Jordan Canonical Form

We have seen that not all matrices can be diagonalized, or equivalently there is not always a basis for $k^n$ consisting of eigenvectors of the matrix.

So, how far off are we from diagonalization? Can any $n \times n$ matrix $A$ be brought to a related form by replacing $A$ with $S^{-1} A S$?

The answer is yes, over $\mathbb{C}$.

Definition. A Jordan block matrix J_n(\lambda) is a $n \times n$ matrix with a single value, $\lambda$, along the diagonal and with $1$’s just above the diagonal. Precisely, $J_n(\lambda)_{ij} = \begin{cases} \lambda & i = j \\ 1 & i + 1 = j \\ 0 & \text{otherwise} \end{cases}$

For example, you saw a $3 \times 3$ Jordan block on Problem 2 from Worksheet 27 and here is another example $J_4(0) = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ which has $\lambda = 0$.

We say that $A$ and $B$ are similar if there is some invertible $S$ with $S^{-1} A S = B$.

Theorem (Jordan Canonical Form). Let $A$ be a $n \times n$ matrix over $\mathbb{C}$. Then $A$ is similar to unique a block diagonal matrix where all the blocks are Jordan blocks.

The Jordan block matrix in the theorem is called the Jordan canonical form of $A$.

It is important to know that this result is not true over $\mathbb{R}$.

Note that a matrix in Jordan Canonical Form is in particular upper triangular - thus it is simple to read off the eigenvalues.

One can also read off the eigenspaces.

Lemma. Each $J_n(\lambda)$ has a one-dimensional $\lambda$- eigenspace.

Proof. (Expand to view)

If $J_n(\lambda) v = \lambda v$ then we have $\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_{n-1} \\ v_n \end{pmatrix} = \begin{pmatrix} v_2 \\ v_3 \\ \vdots \\ v_n \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 0 \end{pmatrix}$ Thus, the eigenspace is $\left\lbrace \begin{pmatrix} x \\ 0 \\ \vdots \\ 0 \\ 0 \end{pmatrix} \mid x \in k \right\rbrace$ which is one-dimensional. ■

Corollary. The sum of the dimensions of the eigenspaces is equal to the number of Jordan blocks in Jordan Canonical Form. Consequently, the $A$ is diagonalizable over $\mathbb{C}$ if and only if all Jordan blocks are $1 \times 1$ in size.

In the other direction, the dimensions of the eigenspaces is not enough to recover the Jordan Canonical Form, though it does limit the possibility strongly.

Example. Note that both $\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \end{pmatrix}$ and $\begin{pmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \end{pmatrix}$ have a two-dimensional $(-1)$-eigenspaces and in Jordan Canonical Form but are not equal.