with the eigenvalues $\lambda_i$ of $A$ via the property $\det(A) = \prod_{i=1}^n \lambda_i$. Since the eigenvalues are the roots of the characteristic polynomial $\det(tI - A)$, this relation follows by setting $t = 0$ in the expression

$\notag \det(tI - A) = t^n + a_{n-1}t^{n-1} + \cdots + a_1 t + a_0 = \displaystyle\prod_{i=1}^n (t - \lambda_i).$

For $n=2$, the determinant is

$\notag \det\biggl( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \biggr) = ad - bc,$

but already for $n=3$ the determinant is tedious to write down. If one must compute $\det(A)$, the formulas (1) and (5) are too expensive unless $n$ is very small: they have an exponential cost. The best approach is to use a factorization of $A$ involving factors that are triangular or orthogonal, so that the determinants of the factors are easily computed. If $PA = LU$ is an LU factorization, with $P$ a permutation matrix, $L$ unit lower triangular, and $U$ upper triangular, then $\det(A) = \det(P) \prod_{i=1}^n u_{ii} = \pm \prod_{i=1}^n u_{ii}$. As this expression indicates, the determinant is prone to overflow and underflow in floating-point arithmetic, so it may be preferable to compute $\log(|\det(A)|) = \sum_{i=1}^n \log|u_{ii}|$.

The determinant features in the formula

$\notag A^{-1} = \displaystyle\frac{\mathrm{adj}(A)}{\det(A)}$

for the inverse, where $\mathrm{adj}(A)$ is the adjugate of $A$ (recall that $\mathrm{adj}(A)$ has $(i,j)$ element $(-1)^{i+j} \det(A_{ji})$). More generally, Cramer’s rule says that the components of the solution to a linear system $Ax = b$ are given by $x_i = \det(A_i(b))/\det(A)$, where $A_i(b)$ denotes $A$ with its $i$th column replaced by $b$. While mathematically elegant, Cramer’s rule is of no practical use, as it is both expensive and numerically unstable in finite precision arithmetic.

## Inequalities

A celebrated bound for the determinant of a Hermitian positive definite matrix $H\in\mathbb{C}^{n\times n}$ is Hadamard’s inequality. Note that for such $H$, $\det(H)$ is real and positive (being the product of the eigenvalues, which are real and positive) and the diagonal elements are also real and positive (since $h_{ii} = e_i^*He_i > 0$).

Theorem 1 (Hadamard’s inequality). For a Hermitian positive definite matrix $H\in\mathbb{C}^{n\times n}$,

$\notag \det(H) \le \displaystyle\prod_{i=1}^n h_{ii},$

with equality if and only if $H$ is diagonal.

Theorem 1 is easy to prove using a Cholesky factorization.

The following corollary can be obtained by applying Theorem 1 to $H = A^*A$ or by using a QR factorization of $A$.

Corollary 2. For $A = [a_1,a_2,\dots,a_n] \in\mathbb{C}^{n\times n}$,

$\notag \det(A) \le \displaystyle\prod_{i=1}^n \|a_i\|_2,$

with equality if and only if the columns of $A$ are orthogonal.

Obviously, one can apply the corollary to $A^*$ and obtain the analogous bound with column norms replaced by row norms.

The determinant of $A\in\mathbb{R}^{n\times n}$ can be interpreted as the volume of the parallelepiped $\{\, \sum_{i=1}^n t_ia_i : 0 \le t_i \le 1, ~ i = 1\colon n\,\}$, whose sides are the columns of $A$. Corollary 2 says that for columns of given lengths the volume is maximized when the columns are orthogonal.

## Nearness to Singularity and Conditioning

The determinant characterizes nonsingularity: $A$ is singular if and only if $\det(A) = 0$. It might be tempting to use $|\det(A)|$ as a measure of how close a nonsingular matrix $A$ is to being singular, but this measure is flawed, not least because of the sensitivity of the determinant to scaling. Indeed if $Q$ is unitary then $\det(\alpha Q) = \alpha^n \det(Q)$ can be given any value by a suitable choice of $\alpha$, yet $\alpha Q$ is perfectly conditioned: $\kappa_2(\alpha Q) = 1$, where $\kappa(A) = \|A\| \|A^{-1}\|$ is the condition number.

To deal with the poor scaling one might normalize the determinant: in view of Corollary 2,

$\notag \psi(A) = \displaystyle\frac{\prod_{i=1}^n \|a_i\|_2} {\det(A)}$

satisfies $\psi(A) \ge 1$ and $\psi(A) = 1$ if and only if the columns of $A$ are orthogonal. Birkhoff (1975) calls $\psi$ the Hadamard condition number. In general, $\psi$ is not related to the condition number $\kappa$, but if $A$ has columns of unit $2$-norm then it can be shown that $\kappa_2(A) < 2\psi(A)$ (Higham, 2002, Prob. 14.13). Dixon (1984) shows that for classes of $n\times n$ random matrices $A_n$ that include matrices with elements independently drawn from a normal distribution with mean $0$, the probability that the inequality

$n^{1/4 - \epsilon} \mathrm{e}^{n/2} < \psi(A_n) < n^{1/4 + \epsilon} \mathrm{e}^{n/2}$

holds tends to $1$ as $n\to\infty$ for any $\epsilon > 0$, so $\psi(A_n) \approx n^{1/4}\mathrm{e}^{n/2}$ for large $n$. This exponential growth is much faster than the growth of $\kappa$, for which Edelman (1998) showed that for the standard normal distribution, $\mathbb{E}(\log(\kappa_2(A_n))) \approx \log n + 1.537$, where $\mathbb{E}$ denotes the mean value. This MATLAB example illustrates these points.

>> rng(1); n = 50; A = randn(n);
>> psi = prod(sqrt(sum(A.*A)))/abs(det(A)), kappa2 = cond(A)
psi =
5.3632e+10
kappa2 =
1.5285e+02
>> ratio = psi/(n^(0.25)*exp(n/2))
ratio =
2.8011e-01


The relative distance from $A$ to the set of singular matrices is equal to the reciprocal of the condition number.

Theorem 3 (Gastinel, Kahan). For $A\in\mathbb{C}^{n\times n}$ and any subordinate matrix norm,

$\notag \min \left\{ \displaystyle\frac{\|\Delta A\|}{\|A\|} : A+\Delta A\mathrm{~singular} \right\} = \displaystyle\frac{1}{\kappa(A)}.$

## Notes

Determinants came before matrices, historically. Most linear algebra textbooks make significant use of determinants, but a lot can be done without them. Axler (1995) shows how the theory of eigenvalues can be developed without using determinants.

Determinants have little application in practical computations, but they are a useful theoretical tool in numerical analysis, particularly for proving nonsingularity.

There is a large number of formulas and identities for determinants. Sir Thomas Muir collected many of them in his five-volume magnum opus The Theory of Determinants in the Historical Order of Development, published between 1890 and 1930. Brualdi and Schneider (1983) give concise derivations of many identities using Gaussian elimination, bringing out connections between the identities.

The quantity obtained by modifying the definition (1) of determinant to remove the $(-1)^{\mathop{\mathrm{sgn}}j}$ term is the permanent. The permanent arises in combinatorics and quantum mechanics and is much harder to compute than the determinant: no algorithm is known for computing the permanent in $p(n)$ operations for a polynomial $p$.

## References

This is a minimal set of references, which contain further useful references within.

# What Is a Vandermonde Matrix?

A Vandermonde matrix is defined in terms of scalars $x_1$, $x_2$, …, $x_n\in\mathbb{C}$ by

$\notag V = V(x_1,x_2,\dots,x_n) = \begin{bmatrix} 1 & 1 & \dots & 1 \\ x_1 & x_2 & \dots & x_n \\ \vdots &\vdots & & \vdots \\ x_1^{n-1} & x_2^{n-1} & \dots & x_n^{n-1} \end{bmatrix} \in \mathbb{C}^{n\times n}.$

The $x_i$ are called points or nodes. Note that while we have indexed the nodes from $1$, they are usually indexed from $0$ in papers concerned with algorithms for solving Vandermonde systems.

Vandermonde matrices arise in polynomial interpolation. Suppose we wish to find a polynomial $p_{n-1}(x) = a_nx^{n-1} + a_{n-1}x^{n-2} + \cdots + a_1$ of degree at most $n-1$ that interpolates to the data $(x_i,f_i)_{i=1}^n$, that is, $p_{n-1}(x_i) = f_i$, $i=1\colon n$. These equations are equivalent to

$\notag V^Ta = f \quad \mathrm{(dual)},$

where $a = [a_1,a_2,\dots,a_n]^T$ is the vector of coefficients. This is known as the dual problem. We know from polynomial interpolation theory that there is a unique interpolant if the $x_i$ are distinct, so this is the condition for $V$ to be nonsingular.

The problem

$\notag Vy = b \quad \mathrm{(primal)}$

is called the primal problem, and it arises when we determine the weights for a quadrature rule: given moments $b_i$ find weights $y_i$ such that $\sum_{j=1}^n y_j^{} x_j^{\,i-1} = b_i$, $i=1\colon n$.

