Home page for accesible maths 6 Jordan normal form 6.D Jordan chains and Jordan bases 6.F Jordan normal form and the minimal polynomial

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6.E Jordan normal form

In this section we define what it means for a matrix to be in Jordan normal form; such matrices could be thought as a being “almost” diagonal.

Definition 6.35:

A Jordan block of size $r$ , for the eigenvalue $\lambda$ is the $r\times r$ matrix:

J_{\lambda}^{(r)}:=\begin{bmatrix}\lambda&1&0&\cdots&0\\ 0&\lambda&1&\cdots&0\\ &\vdots&\ddots&&\vdots\\ 0&\cdots&0&\lambda&1\\ 0&\cdots&0&0&\lambda\end{bmatrix}.

Example 6.36.

$J_{7}^{(2)}=\begin{bmatrix}7&1\\ 0&7\end{bmatrix}$
$J_{a}^{(3)}=\begin{bmatrix}a&1&0\\ 0&a&1\\ 0&0&a\end{bmatrix}$
$J_{0}^{(4)}=\begin{bmatrix}0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\end{bmatrix}$

Definition 6.37:

The direct sum of two matrices is defined as follows. Let $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , $B\in\operatorname{M}_{{m}}({{\mathbb{C}}})$ . Then we let

A\oplus B:=\begin{bmatrix}A&\vec{0}_{n\times m}\\ \vec{0}_{m\times n}&B\end{bmatrix}\in\operatorname{M}_{{n+m}}({{\mathbb{C}}}).

It is the square matrix made by putting $A$ in the upper-left, $B$ in the lower right, and zeroes elsewhere.

Example 6.38.

J_{2}^{(2)}\oplus J_{5}^{(1)}\oplus J_{-1}^{(2)}=\begin{bmatrix}2&1&0&0&0\\ 0&2&0&0&0\\ 0&0&5&0&0\\ 0&0&0&-1&1\\ 0&0&0&0&-1\end{bmatrix}.

Definition 6.39:

A matrix is said to be in Jordan normal form (JNF), if it is the direct sum of Jordan blocks.

A JNF of a matrix $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , is any matrix which is the direct sum of Jordan blocks and is similar to $A$ . In other words, if $P^{-1}AP$ is in Jordan normal form, then it is a JNF of $A$ .

Exercise 6.40:

Find a matrix in $\operatorname{M}_{{10}}({{\mathbb{C}}})$ which is in Jordan normal form, is the direct sum of exactly 5 Jordan blocks, and has exactly 3 different eigenvalues.

[End of Exercise]

A reliable (if long) method to compute a Jordan normal form of $A$ is as follows: first find a Jordan basis, and then use the corresponding change of basis matrix to put $A$ into the required form. This is the reason Jordan bases are important: When a matrix is written in a Jordan basis, it is the direct sum of Jordan blocks (one for each Jordan chain). A consequence of this statement is:

Theorem 6.41.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , and $\mathcal{B}$ a Jordan basis. For each eigenvalue $\lambda$ , the number of Jordan blocks of size $i$ in JNF is equal to the number of Jordan chains of length $i$ in $\mathcal{B}$ .

Here is the most obvious example of this correspondence: If $A=J_{\lambda}^{(n)}$ is a single Jordan block, then the generalized eigenspaces for $\lambda$ are $V_{\lambda}^{(i)}=\operatorname{span}\{\vec{e_{1}},\cdots,\vec{e_{i}}\}$ , and the standard basis $\vec{e_{1}},\cdots,\vec{e_{n}}$ is a Jordan chain of length $n$ .

Example 6.42.

Find a Jordan normal form of $A=\begin{bmatrix}3&2&1\\ 0&3&1\\ -1&-4&-1\end{bmatrix}$ .

Solution: In Example 6.30 we found a Jordan basis for $A$ which consisted of one chain of length 1 for $\lambda=1$ and one chain of length 2 for $\lambda=2$ . So by Theorem 6.41 there exists a matrix $P$ :

P^{-1}AP=J_{1}^{(1)}\oplus J_{2}^{(2)}.

In fact, immediately after Example 6.30 we did two additional computations which also produced this matrix, which is in Jordan normal form. So we have found a Jordan normal form of $A$ .

