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)).

A Vandermonde matrix

\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.

This article is part of the “What Is” series, available from https://nhigham.com/category/what-is and in PDF form from the GitHub repository https://github.com/higham/what-is.

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

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This article is part of the “What Is” series, available from https://nhigham.com/category/what-is and in PDF form from the GitHub repository https://github.com/higham/what-is.