## Determinant

The determinant of $V$ is a function of the $n$ points $x_i$. If $x_i = x_j$ for some $i\ne j$ then $V$ has identical $i$th and $j$th columns, so is singular. Hence the determinant must have a factor $x_i - x_j$. Consequently, we have

$\notag \det( V(x_1,x_2,\dots,x_n) ) = c \displaystyle\prod_{i,j = 1\atop i > j}^n (x_i - x_j),$

where, since both sides have degree $n(n-1)/2$ in the $x_i$, $c$ is a constant. But $\det(V)$ contains a term $x_2 x_3^2 \dots x_n^{n-1}$ (from the main diagonal), so $c = 1$. Hence

$\notag \det(V) = \displaystyle\prod_{i,j = 1\atop i > j}^n (x_i - x_j). \qquad (1)$

This formula confirms that $V$ is nonsingular precisely when the $x_i$ are distinct.

## Inverse

Now assume that $V$ is nonsingular and let $V^{-1} = W = (w_{ij})_{i,j=1}^n$. Equating elements in the $i$th row of $WV = I$ gives

$\sum_{j=1}^n w_{ij} x_k^{\mskip1mu j-1} = \delta_{ik}, \quad k=1\colon n,$

where $\delta_{ij}$ is the Kronecker delta (equal to $1$ if $i=j$ and $0$ otherwise). These equations say that the polynomial $\sum_{j=1}^n w_{ij} x^{\mskip1mu j-1}$ takes the value $1$ at $x = x_i$ and $0$ at $x = x_k$, $k\ne i$. It is not hard to see that this polynomial is the Lagrange basis polynomial:

$\notag \sum_{j=1}^n w_{ij} x^{j-1} = \displaystyle\prod_{k=1\atop k\ne i}^n \left( \frac{x-x_k}{x_i-x_k} \right) =: \ell_i(x). \qquad (2)$

We deduce that

$\notag w_{ij} = \displaystyle\frac{ (-1)^{n-j} \sigma_{n-j}(x_1,\dots,x_{i-1},x_{i+1},\dots,x_n) } { \displaystyle\prod_{k=1 \atop k\ne i}^n (x_i-x_k) }, \qquad (3)$

where $\sigma_k(y_1,\dots,y_n)$ denotes the sum of all distinct products of $k$ of the arguments $y_1,\dots,y_n$ (that is, $\sigma_k$ is the $k$th elementary symmetric function).

From (1) and (3) we see that if the $x_i$ are real and positive and arranged in increasing order $0 < x_1 < x_2 < \cdots 0$ and $V^{-1}$ has a checkerboard sign pattern: the $(i,j)$ element has sign $(-1)^{i+j}$.

Note that summing (2) over $i$ gives

$\notag \displaystyle\sum_{j=1}^n x^{j-1} \sum_{i=1}^n w_{ij} = \sum_{i=1}^n \ell_i(x) = 1,$

where the second equality follows from the fact that $\sum_{i=1}^n \ell_i(x)$ is a degree $n-1$ polynomial that takes the value $1$ at the $n$ distinct points $x_i$. Hence

$\notag \displaystyle\sum_{i=1}^n w_{ij} = \delta_{j1},$

so the elements in the $j$th column of the inverse sum to $1$ for $j = 1$ and $0$ for $j\ge 2$.

## Example

To illustrate the formulas above, here is an example, with $x_i = (i-1)/(n-1)$ and $n = 5$:

$\notag V = \left[\begin{array}{ccccc} 1 & 1 & 1 & 1 & 1\\ 0 & \frac{1}{4} & \frac{1}{2} & \frac{3}{4} & 1\\[\smallskipamount] 0 & \frac{1}{16} & \frac{1}{4} & \frac{9}{16} & 1\\[\smallskipamount] 0 & \frac{1}{64} & \frac{1}{8} & \frac{27}{64} & 1\\[\smallskipamount] 0 & \frac{1}{256} & \frac{1}{16} & \frac{81}{256} & 1 \end{array}\right], \quad V^{-1} = \left[\begin{array}{ccccc} 1 & -\frac{25}{3} & \frac{70}{3} & -\frac{80}{3} & \frac{32}{3}\\[\smallskipamount] 0 & 16 & -\frac{208}{3} & 96 & -\frac{128}{3}\\ 0 & -12 & 76 & -128 & 64\\[\smallskipamount] 0 & \frac{16}{3} & -\frac{112}{3} & \frac{224}{3} & -\frac{128}{3}\\[\smallskipamount] 0 & -1 & \frac{22}{3} & -16 & \frac{32}{3} \end{array}\right],$

for which $\det(V) = 9/32768$.

## Conditioning

Vandermonde matrices are notorious for being ill conditioned. The ill conditioning stems from the monomials being a poor basis for the polynomials on the real line. For arbitrary distinct points $x_i$, Gautschi showed that $V_n = V(x_1, x_2, \dots, x_n)$ satisfies

$\notag \displaystyle\max_i \displaystyle\prod_{j\ne i} \frac{ \max(1,|x_j|) }{ |x_i-x_j| } \le \|V_n^{-1}\|_{\infty} \le \displaystyle\max_i \prod_{j\ne i} \frac{ 1+|x_j| }{ |x_i-x_j| },$

with equality on the right when $x_j = |x_j| e^{\mathrm{i}\theta}$ for all $j$ with a fixed $\theta$ (in particular, when $x_j\ge0$ for all $j$). Note that the upper and lower bounds differ by at most a factor $2^{n-1}$. It is also known that for any set of real points $x_i$,

$\notag \kappa_2(V_n) \ge \Bigl(\displaystyle\frac{2}{n}\Bigr)^{1/2} \, (1+\sqrt{2})^{n-2}$

and that for $x_i = 1/i$ we have $\kappa_{\infty}(V_n) > n^{n+1}$, where the lower bound is an extremely fast growing function of the dimension!

These exponential lower bounds are alarming, but they do not necessarily rule out the use of Vandermonde matrices in practice. One of the reasons is that there are specialized algorithms for solving Vandermonde systems whose accuracy is not dependent on the condition number $\kappa$, and which in some cases can be proved to be highly accurate. The first such algorithm is an $O(n^2)$ operation algorithm for solving $V_ny =b$ of Björck and Pereyra (1970). There is now a long list of generalizations of this algorithm in various directions, including for confluent Vandermonde-like matrices (Higham, 1990), as well as for more specialized problems (Demmel and Koev, 2005) and more general ones (Bella et al., 2009). Another important observation is that the exponential lower bounds are for real nodes. For complex nodes $V_n$ can be much better conditioned. Indeed when the $x_i$ are the roots of unity, $V_n/\sqrt{n}$ is the unitary Fourier matrix and so $V_n$ is perfectly conditioned.

## Generalizations

Two ways in which Vandermonde matrices have been generalized are by allowing confluency of the points $x_i$ and by replacing the monomials by other polynomials. Confluency arises when the $x_i$ are not distinct. If we assume that equal $x_i$ are contiguous then a confluent Vandermonde matrix is obtained by “differentiating” the previous column for each of the repeated points. For example, with points $x_1, x_1, x_1, x_2, x_2$ we obtain

$\notag \begin{bmatrix} 1 & 0 & 0 & 1 & 0 \\ x_1 & 1 & 0 & x_2 & 1 \\ x_1^2 & 2x_1 & 2 & x_2^2 & 2x_2 \\ x_1^3 & 3x_1^2 & 6x_1 & x_2^3 & 3x_2^2 \\ x_1^4 & 4x_1^3 & 12x_1^2 & x_2^4 & 4x_2^3 \end{bmatrix}. \qquad (4)$

The transpose of a confluent Vandermonde matrix arises in Hermite interpolation; it is nonsingular if the points corresponding to the “nonconfluent columns” are distinct (that is, if $x_1 \ne x_2$ in the case of (4)).

A Vandermonde-like matrix is defined in terms of a set of polynomials $\{p_i(x)\}_{i=0}^n$ with $p_i$ having degree $i$:

$\notag \begin{bmatrix} p_0(x_1) & p_0(x_2) & \dots & p_0(x_n)\\ p_1(x_1) & p_1(x_2) & \dots & p_1(x_n)\\ \vdots & \vdots & \dots & \vdots\\ p_{n-1}(x_1) & p_{n-1}(x_2) & \dots & p_{n-1}(x_n)\\ \end{bmatrix}.$

Of most interest are polynomials that satisfy a three-term recurrence, in particular, orthogonal polynomials. Such matrices can be much better conditioned than general Vandermonde matrices.

## Notes

Algorithms for solving confluent Vandermonde-like systems and their rounding error analysis are described in the chapter “Vandermonde systems” of Higham (2002).

Gautschi has written many papers on the conditioning of Vandermonde matrices, beginning in 1962. We mention just his most recent paper on this topic: Gautschi (2011).

## References

This is a minimal set of references, which contain further useful references within.

# What Is the Wilson Matrix?

The $4\times 4$ matrix

$\notag W = \begin{bmatrix} 5 & 7 & 6 & 5 \\ 7 & 10 & 8 & 7 \\ 6 & 8 & 10 & 9 \\ 5 & 7 & 9 & 10 \end{bmatrix}$

appears in a 1946 paper by Morris, in which it is described as having been “devised by Mr. T. S. Wilson.” The matrix is symmetric positive definite with determinant $1$ and inverse

$\notag W^{-1} = \begin{bmatrix} 68 & -41 & -17 & 10\\ -41 & 25 & 10 & -6\\ -17 & 10 & 5 & -3\\ 10 & -6 & -3 & 2 \end{bmatrix},$

so it is moderately ill conditioned with $\kappa_2(W) = \|W\|_2 \|W^{-1}\|_2 \approx 2.98409\times 10^3$. This little matrix has been used as an example and for test purposes in many research papers and books over the years, in particular by John Todd, who described it as “the notorious matrix $W$ of T. S. Wilson”.

