# What Is the Jordan Canonical Form?

How close can similarity transformations take a matrix towards diagonal form? The answer is given by the Jordan canonical form, which achieves the largest possible number of off-diagonal zero entries (Brualdi, Pei, and Zhan, 2008).

Theorem (Jordan canonical form). Any matrix $A\in\mathbb{C}^{n\times n}$ can be expressed as

\notag \begin{aligned} A &= ZJZ^{-1}, \quad J = \mathrm{diag}(J_1, J_2, \dots, J_p), \\ J_k &= J_k(\lambda_k) = \begin{bmatrix} \lambda_k & 1 & & \\ & \lambda_k & \ddots & \\ & & \ddots & 1 \\ & & & \lambda_k \end{bmatrix} \in \mathbb{C}^{m_k\times m_k}, \label{Jk} \end{aligned}

where $Z$ is nonsingular and $m_1 + m_2 + \cdots + m_p = n$. The matrix $J$ is unique up to the ordering of the blocks $J_k$.

The matrix $J$ is (up to reordering of the diagonal blocks) the Jordan canonical form of $A$ (or the Jordan form, for short).

The bidiagonal matrices $J_k$ are called Jordan blocks. Clearly, the eigenvalues of $J_k$ are $\lambda_k$ repeated $m_k$ times and $J_k$ has a single eigenvector, $e_1\in\mathbb{R}^{m_k}$. Two different Jordan blocks can have the same eigenvalues.

In total, $J$ has $p$ linearly independent eigenvectors, and the same is true of $A$.

The Jordan canonical form is an invaluable tool in matrix analysis, as it provides a concrete way to prove and understand many results. However, the Jordan form can not be reliably computed in finite precision arithmetic, so it is of little use computationally, except in special cases such as when $A$ is Hermitian or normal.

For a Jordan block $J_k = J_k(\lambda_k)\in\mathbb{C}^{m_k\times m_k}$ we have

\notag \begin{aligned} J_k - \lambda_k I &= \begin{bmatrix} 0 & 1 & & \\ & 0 & \ddots & \\ & & \ddots & 1 \\ & & & 0 \end{bmatrix}, \quad (J_k - \lambda_k I)^2 = \begin{bmatrix} 0 & 0 & 1 & & \\ & 0 & 0 & \ddots & \\ & & \ddots & \ddots& 1 \\ & & & \ddots & 0 \\ & & & & 0 \end{bmatrix},\\ \dots,\quad (J_k - \lambda_k I)^{m_k-1} &= \begin{bmatrix} 0 & 0 & \dots & 1 \\ & 0 & \dots & 0 \\ & & \ddots & \vdots \\ & & & 0 \end{bmatrix}, \quad (J_k - \lambda_k I)^{m_k} = 0. \qquad (*)\notag \end{aligned}

The superdiagonal of ones moves up to the right with each increase in the index of the power, until it disappears off the top corner of the matrix.

It is easy to see that $(A - \lambda I_n)^{j} = Z(J - \lambda I_n)^{j} Z^{-1} = Z\mathrm{diag}\bigl((J_k(\lambda_k) - \lambda I_{m_k})^{j}\bigr) Z^{-1}$, and so

$\mathrm{rank}( (A - \lambda I_n)^{j} ) = \sum_{k = 1}^p\mathrm{rank}\bigl( (J_k(\lambda_k) - \lambda I_{m_k})^{j} \bigr).$

For $\lambda = \lambda_k$, these quantities provide information about the size of the Jordan blocks associated with $\lambda_k$. To be specific, let

$\notag d_j = \mathrm{rank}( (A - \lambda_kI_n)^{j}), \quad j\ge 1, \quad \quad d_0 = n$

and

$\notag \omega_j = d_{j-1} - d_j, \quad j \ge 1.$

By considering the equations $(*)$ above, it can be shown that $\omega_j$ is the number of Jordan blocks of size at least $j$ in which $\lambda_k$ appears. Moreover, the number of Jordan blocks of size $j$ is $\omega_j - \omega_{j+1} = d_{j-1} - 2d_j + d_{j+1}$. Therefore if we know the eigenvalues and the ranks of $(A - \lambda_k I_n)^j$ for each eigenvalue $\lambda_k$ and appropriate $j$ then we can determine the Jordan structure. As an important special case, if $\mathrm{rank}(A - \lambda_k I_n) = n-1$ then we know that $\lambda_k$ appears in a single Jordan block. The sequence of $\omega_j$ is known as the Weyr characteristic, and it satisfies $\omega_1 \ge \omega_2 \ge \cdots$.

