Jordan Canonical Form

Jordan Blocks

$$J_k(\lambda) = \begin{bmatrix} \lambda & 1 & & 0 \\ & \lambda & \ddots & \\ & & \ddots & 1 \\ 0 & & & \lambda \end{bmatrix} \in M_k$$

has the single eigenvalue $\lambda$, and its eigenvectors are in $\mathrm{Span}\{e_1\}$.

$J_k(0)$ shifts entries up by one, and its powers move the ones further up,

$$\big(J_k(0)^L\big)_{ij} = \begin{cases} 1, & j - i = L \\ 0, & \text{else} \end{cases}$$

so $\mathrm{rank}\, J_k(0)^L = \max(k - L, 0)$.

Jordan Matrices

A Jordan matrix is the direct sum of Jordan blocks,

$$J = \begin{bmatrix} J_{n_1}(\lambda_1) & & 0 \\ & J_{n_2}(\lambda_2) & \\ 0 & & \ddots \end{bmatrix} = \bigoplus_i J_{n_i}(\lambda_i)$$

Permuting the blocks of the direct sum gives a similar matrix, via $PJP^T$.

Jordan's Theorem For every $A \in M_n$ there is a Jordan matrix $J \sim A$, unique up to rearrangement of blocks.

• $A$ is diagonalizable $\iff$ $n_i = 1$ for all blocks.
• If $A = SJS^{-1}$, $\mathrm{rank}(A) = \mathrm{rank}(J) = \sum \mathrm{rank}(\text{blocks of } J)$.

Dual Sequence

Any non-increasing sequence of natural numbers has a dual sequence, created by transposing its Ferrers diagram. The $\ell$th entry of the dual counts the entries that are $\ge \ell$.

E.g. $5, 3, 3, 2 \to 4, 4, 3, 1, 1$. The diagram has rows of lengths $5, 3, 3, 2$, so its columns have lengths $4, 4, 3, 1, 1$. Taking the dual twice gives back the original sequence.

Block Sizes from Ranks

First let $A \sim \bigoplus_i J_{n_i}(0)$, e.g. $J = J_5(0) \oplus J_3(0) \oplus J_3(0) \oplus J_2(0)$. Since $\mathrm{rank}\, A^\ell = \sum_i \max(n_i - \ell, 0)$,

$$s_\ell := \mathrm{rank}\, A^{\ell - 1} - \mathrm{rank}\, A^\ell = \#\{i : n_i \ge \ell\}$$

which is the $\ell$th entry of the dual sequence of the block sizes. So the block sizes are the dual of the rank differences,

$$(t_1, t_2, \ldots) = (s_1, s_2, \ldots)^* \implies A \sim \bigoplus_i J_{t_i}(0)$$

In general let $A \sim \bigoplus_{i=1}^{k} J_{n_i}(\lambda_i)$ and $\gamma \in \mathbb{C}$. Then

$$A - \gamma I = SJS^{-1} - \gamma I = S(J - \gamma I)S^{-1} = S\Big[\bigoplus_i J_{n_i}(\lambda_i - \gamma)\Big]S^{-1}$$

The blocks with $\lambda_i \neq \gamma$ are invertible and keep full rank under powers, so they are dead weight in the rank differences. For $\gamma = \lambda_j$ only the blocks of $\lambda_j$ lose rank, so

$$s_\ell = \mathrm{rank}(A - \lambda_j I)^{\ell - 1} - \mathrm{rank}(A - \lambda_j I)^\ell$$

and taking the dual gives the Jordan block sizes of $\lambda_j$.

• The first rank difference is $s_1 = \mathrm{gemu}(\lambda_j)$, the number of Jordan blocks of $\lambda_j$.
• The dual of the rank differences gives the Jordan blocks.
• The size of the largest Jordan block gives the minimal polynomial.

$A$ is Similar to $A^T$

For all $A \in M_n$, $A \sim A^T$.

Proof

1. $p_A(t) = \det(tI - A) = \det(tI - A^T) = p_{A^T}(t)$, so they have the same eigenvalues.
2. $\mathrm{rank}(A - \lambda I)^\ell = \mathrm{rank}\big[(A - \lambda I)^\ell\big]^T = \mathrm{rank}(A^T - \lambda I)^\ell$ for all $\lambda \in \sigma(A)$ and all $\ell$.
3. So they have the same Jordan matrix.