Rutishauser (1968) stated that “the famous Wilson matrix is not a very striking example of an ill-conditioned matrix”, on the basis that $\kappa_2(A)\le 40{,}000$ for a “positive definite symmetric $4\times 4$ matrix with integer elements not exceeding $10$” and he gave the positive definite matrix

$\notag A_0 = \begin{bmatrix} 10 & 1 & 4 & 0 \\ 1 & 10 & 5 & -1 \\ 4 & 5 & 10 & 7 \\ 0 & -1 & 7 & 9 \end{bmatrix}, \quad A_0^{-1} =\begin{bmatrix} 105& 167 & -304 & 255\\ 167 & 266 & -484 & 406\\ -304 & -484 & 881 & -739\\ 255 & 406 & -739 & 620 \end{bmatrix},$

for which $\kappa_2(A_0) = 3.57924\times 10^4$. The matrix $A_0$ is therefore a factor 12 more ill conditioned than $W$. Rutishauser did not give a proof of the stated bound.

Moler (2018) asked how ill-conditioned $W$ is relative to matrices in the set

\notag \begin{aligned} \mathcal{S} &= \{\, A\in\mathbb{R}^{4\times 4}: A=A^T \mathrm{~is~nonsingular~with~integer~entries}\nonumber\\ & \hspace{2.9cm} \mathrm{between~1~and~10} \,\}. \end{aligned}

He generated one million random matrices from $\mathcal{S}$ and found that about 0.21 percent of them had a larger condition number than $W$. The matrix with the largest condition number was the indefinite matrix

$\notag A_1 = \begin{bmatrix} 1 & 3 & 10 & 10\\ 3 & 4 & 8 & 9\\ 10 & 8 & 3 & 9\\ 10 & 9 & 9 & 3 \end{bmatrix}, \quad A_1^{-1} = \begin{bmatrix} 573 & -804 & 159 & 25\\ -804 & 1128 & -223 & -35\\ 159 & -223 & 44 & 7\\ 25 & -35 & 7 & 1 \end{bmatrix},$

for which $\kappa_2(A_1) \approx 4.80867\times 10^4$. How far is this matrix from being a worst case?

As the Wilson matrix is positive definite, we are also interested in how ill conditioned a matrix in the set

\notag \begin{aligned} \mathcal{P} &= \{\, A\in\mathbb{R}^{4\times 4}: A=A^T \mathrm{~is~symmetric~positive~definite ~with~integer~entries}\nonumber\\ & \hspace{2.9cm} \mathrm{between~1~and~10} \,\} \end{aligned}

can be.

## Condition Number Bounds

We first consider bounds on $\kappa_2(A)$ for $A \in \mathcal{S}$. It is possible to obtain a bound from first principles by using the relation $A^{-1} = \mathrm{adj}(A)/\det(A)$, where $\mathrm{adj}(A)$ is the adjugate matrix, along with the fact that $|\det(A)| \ge 1$ since $A$ has integer entries. Higham and Lettington (2021) found that the smallest bound they could obtain came from a bound of Merikoski et al. (1997): for nonsingular $B\in\mathbb{R}^{n\times n}$,

$\notag \kappa_2(B) \le \left(\displaystyle\frac{1+x}{1-x}\right)^{1/2}, \quad x = \sqrt{1 - (n/\|B\|_F^2)^n |\det(B)|^2 }.$

Applying this bound to $A\in\mathcal{S}$, using the fact that $(1+x)/(1-x)$ is monotonically increasing for $x\in(0,1)$, gives

$\notag \kappa_2(A) \le 2.97606\dots \times 10^5 =: \beta_S, \quad A\in\mathcal{S}. \qquad (1)$

Another result from Merikoski et al. (1997) gives, for symmetric positive definite $C\in\in\mathbb{R}^{n\times n}$,

$\notag \kappa_2(C) \le \displaystyle\frac{1+x}{1-x}, \quad x = \sqrt{1 - (n/\mathrm{trace}(C))^n \det(C) }.$

For $A\in\mathcal{P}$, since $\det(A) \ge 1$ we have $x \le \sqrt{1 - (1/10)^4}$, and hence

$\notag \kappa_2(A) \le 3.99980 \times 10^4 =: \beta_P, \quad A\in\mathcal{P}. \qquad (2)$

Recall that Rutishauser’s bound is $4\times 10^4$. The bounds (1) and (2) remain valid if we modify the definitions of $\mathcal{S}$ and $\mathcal{P}$ to allow zero elements (note that Rutishauser’s matrix $A_0$ has a zero element).

## Experiment

The sets $\mathcal{S}$ and $\mathcal{P}$ are large: $\mathcal{S}$ has on the order of $10^{10}$ elements. Exhaustively searching over the sets in reasonable time is possible with a carefully optimized code. Higham and Lettington (2021) use a MATLAB code that loops over all symmetric matrices with integer elements between $1$ and $10$ and

• evaluates $\det(A)$ from an explicit expression (exactly computed for such matrices) and discards $A$ if the matrix is singular;
• computes the eigenvalues $\lambda_i$ of $A$ and obtains the condition number as $\kappa_2(A) = \max_i |\lambda_i|/\min_i |\lambda_i|$ (since $A$ is symmetric); and
• for $\mathcal{P}$, checks whether $A$ is positive definite by checking whether the smallest eigenvalue is positive.

The code is available at https://github.com/higham/wilson-opt.

The maximum over $\mathcal{S}$ is attained for

$\notag A_2 = \begin{bmatrix} 2 & 7 & 10 & 10\\ 7 & 10 & 10 & 9\\ 10 & 10 & 10 & 1\\ 10 & 9 & 1 & 9 \end{bmatrix}, \quad A_2^{-1} = \begin{bmatrix} 640 & -987 & 323 & 240\\ -987 & 1522 & -498 & -370\\ 323 & -498 & 163 & 121\\ 240 & -370 & 121 & 90 \end{bmatrix},$

which has $\kappa_2(A_2) \approx 7.6119 \times 10^4$. and determinant $-1$. The maximum over $\mathcal{P}$ is attained for

$\notag A_3 = \begin{bmatrix} 9 & 1 & 1 & 5\\ 1 & 10 & 1 & 9\\ 1 & 1 & 10 & 1\\ 5 & 9 & 1 & 10 \end{bmatrix}, \quad A_3^{-1} = \begin{bmatrix} 188 & 347 & -13 & -405\\ 347 & 641 & -24 & -748\\ -13 & -24 & 1 & 28\\ -405 & -748 & 28 & 873 \end{bmatrix}.$

which has $\kappa_2(A_3) \approx 3.5529 \times 10^4$ and determinant $1$. Obviously, symmetric permutations of these matrices are also optimal.

The following table summarizes the condition numbers of the matrices discussed and how close they are to the bounds.

Matrix $A$ Comment $\kappa_2(A)$ $\beta_S/\kappa_2(A)$ $\beta_P/\kappa_2(A)$
$W$ Wilson matrix $2.98409\times 10^3$ 99.73 13.40
$A_9$ Rutishauser’s matrix $3.57924\times 10^4$ 8.31 1.12
$A_1$ By random sampling $4.80867\times 10^4$ 6.19
$A_2$ Optimal matrices in $\mathcal{S}$ $7.61190\times 10^4$ 3.91
$A_3$ Optimal matrices in $\mathcal{P}$ $3.55286\times 10^4$ 8.38 1.13

Clearly, the bounds are reasonably sharp.

We do not know how Wilson constructed his matrix or to what extent he tried to maximize the condition number subject to the matrix entries being small integers. One possibility is that he constructed it via the factorization in the next section.

## Integer Factorization

The Cholesky factor of the Wilson matrix is

$\notag R = \begin{bmatrix} \sqrt{5} & \frac{7\,\sqrt{5}}{5} & \frac{6\,\sqrt{5}}{5} & \sqrt{5}\\[\smallskipamount] 0 & \frac{\sqrt{5}}{5} & -\frac{2\,\sqrt{5}}{5} & 0\\[\smallskipamount] 0 & 0 & \sqrt{2} & \frac{3\,\sqrt{2}}{2}\\[\smallskipamount] 0 & 0 & 0 & \frac{\sqrt{2}}{2} \end{bmatrix} \quad (W = R^TR).$

Apart from the zero $(2,4)$ element, it is unremarkable. If we factor out the diagonal then we obtain the $LDL^T$ factorization, which has rational elements:

$\notag L = \begin{bmatrix}1 & 0 & 0 & 0\\ \frac{7}{5} & 1 & 0 & 0\\ \frac{6}{5} & -2 & 1 & 0\\ 1 & 0 & \frac{3}{2} & 1 \end{bmatrix}, \quad D = \begin{bmatrix}5 & 0 & 0 & 0\\ 0 & \frac{1}{5} & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & \frac{1}{2} \end{bmatrix} \quad (W = LDL^T).$

Suppose we drop the requirement of triangularity and ask whether the Wilson matrix has a factorization $W = Z^T\!Z$ with a $4\times4$ matrix $Z$ of integers. It is known that every symmetric positive definite $n\times n$ matrix $A$ of integers with determinant $1$ has a factorization $A = Z^T\!Z$ with $Z$ an $n\times n$ matrix of integers as long as $n \le 7$, but examples are known for $n = 8$ for which the factorization does not exist. This result is mentioned by Taussky (1961) and goes back to Hermite, Minkowski, and Mordell. Higham and Lettington (2021) found the integer factor