As an example of a matrix for which we can easily deduce the Jordan form consider the nilpotent matrix $B = \bigl[\begin{smallmatrix} 0_r & I_r\\0_r & 0_r \end{smallmatrix}\bigr]$, for which $B^2 = 0$ and all the eigenvalues are zero. Since $\mathrm{rank}(B) = r$, we have $d_0 = 2r$, $d_1 = r$, and $d_2 = 0$. Hence $\omega_1 = 0$ and $\omega_2 = r$, so there are $r$ $2\times 2$ Jordan blocks. (In fact, $A$ can be permuted into Jordan form by a similarity transformation.)

Here is an example with $A$ the $11\times 11$ matrix anymatrix('core/collatz',11).

$\notag A = \left[\begin{array}{ccccccccccc} 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right]$

We have $\mathrm{rank}(A) = 10$ and $\mathrm{rank}(A-2I) = 10$, so $0$ and $2$ are simple eigenvalues. All the other eigenvalues are $1$ and they have the following $d_j$ and $\omega_j$ values:

$\notag \begin{array}{cccc} j & d_j &\omega_j &\omega_j - \omega_{j+1}\\\hline 0 & 11 & & \\ 1 & 7 & 4 & 2 \\ 2 & 5 & 2 & 1 \\ 3 & 4 & 1 & 0 \\ 4 & 3 & 1 & 0 \\ 5 & 2 & 1 & 1 \\ 6 & 2 & 0 & 0 \\ \end{array}$

We conclude that the eigenvalue $1$ occurs in one block of order $5$, one block of order $2$, and two blocks of order $1$.

A matrix and its transpose have the same Jordan form. One way to see this is to note that $A = ZJZ^{-1}$ implies

$A^T = Z^{-T}J^TZ^T = Z^{-T}P \cdot PJ^TP \cdot PZ^T = (ZP)^{-T}J \,(ZP)^T,$

where $P$ is the identity matrix with the its columns reversed. A consequence is that $A$ and $A^T$ are similar.

## Real Jordan Form

A version of the Jordan form with $Z$ and $J$ real exists for $A\in\mathbb{R}^{n\times n}$. The main change is how complex eigenvalues are represented. Since the eigenvalues now occur in complex conjugate pairs $\lambda$ and $\overline{\lambda}$, and each of the pair has the same Jordan structure (which follows from the fact that a matrix and its complex conjugate have the same rank), pairs of Jordan blocks corresponding to $\lambda$ and $\overline{\lambda}$ are combined into a real block of twice the size. For example, Jordan blocks

$\notag \begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}, \; \begin{bmatrix}\,\overline{\lambda} & 1 \\ 0 & \overline{\lambda} \end{bmatrix} \in\mathbb{C}^{2 \times 2}$

become

$\notag \left[\begin{array}{@{\mkern3mu}rr|rr@{\mkern7mu}} a & b & 1 & 0 \\ -b & a & 0 & 1 \\\hline 0 & 0 & a & b \\ 0 & 0 &-b & a \end{array}\right] \in\mathbb{R}^{4 \times 4}$

in the real Jordan form, where $\lambda = a + \mathrm{i} b$. Note that the eigenvalues of $\bigl[\begin{smallmatrix} a & b \\ -b & a \end{smallmatrix}\bigr]$ are $a \pm \mathrm{i} b$.

## Notes

Proofs of the Jordan canonical form and its real variant can be found in many textbooks. See also Brualdi (1987) and Fletcher and Sorensen (1983), who give proofs that go via the Schur decomposition.