$\notag Z_0 = \begin{bmatrix} 2 & 3 & 2 & 2\\ 1 & 1 & 2 & 1\\ 0 & 0 & 1 & 2\\ 0 & 0 & 1 & 1 \end{bmatrix}$

of $W$, which is block upper triangular so can be thought of as a block Cholesky factor. Higham, Lettington, and Schmidt (2021) draw on recent research that links the existence of such factorizations to number-theoretic considerations of quadratic forms to show that for the existence of an integer solution $Z$ to $A = Z^TZ$ it is necessary that a certain quadratic equation in $n$ variables has an integer solution. In the case of the Wilson matrix the equation is

$2 w^2+x_1^2+x_1 x_2+x_1 x_3+x_2^2+x_2 x_3+x_3^2=952.$

The authors solve this equation computationally and find $Z_1$ and two rational factors:

$\notag Z_1=\left[ \begin{array}{cccc} \frac{1}{2} & 1 & 0 & 1 \\ \frac{3}{2} & 2 & 3 & 3 \\ \frac{1}{2} & 1 & 0 & 0 \\ \frac{3}{2} & 2 & 1 & 0 \\ \end{array} \right], \quad Z_2=\left[ \begin{array}{@{\mskip2mu}rrrr} \frac{3}{2} & 2 & 2 & 2 \\ \frac{3}{2} & 2 & 2 & 1 \\ \frac{1}{2} & 1 & 1 & 2 \\ -\frac{1}{2} & -1 & 1 & 1 \\ \end{array} \right].$

They show that these matrices are the only factors $Z\in\frac{1}{16}\mathbb{Z}$ of $W$ up to left multiplication by integer orthogonal matrices.

## Conclusions

The Wilson matrix has provided sterling service throughout the digital computer era as a convenient symmetric positive definite matrix for use in textbook examples and for testing algorithms. The recent discovery of its integer factorization has led to the development of new theory on when general $n\times n$ integer matrices $A$ can be factored as $A = Z^TZ$ (when $A$ is symmetric positive definite) or $A = Z^2$ (a problem also considered in Higham, Lettington, and Schmidt (2021)), with integer $Z$.

Olga Taussky Todd wrote in 1961 that “matrices with integral elements have been studied for a very long time and an enormous number of problems arise, both theoretical and practical.” We wonder what else can be learned from the Wilson matrix and other integer test matrices.

## References

This is a minimal set of references, which contain further useful references within.

# What Is a Rank-Revealing Factorization?

In many applications a matrix $A\in\mathbb{R}^{m\times n}$ has less than full rank, that is, $r = \mathrm{rank}(A) < \min(m,n)$. Sometimes, $r$ is known, and a full-rank factorization $A = GH$ with $G\in\mathbb{R}^{m \times r}$ and $H\in\mathbb{R}^{r \times n}$, both of rank $r$, is given—especially when $r = 1$ or $r = 2$. Often, though, the rank $r$ is not known. Moreover, rather than being of exact rank $r$, $A$ is merely close to a rank $r$ matrix because of errors from various possible sources.

What is usually wanted is a factorization that displays how close $A$ is to having particular ranks and provides an approximation to the range space of a lower rank matrix. The ultimate tool for providing this information is the singular value decomposition (SVD)

$\notag A = U\Sigma V^T, \quad \Sigma = \mathrm{diag}(\sigma_1,\dots, \sigma_p)\in\mathbb{R}^{m\times n},$

where $p = \min(m,n)$, $\sigma_1\ge \sigma_2\ge \cdots \ge \sigma_p \ge 0$, and $U\in\mathbb{R}^{m\times m}$ and $V\in\mathbb{R}^{n\times n}$ are orthogonal. The Eckart–Young theorem says that

$\notag \min_{\mathrm{rank}(B) = k} \|A-B\|_q = \begin{cases} \sigma_{k+1}, & q = 2, \\ \Bigl(\sum_{i=k+1}^r \sigma_i^2\Bigr)^{1/2}, & q = F, \end{cases}$

and that the minimum is attained at

$\notag A_k = U \Sigma_k V^T, \quad \Sigma_k = \mathrm{diag}(\sigma_1, \dots, \sigma_k, 0, \dots, 0),$

so $A_k$ is the best rank-$k$ approximation to $A$ in both the $2$-norm and the Frobenius norm.

Although the SVD is expensive to compute, it may not be significantly more expensive than alternative factorizations. However, the SVD is expensive to update when a row or column is added to or removed from the matrix, as happens repeatedly in signal processing applications.

Many different definitions of a rank-revealing factorization have been given, and they usually depend on a particular matrix factorization. We will use the following general definition.

Definition 1. A rank-revealing factorization (RRF) of $A\in\mathbb{R}^{m\times n}$ is a factorization

$\notag A = XDY^T, \quad X\in\mathbb{R}^{m\times p}, \quad D\in\mathbb{R}^{p\times p}, \quad Y\in\mathbb{R}^{n\times p},$

where $p \le \min(m,n)$, $D$ is diagonal and nonsingular, and $X$ and $Y$ are well conditioned.

An RRF concentrates the rank deficiency and ill condition of $A$ into the diagonal matrix $D$. An RRF clearly exists, because the SVD is one, with $X$ and $Y$ having orthonormal columns and hence being perfectly conditioned. Justification for this definition comes from a version of Ostrowski’s theorem, which shows that

$\notag \sigma_i(A) = \theta_i \sigma_i(D), \quad i = 1\colon \min(m,n), \qquad (1)$

where $\sigma_p(X)\sigma_p(Y) \le \theta_i \le \sigma_1(X) \sigma_1(Y)$. Hence as long as $X$ and $Y$ are well conditioned, the singular values are good order of magnitude approximations to those of $A$ up a scale factor.

Without loss of generality we can assume that

$\notag D = \mathrm{diag}(d_i), \quad |d_1| \ge |d_2| \ge \cdots \ge |d_p|$

(since $XDY^T = XP\cdot P^TDP \cdot P^T Y^T$ for any permutation matrix $P$ and the second expression is another RRF). For $\widetilde{A}_k = X \mathrm{diag}(d_1,\dots,d_k,0,\dots,0)Y^T$ we have

$\notag \|A - \widetilde{A}_k\| \le \|X\| \|Y\| \|\mathrm{diag}(0,\dots,0,d_{k+1},\dots,d_p)\|,$

so $A$ is within distance of order $|d_{k+1}|$ from the rank-$k$ matrix $\widetilde{A}_k$, which is the same order as the distance to the nearest rank-$k$ matrix if $|d_{k+1}| \approx \sigma_{k+1}$.

Definition 2 is a strong requirement, since it requires all the singular values of $A$ to be well approximated by the (scaled) diagonal elements of $D$. We will investigate below how it compares with another definition of RRF.

## Numerical Rank

An RRF helps to determine the numerical rank, which we now define.

Definition 2. For a given $\epsilon > 0$ the numerical rank of $A$ is the largest integer $k$ such that $\sigma_k > \epsilon$.

By the Eckart–Young theorem, the numerical rank is the smallest rank attained over all $A+E$ with $\|E\|_2 \le \epsilon$. For the numerical rank to be meaningful in the sense that it is unchanged if $\epsilon$ is perturbed slightly, we need $\epsilon$ not to be too close to $\sigma_k$ or $\sigma_{k+1}$, which means that there must be a significant gap between these two singular values.

## QR Factorization

One might attempt to compute an RRF by using a QR factorization $A = QR$, where $Q\in\mathbb{R}^{m\times n}$ has orthonormal columns, $R\in\mathbb{R}^{n\times n}$ is upper triangular, and we assume that $m\ge n$. In Definition 1, we can take

$\notag X = I, \quad D = \mathrm{diag}(R), \quad Y^T = D^{-1}R. \qquad (*)$

However, it is easy to see that QR factorization in its basic form is flawed as a means for computing an RRF. Consider the matrix

$\notag A = \begin{bmatrix} 0 & 1 & 0 &0\\ 0 & 0 & 1 &0 \\ 0 & 0 & 0 &1\\ 0 & 0 & 0 &0 \end{bmatrix},$

which is a Jordan block with zero eigenvalue. This matrix is its own QR factorization ($R = A$), and the prescription $(*)$ gives $D = 0$, so $A \ne XDY^T$. The essential problem is that the diagonal of $R$ has no connection with the nonzero singular values of $A$. What is needed are column permutations: $A\Pi = \mathrm{diag}(1,1,1,0)$ for the permutation matrix $\Pi$ that reorders $[a_1,a_2,a_3,a_4]$ to $[a_2,a_3,a_4,a_1]$, and this is a perfect RRF with $X = Y = I$.

For a less trivial example, consider the matrix

$\notag A = \left[\begin{array}{rrrr} 1 & 1 &\theta &0\\ 1 & -1 & 2 &1 \\ 1 & 0 &1+\theta &-1\\ 1 &-1 & 2 &-1 \end{array}\right], \quad \theta = 10^{-8}. \qquad (\dagger)$

Computing the QR factorization we obtain

R =
-2.0000e+00   5.0000e-01  -2.5000e+00   5.0000e-01
0   1.6583e+00  -1.6583e+00  -1.5076e-01
0            0  -4.2640e-09   8.5280e-01
0            0            0  -1.4142e+00


The $(3,3)$ element tells us that $A$ is within distance about $4\times 10^{-9}$ of being rank deficient and so has a singular value bounded above by this quantity, but it does not provide any information about the next larger singular value. Moreover, in $(*)$, $\kappa_2(Y)$ is of order $10^{16}$ for this factorization. We need any small diagonal elements to be in the bottom right-hand corner, and to achieve this we need to introduce column permutations to move the “dependent columns” to the end.

## QR Factorization With Column Pivoting

A common method for computing an RRF is QR factorization with column pivoting, which for a matrix $A\in\mathbb{R}^{m\times n}$ with $m\ge n$ computes a factorization $A\Pi = QR$, where $\Pi$ is a permutation matrix, $Q\in\mathbb{R}^{m\times n}$ has orthonormal columns, and $R\in\mathbb{R}^{n\times n}$ is upper triangular and satisfies the inequalities

$\notag |r_{kk}|^2 \ge \displaystyle\sum_{i=k}^j |r_{ij}|^2, \quad j=k+1\colon n, \quad k=1\colon n. \qquad (2)$

In particular,

$\notag |r_{11}| \ge |r_{22}| \ge \cdots \ge |r_{nn}|. \qquad(3)$

If $|r_{kk}| \ge \epsilon \ge |r_{k+1,k+1}|$ with $\epsilon > 0$ then we can write

$\notag R = \begin{array}[b]{@{\mskip33mu}c@{\mskip-16mu}c@{\mskip-10mu}c@{}} \scriptstyle k & \scriptstyle n-k & \\ \multicolumn{2}{c}{ \left[\begin{array}{c@{~}c@{~}} R_{11}& R_{12} \\ 0 & R_{22} \\ \end{array}\right]} & \mskip-12mu\ \begin{array}{c} \scriptstyle k \\ \scriptstyle n-k \end{array} \end{array}, \qquad(4)$

with

$\notag \|R_{22}\|_2 \le \|R_{22}\|_F \le 2^{-1/2}(n-k+1)\epsilon.$

Hence $R$ is within $2$-norm distance $2^{-1/2}(n-k+1)\epsilon$ of the rank-$k$ matrix $\left[\begin{smallmatrix} R_{11} & R_{12} \\ 0 & 0 \end{smallmatrix}\right]$. Note that if $Q = [Q_1~Q_2]$ is partitioned conformally with $Q$ in (4) then

$\notag A\Pi = \begin{bmatrix} Q_1 & Q_2 \end{bmatrix} \begin{bmatrix} R_{11} & R_{12} \\ 0 & R_{22} \\ \end{bmatrix} = Q_1 \begin{bmatrix} R_{11} & R_{12} \end{bmatrix} + \begin{bmatrix} 0 & Q_2 R_{22} \end{bmatrix},$

so $\| A\Pi - Q_1 [R_{11}~R_{12}]\|_2 \le \|R_{22}\|_2$, which means that $Q_1$ provides an $O(\epsilon)$ approximation to the range of $A$.

To assess how good an RRF this factorization is (with $p = n$) we write it as

$\notag A = QR\Pi^T = Q D Y^T, \quad D = \mathrm{diag}(r_{ii}), \quad Y^T = D^{-1}R \Pi^T. \quad (\#)$

Applying (1) gives

$\notag \sigma_i(A) = \theta_i \sigma_i(D), \quad i = 1\colon p, \qquad (5)$

where $\sigma_n(Y)\le \theta_i \le \sigma_1(Y)$, since $Q$ has orthonormal columns and so has unit singular values. Now $D^{-1}R$ has unit diagonal and, in view of (2), its off-diagonal elements are bounded by $1$. Therefore $\sigma_1(Y) = \|Y\|_2 \le \|Y\|_F \le (n(n+1)/2)^{1/2}$. On the other hand, $\sigma_p(Y)^{-1} \le 2^{n-1}$ by Theorem 1 in Bounds for the Norm of the Inverse of a Triangular Matrix. Therefore

$\notag 2^{1-n} \le \theta_i \le (n(n+1)/2)^{1/2}.$

The lower bound is an approximate equality for small $\tau$ for the triangular matrix

$\notag R_n(\theta) = \mathrm{diag}(1,s,\dots,s^{n-1}) \begin{bmatrix} 1 & -c & -c & \dots & -c \\ & 1 & -c & \dots & -c \\ & & \ddots &\ddots & \vdots \\ & & &\ddots & -c \\ & & & & 1 \end{bmatrix}, \quad c=\cos\tau, \quad s=\sin\tau,$

devised by Kahan, which is invariant under QR factorization with column pivoting. Therefore QR factorization with column pivoting is not guaranteed to reveal the rank, and indeed it can fail to do so by an exponentially large factor.

For the matrix $(\dagger)$, QR with column pivoting reorders $A$ to $A\Pi = [a_3,~a_4,~a_2,~a_1]$ and yields

R =
-3.0000e+00   3.3333e-01   1.3333e+00  -1.6667e+00
0  -1.6997e+00   2.6149e-01   2.6149e-01
0            0   1.0742e+00   1.0742e+00
0            0            0   3.6515e-09


This $R$ suggests a numerical rank of $3$ for $\epsilon = 10^{-8}$ (say). In fact, this factorization provides a very good RRF, as in $(\#)$ we have $\kappa_2(Y) \approx 3.4$.

## QR Factorization with Other Pivoting Choices

Consider a $QR$ factorization $A\Pi = QR$ with triangular factor partitioned as

$\notag R = \begin{array}[b]{@{\mskip33mu}c@{\mskip-16mu}c@{\mskip-10mu}c@{}} \scriptstyle k & \scriptstyle n-k & \\ \multicolumn{2}{c}{ \left[\begin{array}{c@{~}c@{~}} R_{11}& R_{12} \\ 0 & R_{22} \\ \end{array}\right]} & \mskip-12mu\ \begin{array}{c} \scriptstyle k \\ \scriptstyle n-k \end{array} \end{array}. \qquad (6)$

We have

\notag \begin{aligned} \sigma_{\min}(R_{11}) &\le \sigma_k(A), \quad \qquad (7)\\ \sigma_{\max}(R_{22}) &\ge \sigma_{k+1}(A), ~\qquad (8) \end{aligned}

where (7) is from singular value interlacing inequalities and (8) follows from the Eckart-Young theorem, since setting $R_{22}$ to zero gives a rank-$k$ matrix. Suppose $A$ has numerical rank $k$ and $\sigma_{k+1} \ll \sigma_k$. We would like to be able to detect this situation from $R$, so clearly we need

$\notag \sigma_{\min}(R_{11}) \approx \sigma_k(A), \quad \sigma_{\max}(R_{22}) \approx \sigma_{k+1}(A). \qquad (9)$

In view of the inequalities (7) and (8) this means that we wish to choose $\Pi$ maximize $\sigma_{\min}(R_{11})$ and minimize $\sigma_{\max}(R_{22})$.

Some theoretical results are available on the existence of such QR factorizations. First, we give a result that shows that for $k = n-1$ the approximations in (9) can hold to within a factor $n^{1/2}$.

Theorem 1. For $A\in\mathbb{R}^{m\times n}$ with $m\ge n$ there exists a permutation $\Pi$ such that $A$ has the QR factorization $A\Pi = QR$ with $|r_{nn}| \le n^{1/2}\sigma_n(A)$ and $\sigma_{\min}(R_{11}) \ge n^{-1/2} \sigma_{n-1}(A)$, where $R_{11} = R(1\colon n-1, 1\colon n-1)$.

Proof. Let $Av = \sigma_n u$, with $\|v\|_2 = \|u\|_2 = 1$ and let $\Pi^T$ be such that $\widetilde{v} = \Pi^Tv$ satisfies $|\widetilde{v}_n| = \|\widetilde{v}\|_{\infty}$. Then if $A\Pi = QR$ is a QR factorization,

\notag \begin{aligned} \sigma_n = \| \sigma_n u \|_2 = \| Av \|_2 = \| QR\Pi^Tv \|_2 = \| R\mskip1mu \widetilde{v} \|_2 \ge | r_{nn} \widetilde{v}_n | \ge n^{-1/2} | r_{nn} |, \end{aligned}

since $\|\widetilde{v}\|_2 = 1$, which yields the result.

Next, we write $\Pi = [\Pi_1~\pi]$, where $\pi\in\mathbb{R}^n$, and partition

$\notag R = \begin{bmatrix} R_{11} & R_{12} \\ 0 & R_{22} \\ \end{bmatrix}$

with $R_{11}\in\mathbb{R}^{(n-1)\times (n-1)}$. Then

$A\Pi_1 = Q \begin{bmatrix} R_{11} \\ 0 \end{bmatrix}$

implies $\sigma_{\min}(A\Pi_1) = \sigma_{\min}(R_{11})$. On the other hand, if $A = U\Sigma V^T$ is an SVD with $U\in\mathbb{R}^{m\times n}$, $\Sigma = \mathrm{diag}(D_1,\sigma_n)\in\mathbb{R}^{n\times n}$, and $V = [V_1~v]$ then

$\notag A\Pi_1 = U\Sigma V^T \Pi_1 = U \begin{bmatrix} D_1 & 0 \\ 0 & \sigma_n \\ \end{bmatrix} \begin{bmatrix} V_1^T \\ v^T \end{bmatrix} \Pi_1 = U \begin{bmatrix} D_1V_1^T \\ \sigma_n v^T \end{bmatrix} \Pi_1,$

so

$\notag \sigma_{\min}(A\Pi_1) = \sigma_{\min}\left( \begin{bmatrix} D_1V_1^T \\ \sigma_n v^T \end{bmatrix} \Pi_1 \right) \ge \sigma_{\min}(D_1V^T\Pi_1) \ge \sigma_{n-1}\sigma_{\min}(V^T\Pi_1).$

Finally, we note that we can partition the orthogonal matrix $V^T\Pi_1$ as

$\notag V^T\Pi = \begin{bmatrix} V_1^T\Pi_1 & V_1^T\pi \\ v^T\Pi_1 & v^T\pi \end{bmatrix},$

and the CS decomposition implies that

$\notag \sigma_{\min}(V_1^T\Pi_1) = \sigma_{\min}(v^T\pi) = |v^T\pi| = |\widetilde{v}_n| \ge n^{-1/2}.$

Hence $\sigma_{\min}(R_{11}) \ge n^{-1/2} \sigma_{n-1}$, as required. $~\square$

Theorem 1 is a special case of the next result of Hong and Pan (1992).

Theorem 2. For $A\in\mathbb{R}^{m\times n}$ with $m\ge n$ and any $k$ there exists a permutation matrix $\Pi$ such that $A$ has the QR factorization $A\Pi = QR$ where, with $R$ partitioned as in (6),

$\notag \sigma_{\max}(R_{22}) \le f(k,n) \sigma_{k+1}(A), \quad \sigma_{\min}(R_{11}) \ge f(k,n)^{-1} \sigma_k(A),$

where $f(k,n) = (k(n-k) + \min(k,n-k))^{1/2}$.

The proof of Theorem 2 is constructive and chooses $\Pi$ to move a submatrix of maximal determinant of $V_2$ to the bottom of $V_2$, where $V_2$ comprises the last $n-k$ columns of the matrix of right singular vectors.

Theorem 2 shows the existence of an RRF up to the factor $f(k,n) \le (n+1)/2$, but it does not provide an efficient algorithm for computing one.

How do the conditions (9) relate to Definition 1? Since algorithms for computing an RRF are usually not specialized to any particular $k$, it is reasonable to ask that (9) holds for all $k$. We consider what can be said if the first condition in (9) holds as an equality for all $k$.

Lemma 3. If $R\in\mathbb{R}^{m\times n}$ is upper triangular and satisfies $\sigma_{\min}(R_{11}) = \sigma_k(R)$ for $k=1\colon n-1$ in the partitioning (6) then $R$ is diagonal.

Proof. The proof is by induction. Assume that $R_{k-1} = \mathrm{diag}(\sigma_1(R),\dots,\sigma_{k-1}(R))$. This is clearly true for $k = 2$, since $|r_{11}| = \sigma_1(R)$. Write

$\notag R_k = \begin{bmatrix} R_{k-1} & z \\ 0 & \rho \end{bmatrix}.$

Then $|\rho| \ge \sigma_{\min}(R_k)$ by (8). Also $\sigma_i(R_k) \le \sigma_i(R)$ for $i = 1\colon k$ by standard singular value inequalities. We therefore have

\notag \begin{aligned} \sum_{i=1}^k \sigma_i(R)^2 \ge \sum_{i=1}^k \sigma_i(R_k)^2 = \|R_k\|_F^2 = \|R_{k-1}\|_F^2 + \|z\|_2^2 + \rho^2\\ = \sum_{i=1}^{k-1} \sigma_i(R)^2 + \|z\|_2^2 + \rho^2 \ge \sum_{i=1}^k \sigma_i(R)^2 + \|z\|_2^2. \end{aligned}

It follows that $z = 0$, and so $R_k$ is diagonal, which completes the induction.

We conclude from the lemma that if the first condition in (9) is an equality for all $k$ then we have a perfect RRF $A\Pi = QR$ with $R$ diagonal. Therefore if the approximations in (9) are reasonably good for all $k$ we should have a reasonably good RRF.

Much work has been done on algorithms that choose the permutation matrix $\Pi$ in a different way to column pivoting or post-process a QR factorization with column pivoting, with the aim of satisfying (9) at reasonable cost. Typically, these algorithms involve estimating singular values and singular vectors. We are not aware of any algorithm that is guaranteed to satisfy (9) and requires only $O(n^3)$ flops.

## UTV Decomposition

By applying Householder transformations on the right, a QR factorization with column pivoting can be turned into a complete orthogonal decomposition of $A\in\mathbb{R}^{m\times n}$, which has the form

$\notag A = U \begin{bmatrix} T & 0 \\ 0 & 0 \end{bmatrix} V^T, \qquad (10)$

where $T\in\mathbb{R}^{r \times r}$ is upper triangular and $U\in\mathbb{R}^{m\times m}$ and $V\in\mathbb{R}^{n\times n}$ are orthogonal. Stewart (1998) calls (6) with $T$ upper triangular or lower triangular a UTV decomposition and he defines a rank-revealing UTV decomposition of numerical rank $r$ by

\notag \begin{aligned} A &= U \begin{bmatrix} T & F \\ 0 & G \end{bmatrix} V^T, \qquad T\in\mathbb{R}^{r \times r}, \\ & \sigma_r(T) \approx \sigma_r(A), \quad \|F\|_F^2 + \|G\|_F^2 \approx \sigma_{r+1}^2 + \cdots + \sigma_n^2. \end{aligned}

The UTV decomposition is easy to update (when a row is added) and downdate (when a row is removed) using Givens rotations and it is suitable for parallel implementation. Initial determination of the UTV decomposition can be done by applying the updating algorithm as the rows are brought in one at a time.

## LU Factorization

Instead of QR factorization we can build an RRF from an LU factorization with pivoting. For $A\in\mathbb{R}^{m\times n}$ with $m\ge n$, let

$\notag \Pi_1 A \Pi_2 = LU = \begin{bmatrix} L_{11} & 0 \\ L_{12} & L_{22} \end{bmatrix} \begin{bmatrix} U_{11} & U_{12}\\ 0 & U_{22} \end{bmatrix},$

where $\Pi_1$ and $\Pi_2$ are permutation matrices, $L$ and $U$ are $m\times n$ lower and $n\times n$ upper triangular, respectively, and $L_{11}$ and $U_{11}$ are $k\times k$. Analogously to (7) and (8), we always have $\sigma_{\min}(L_{11}U_{11}) \le \sigma_k(A)$ and $\sigma_{\max}(L_{22}U_{22}) \ge \sigma_{k+1}(A)$. With a suitable pivoting strategy we can hope that $\sigma_{\min}(L_{11}U_{11}) \approx \sigma_k(A)$ and $\sigma_{\max}(L_{22}U_{22}) \approx \sigma_{k+1}(A)$.

A result of Pan (2000) shows that an RRF based on LU factorization always exists up to a modest factor $f(k,n)$. This is analogue for LU factorization of Theorem 2.

Theorem 3 For $A\in\mathbb{R}^{m\times n}$ with $m\ge n$ and any $k$ there exist permutation matrices $\Pi_1$ and $\Pi_2$ such that

$\notag \Pi_1 A \Pi_2 = LU = \begin{bmatrix} L_{11} & 0 \\ L_{12} & I_{m-k,n-k} \end{bmatrix} \begin{array}[b]{@{\mskip33mu}c@{\mskip-16mu}c@{\mskip-10mu}c@{}} \scriptstyle k & \scriptstyle n-k & \\ \multicolumn{2}{c}{ \left[\begin{array}{c@{~}c@{~}} U_{11}& U_{12} \\ 0 & U_{22} \\ \end{array}\right]} & \mskip-12mu\ \begin{array}{c} \scriptstyle k \\ \scriptstyle n-k \end{array} \end{array},$

where $L_{11}$ is unit lower triangular, $U_{11}$ is upper triangular, and

$\notag \sigma_{\max}(U_{22}) \le f(k,n) \sigma_{k+1}(A), \quad \sigma_{\min}(L_{11}U_{11}) \ge f(k,n)^{-1} \sigma_k(A),$

where $f(k,n) = k(n-k) + 1$.

Again the proof is constructive, but the permutations it chooses are too expensive to compute. In practice, complete pivoting often yields a good RRF.

In terms of Definition 1, an RRF has

$\notag X = \Pi_1^TL D, \quad D = \mathrm{diag}(u_{ii}), \quad Y^T = D^{-1}U\Pi_2. \qquad (\ddagger)$

For the matrix ($\dagger$), the $U$ factor for LU factorization without pivoting is

U =
1.0000e+00   1.0000e+00   1.0000e-08            0
0  -2.0000e+00   2.0000e+00   1.0000e+00
0            0   5.0000e-09  -1.5000e+00
0            0            0  -2.0000e+00


As for QR factorization without pivoting, an RRF is not obtained from $(\ddagger)$.. However, with complete pivoting we obtain

U =
2.0000e+00   1.0000e+00  -1.0000e+00   1.0000e+00
0  -2.0000e+00            0            0
0            0   1.0000e+00   1.0000e+00
0            0            0  -5.0000e-09


which yields a very good RRF $(\ddagger)$ with $\kappa_2(X) = 3.5$ and $\kappa_2(Y) = 3.4$.

## Notes

QR factorization with column pivoting is difficult to implement efficiently, as the criterion for choosing the pivots requires the norms of the active parts of the remaining columns and this requires a significant amount of data movement. In recent years, randomized RRF algorithms have been developed that use projections with random matrices to make pivot decisions based on small sample matrices and thereby reduce the amount of data movement. See, for example, Martinsson et al. (2019).

## References

This is a minimal set of references, which contain further useful references within.

# What Is a Pseudo-Orthogonal Matrix?

A matrix $Q\in\mathbb{R}^{n\times n}$ is pseudo-orthogonal if

$\notag Q^T \Sigma Q = \Sigma, \qquad (1)$

where $\Sigma = \mathrm{diag}(\pm 1)$ is a signature matrix. A matrix $Q$ satisfying (1) is also known as a $J$-orthogonal matrix, where $J$ is another notation for a signature matrix. Of course, if $\Sigma = I$ then $Q$ is orthogonal.

It is easy to show that $Q^T$ is also pseudo-orthogonal. Furthermore, $Q$ is clearly nonsingular and it satisfies

$\notag Q = \Sigma Q^{-T}\Sigma. \qquad (2)$

Since $\Sigma$ is orthogonal, this equation implies that $\|Q\|_\ell = \|Q^{-T}\|_\ell = \|Q^{-1}\|_\ell$ and hence that

$\notag \kappa_p(Q) = \|Q\|_\ell^2, \quad \ell = 2,F. \qquad(3).$

What are some examples of pseudo-orthogonal matrices? For $n = 2$ and $\Sigma = \left[\begin{smallmatrix}1 & 0 \\ 0 & -1 \end{smallmatrix}\right]$, $Q$ is of the form

$\notag Q = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \quad ab - cd = 0, \quad a^2 - c^2 = 1, \quad b^2 - d^2 = -1,$

which includes the matrices

$\notag Q = \begin{bmatrix} \cosh \theta & -\sinh\theta \\ -\sinh\theta & \cosh\theta \end{bmatrix}, \quad \theta\in\mathbb{R}. \qquad (4)$

The Lorentz group, representing symmetries of the spacetime of special relativity, corresponds to $4\times 4$ matrices with $\Sigma = \mathrm{diag}(1,1,1,-1)$.

Equation (2) shows that $Q$ is similar to the inverse of its transpose and hence (since every matrix is similar to its transpose) similar to its inverse. It follows that if $\lambda$ is an eigenvalue of $Q$ then $\lambda^{-1}$ is also an eigenvalue and it has the same algebraic and geometric multiplicities as $\lambda$.

By permuting rows and columns in (1) we can arrange that

$\notag \Sigma = \Sigma_{p,q} := \begin{bmatrix} I_p & 0 \\ 0 & -I_q \end{bmatrix}. \qquad (5)$

We assume that $\Sigma$ has this form throughout the rest of this article. This form of $\Sigma$ allows us to conveniently characterize matrices that are both orthogonal and pseudo-orthogonal. Such a matrix must satisfy $\Sigma Q = Q\Sigma$, which means that $Q = \mathrm{diag}(Q_{11},Q_{22})$, and any such orthogonal matrix is pseudo-orthogonal.

## Applications

Pseudo-orthogonal matrices arise in hyperbolic problems, that is, problems where there is an underlying indefinite scalar product or weight matrix. An example is the problem of downdating the Cholesky factorization, where in the simplest case we have the Cholesky factorization $C = R^T\!R$ of a symmetric positive definite $C\in\mathbb{R}^{n\times n}$ and want the Cholesky factorization of $\widetilde{C} = C - zz^T$, which is assumed to be symmetric positive definite. A more general downdating problem is that we are given

$\notag A = \begin{array}[b]{cc} \left[\begin{array}{@{}c@{}} A_1\\ A_2 \end{array}\right] & \mskip-22mu\ \begin{array}{l} \scriptstyle p \\ \scriptstyle q \end{array} \end{array}, \quad p\ge n,$

and the Cholesky factorization $A^T\!A = R^T\!R$ and wish to obtain the Cholesky factor $S$ of $A_1^TA_1 = R^T\!R - A_2^TA_2$. Note that $R$ and $S$ are $n\times n$. This task arises when we solve a regression problem after the observations corresponding to $A_2$ have been removed. The simple case above corresponds to removing one row ($q = 1$). Assuming that $q \ll p$, we would like to obtain $S$ cheaply from $R$, and numerical stability considerations dictate that we should avoid explicitly forming $A_1^TA_1$. If we can find a pseudo-orthogonal matrix $Q$ such that

$\notag Q \begin{bmatrix} R \\ A_2 \end{bmatrix} = \begin{bmatrix} S \\ 0 \end{bmatrix}, \qquad (6)$

with $\Sigma$ given by (5) and $S\in\mathbb{R}^{n\times n}$ upper triangular, then

$\notag A_1^TA_1 = \begin{bmatrix} R \\ A_2 \end{bmatrix}^T \Sigma \begin{bmatrix} R \\ A_2 \end{bmatrix} = \begin{bmatrix} R \\ A_2 \end{bmatrix}^T Q^T \Sigma Q \begin{bmatrix} R \\ A_2 \end{bmatrix} = \begin{bmatrix} S \\ 0 \end{bmatrix}^T \Sigma \begin{bmatrix} S \\ 0 \end{bmatrix} = S^T\!S,$

so $S$ is the desired Cholesky factor.

The factorization (6) is called a hyperbolic QR factorization and it can be computed by using hyperbolic rotations to zero out the elements of $A_2$. A $2\times2$ hyperbolic rotation has the form (4), and an $n\times n$ hyperbolic rotation is an identity matrix with a $2\times 2$ hyperbolic rotation embedded in it at the intersection of rows and columns $i$ and $j$, for some $i$ and $j$.

In general, a hyperbolic QR factorization of $A\in\mathbb{R}^{m\times n}$ with $m = p+q$ and $p\ge n$ has the form $QA = \left[\begin{smallmatrix} R \\ 0 \end{smallmatrix}\right]$ with $Q$ pseudo-orthogonal with respect to $\Sigma = \Sigma_{p,q}$ and $R \in\mathbb{R}^{n\times n}$ upper triangular. The factorization exists if $A^T\Sigma A$ is positive definite.

Another hyperbolic problem is the indefinite least squares problem

$\notag \min_x \,(b-Ax)^T \Sigma (b-Ax), \qquad (7)$

where $A\in\mathbb{R}^{m\times n}$, $m\ge n$, and $b\in\mathbb{R}^m$ are given, and $\Sigma = \Sigma_{p,q}$ with $m = p + q$. For $p=0$ or $q=0$ we have the standard least squares (LS) problem and the quadratic form is definite, while for $pq>0$ the problem is to minimize a genuinely indefinite quadratic form. This problem arises, for example, in the area of optimization known as $H^{\infty}$ smoothing.

The normal equations for (7) are $A^T\Sigma Ax = A^T\Sigma b$, and since the Hessian matrix of the quadratic objective function in (7) is $A^T\Sigma A$ it follows that the indefinite least squares problem has a unique solution if and only if $A^T\Sigma A$ is positive definite. To solve the problem we can use a hyperbolic QR factorization $QA = \left[\begin{smallmatrix} R \\ 0 \end{smallmatrix}\right]$ to write

\notag \begin{aligned} A^T\Sigma A &= A^T Q^T \Sigma Q A = \begin{bmatrix} R \\ 0 \end{bmatrix}^T \begin{bmatrix} I_p & 0 \\ 0 & -I_q \end{bmatrix} \begin{bmatrix} R \\ 0 \end{bmatrix} = R^T\!R, \\ A^T\Sigma b &= A^T Q^T\Sigma Q b = \begin{bmatrix} R \\ 0 \end{bmatrix}^T \! \Sigma Q b. \end{aligned}

Solving the problem now reduces to solving the triangular system $Rx = d$, where $d$ comprises the first $n$ components of $Qb$. The same equation can also be obtained without using the normal equations by substituting the hyperbolic QR factorization into (7).

## The Exchange Operator

A simple technique exists for converting pseudo-orthogonal matrices into orthogonal matrices and vice versa. Let $A\in\mathbb{R}^{n\times n}$ with $n = p + q$, partition

$\notag A = \mskip5mu \begin{array}[b]{@{\mskip-20mu}c@{\mskip0mu}c@{\mskip-1mu}c@{}} & \mskip10mu\scriptstyle p & \scriptstyle q \\ \mskip15mu \begin{array}{r} \scriptstyle p \\ \scriptstyle q \end{array}~ & \multicolumn{2}{c}{\mskip-15mu \left[\begin{array}{c@{~}c@{~}} A_{11} & A_{12}\\ A_{21} & A_{22} \end{array}\right] } \end{array}, \qquad (8)$

and assume $A_{11}$ is nonsingular. The exchange operator is defined by

$\notag \mathrm{exc}(A) = \begin{bmatrix} A_{11}^{-1} & -A_{11}^{-1}A_{12} \\ A_{21}A_{11}^{-1} & A_{22} -A_{21}A_{11}^{-1}A_{12} \end{bmatrix}.$

It is easy to see that the exchange operator is involutory, that is,

$\notag \mathrm{exc}(\mathrm{exc}(A)) = A,$

and moreover (recalling that $\Sigma$ is given by (5)) that

$\notag \mathrm{exc}(\Sigma A\Sigma) = \Sigma \mathrm{exc}(A)\Sigma = \mathrm{exc}(A^T)^T. \qquad(9)$

The next result gives a formula for the inverse of $\mathrm{exc}(A)$.

Lemma 1. Let $A\in\mathbb{R}^{n\times n}$ with $A_{11}$ nonsingular. If $A$ is nonsingular and $\mathrm{exc}(A^{-1})$ exists then $\mathrm{exc}(A)$ is nonsingular and $\mathrm{exc}(A)^{-1} = \mathrm{exc}(A^{-1})$.

Proof. Consider the equation

$\notag y = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = Ax.$

By solving the first equation for $x_1$ and then eliminating $x_1$ from the second equation we obtain

$\notag \begin{bmatrix} x_1 \\ y_2 \end{bmatrix} = \mathrm{exc}(A) \begin{bmatrix} y_1 \\ x_2 \end{bmatrix}. \qquad (10)$

By the same argument applied to $x = A^{-1}y$, we have

$\notag \begin{bmatrix} y_1 \\ x_2 \end{bmatrix} = \mathrm{exc}(A^{-1}) \begin{bmatrix} x_1 \\ y_2 \end{bmatrix}.$

Hence for any $x_1$ and $y_2$ there is a unique $y_1$ and $x_2$, which implies by (10) that $\mathrm{exc}(A)$ is nonsingular and that $\mathrm{exc}(A)^{-1} = \mathrm{exc}(A^{-1})$. $~\square$

Now we will show that the exchange operator maps pseudo-orthogonal matrices to orthogonal matrices and vice versa.

Theorem 2. Let $A\in\mathbb{R}^{n\times n}$. If $A$ is pseudo-orthogonal then $\mathrm{exc}(A)$ is orthogonal. If $A$ is orthogonal and $A_{11}$ is nonsingular then $\mathrm{exc}(A)$ is pseudo-orthogonal.

Proof. If $A$ is pseudo-orthogonal then $A_{11}^TA_{11} = I + A_{21}^TA_{21}$, which implies that $A_{11}$ is nonsingular. Since $\Sigma A^T\Sigma = A^{-1}$, it follows that $A^{-1}$ also has a nonsingular $(1,1)$ block and so $\mathrm{exc}(A^{-1})$ exists. Furthermore, using Lemma 1, $\mathrm{exc}(\Sigma A^T\Sigma) = \mathrm{exc}(A^{-1}) = \mathrm{exc}(A)^{-1}$. But (9) shows that $\mathrm{exc}(\Sigma A^T\Sigma) = \mathrm{exc}(A)^T$, and we conclude that $\mathrm{exc}(A)$ is orthogonal.

Assume now that $A$ is orthogonal with $A_{11}$ nonsingular. Then $\mathrm{exc}(A^T) = \mathrm{exc}(A^{-1})$ exists and Lemma 1 shows that $\mathrm{exc}(A)$ is nonsingular and $\mathrm{exc}(A)^{-1} = \mathrm{exc}(A^{-1}) = \mathrm{exc}(A^T)$. Hence, using (9),

$I = \mathrm{exc}(A^T) \mathrm{exc}(A) = \Sigma\mathrm{exc}(A)^T\Sigma \cdot \mathrm{exc}(A),$

which shows that $\mathrm{exc}(A)$ is pseudo-orthogonal. $~\square$

This MATLAB example uses the exchange operator to convert an orthogonal matrix obtained from a Hadamard matrix into a pseudo-orthogonal matrix.

>> p = 2; n = 4;
>> A = hadamard(n)/sqrt(n), Sigma = blkdiag(eye(p),-eye(n-p))
A =
5.0000e-01   5.0000e-01   5.0000e-01   5.0000e-01
5.0000e-01  -5.0000e-01   5.0000e-01  -5.0000e-01
5.0000e-01   5.0000e-01  -5.0000e-01  -5.0000e-01
5.0000e-01  -5.0000e-01  -5.0000e-01   5.0000e-01
Sigma =
1     0     0     0
0     1     0     0
0     0    -1     0
0     0     0    -1
>> Q = exc(A,p), Q'*Sigma*Q
Q =
1     1    -1     0
1    -1     0    -1
1     0    -1    -1
0     1    -1     1
ans =
1     0     0     0
0     1     0     0
0     0    -1     0
0     0     0    -1


The code uses the function

function X = exc(A,p)
%EXC     Exchange operator.
%   EXC(A,p) is the result of applying the exchange operator to
%   the square matrix A, which is regarded as a block 2-by-2
%   matrix with leading block of dimension p.
%   p defaults to floor(n)/2.

[m,n] = size(A);
if m ~= n, error('Matrix must be square.'), end
if nargin < 2, p = floor(n/2); end

A11 = A(1:p,1:p);
A12 = A(1:p,p+1:n);
A21 = A(p+1:n,1:p);
A22 = A(p+1:n,p+1:n);

X21 = A11\A12;
X = [inv(A11) -X21;
A21/A11  A22-A21*X21];


## Hyperbolic CS Decomposition

For an orthogonal matrix expressed in block $2\times 2$ form there is a close relationship between the singular value decompositions (SVDs) of the blocks, as revealed by the CS decomposition (see What Is the CS Decomposition?). An analogous decomposition holds for a pseudo-orthogonal matrix. Let $Q\in\mathbb{R}^{n \times n}$ be pseudo-orthogonal with respect to $\Sigma$ in (5), and suppose that $Q$ is partitioned as

$\notag Q = \begin{array}[b]{@{\mskip33mu}c@{\mskip-16mu}c@{\mskip-10mu}c@{}} \scriptstyle p & \scriptstyle n-p & \\ \multicolumn{2}{c}{ \left[\begin{array}{c@{~}c@{~}} Q_{11}& Q_{12} \\ Q_{21}& Q_{22} \\ \end{array}\right]} & \mskip-12mu\ \begin{array}{c} \scriptstyle p \\ \scriptstyle n-p \end{array} \end{array}, \quad p \le \displaystyle\frac{n}{2}.$

Then there exist orthogonal matrices $U_1,V_1\in\mathbb{R}^{p \times p}$ and $U_2,V_2\in\mathbb{R}^{q \times q}$ such that

$\notag \begin{bmatrix} U_1^T & 0\\ 0 & U_2^T \end{bmatrix} \begin{bmatrix} Q_{11} & Q_{12}\\ Q_{21} & Q_{22} \end{bmatrix} \begin{bmatrix} V_1 & 0\\ 0 & V_2 \end{bmatrix} = \begin{array}[b]{@{\mskip35mu}c@{\mskip30mu}c@{\mskip-10mu}c@{}c} \scriptstyle p & \scriptstyle p & \scriptstyle n-2p & \\ \multicolumn{3}{c}{ \left[\begin{array}{c@{~}|c@{~}c} C & -S & 0 \\ \hline -S & C & 0 \\ 0 & 0 & I_{n-2p} \end{array}\right]} & \mskip-12mu \begin{array}{c} \scriptstyle p \\ \scriptstyle p \\ \scriptstyle n-2p \end{array} \end{array}, \qquad (11)$

where $C = \mathrm{diag}(c_i)$, $S = \mathrm{diag}(s_i)$, and $C^2 - S^2 = I$, with $c_i > s_i \ge 0$ for all $i$. This is the hyperbolic CS decomposition, and it can be proved by applying the CS decomposition of an orthogonal matrix to $\mathrm{exc}(Q)$.

The leading principal submatrix $\left[\begin{smallmatrix}C & -S \\ -S & C \end{smallmatrix}\right]$ in (11) generalizes the $2\times 2$ matrix (4), and in fact it can be permuted into a direct sum of such matrices.

Note that the matrix on the right in (11) is symmetric positive definite. Therefore the singular values of $Q$ are the eigenvalues of that matrix, namely

$\notag c_1 \pm s_1, \dots, c_p \pm s_p; \quad 1~\mathrm{with~multiplicity~}n - 2p.$

Since $c_i^2 - s_i^2 = 1$ for all $i$, the first $2p$ singular values occur in reciprocal pairs, hence the largest and smallest singular values satisfy $\sigma_1 = \sigma_n^{-1}\ge 1$ (with strict inequality unless $p = 0$). This gives another proof of (3).

## Numerical Stability

While an orthogonal matrix is perfectly conditioned, a pseudo-orthogonal matrix can be arbitrarily ill conditioned, as follows from (3). For example, the MATLAB function gallery('randjorth') produces a random pseudo-orthogonal matrix with a default condition number of sqrt(1/eps).

>> rng(1); A = gallery('randjorth',2,2) % p = 2, n = 4
A =
2.9984e+03  -4.2059e+02   1.5672e+03  -2.5907e+03
1.9341e+03  -2.6055e+03   3.1565e+03  -7.5210e+02
3.1441e+03  -6.2852e+02   1.8157e+03  -2.6427e+03
1.6870e+03  -2.5633e+03   3.0204e+03  -5.4157e+02
>> cond(A)
ans =
6.7109e+07


This means that algorithms that use pseudo-orthogonal matrices are potentially numerically unstable. Therefore algorithms need to be carefully constructed and rounding error analysis must be done to ensure that an appropriate form of numerical stability is obtained.

## Notes

Pseudo-orthogonal matrices form the automorphism group of the scalar product defined by $\langle x,y\rangle = x^T\Sigma y$ for $x,y\in\mathbb{R}^n$. More results for pseudo-orthogonal matrices can be obtained as special cases of results for automorphism groups of general scalar products. See, for example, Mackey, Mackey, and Tisseur (2006).

For $\Sigma \ne \pm I$ the set of pseudo-orthogonal matrices is known to have four connected components, a topological property that can be proved using the hyperbolic CS decomposition (Motlaghian, Armandnejad, and Hall, 2018).

One can define pseudo-unitary matrices in an analogous way, as $Q\in\mathbb{C}^{n\times n}$ such that $Q^*\Sigma Q = \Sigma$. These correspond to the automorphism group of the scalar product $\langle x,y\rangle = x^*\Sigma y$ for $x,y\in\mathbb{C}^n$. The results we have discussed generalize in a straightforward way to pseudo-unitary matrices.

The exchange operator is also known as the principal pivot transform and as the sweep operator in statistics. Tsatsomeros (2000) gives a survey of its properties

The hyperbolic CS decomposition was derived by Lee (1948) and, according to Lee, was present in work of Autonne (1912).

## References

This is a minimal set of references, which contain further useful references within.