It is often called “The JNF of $A$ ”, but you should know that it is not quite unique:

Example 6.43.

Consider the following two matrices:

A=J_{\lambda}^{(1)}\oplus J_{\mu}^{(2)}=\begin{bmatrix}\lambda&0&0\\ 0&\mu&1\\ 0&0&\mu\end{bmatrix}\neq\begin{bmatrix}\mu&1&0\\ 0&\mu&0\\ 0&0&\lambda\end{bmatrix}=J_{\mu}^{(2)}\oplus J_{\lambda}^{(1)}=B.

These two matrices are in Jordan normal form, since they are both the direct sum of two Jordan blocks. Also notice that if $P=\begin{bmatrix}0&0&1\\ 1&0&0\\ 0&1&0\end{bmatrix}$ then $P^{-1}AP=B$ . Therefore these two matrices, which are both in JNF, are similar to each other.

Exercise 6.44:

Find the JNF of $A=\begin{bmatrix}0&4\\ -1&-4\end{bmatrix}$ (see also Exercise 6.31).

[End of Exercise]

Exercise 6.45:

Find the JNF of $A=\begin{bmatrix}-1&0\\ 1&-1\end{bmatrix}$ (see also Exercise 6.32).

[End of Exercise]

Exercise 6.46:

Find the JNF of $A=\begin{bmatrix}2&0&0\\ 0&2&0\\ 1&-1&2\end{bmatrix}$ (see also Exercise 6.34).

[End of Exercise]

The next theorem states that the only way two matrices in Jordan normal form can be similar to each other is if they have the same Jordan blocks but are written in a different order, like the previous example. It is closely related to Theorem 6.41.

Theorem 6.47.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ be a matrix with $\mathcal{B}$ a Jordan basis (which exists, by Theorem 6.33). If $P$ is the change of basis matrix from $\mathcal{B}$ to the standard basis, then $P^{-1}AP$ is in Jordan normal form; in other words, it is the direct sum of Jordan blocks.

Furthermore, the Jordan blocks occurring in $P^{-1}AP$ are uniquely determined by $A$ (but the order of the blocks is not determined)

According to this theorem, one might say “The Jordan normal form of a matrix is unique up to a permutation of the Jordan blocks”.

A key difficulty, therefore, is to determine for each eigenvalue, how many Jordan blocks there are, and what size they are. Once you know that, you know the Jordan normal form. To find the size and number of Jordan blocks, the only piece of information you need is the dimensions of the generalized eigenspaces. The next theorem says how this works.

Theorem 6.48.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , and $\lambda\in{\mathbb{C}}$ an eigenvalue. Then the number of Jordan blocks of size $\geq i$ for $\lambda$ is equal to

\dim V_{\lambda}^{(i)}-\dim V_{\lambda}^{(i-1)}

for $i=1,2,\cdots$ .

In particular, the number of Jordan blocks for $\lambda$ equals $\dim V_{\lambda}^{(1)}$ . Recall $V_{\lambda}^{(0)}=\{\vec{0}\}$ .

Furthermore, the sum of the sizes of all the blocks for all eigenvalues, must equal $n$ .

Example 6.49.

Find the Jordan normal form of $A=\begin{bmatrix}-3&0&0&0\\ 0&-3&1&0\\ 0&0&-3&0\\ 1&0&-3&-3\end{bmatrix}$ .

(Solution:) The characteristic polynomial is $c_{A}(x)=\det(A-xI_{4})=(-3-x)^{4}$ , so the only eigenvalue is $\lambda=-3$ . Let’s find the dimensions of the generalized eigenspaces. We have that

A+3I_{4}=\begin{bmatrix}0&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 1&0&-3&0\end{bmatrix},

and

(A+3I_{4})^{2}=\begin{bmatrix}0&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 1&0&-3&0\end{bmatrix}\begin{bmatrix}0&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 1&0&-3&0\end{bmatrix}=\vec{0}.

Since $A+3I_{4}$ clearly has rank 2 (if we swap the first and fourth rows it is in echelon form, in which case the rank is the number of non-zero rows), by the dimension theorem, the dimension of its kernel is also 2. So

$\dim V_{-3}^{(1)}=2$
$\dim V_{-3}^{(2)}=4$
$\dim V_{-3}^{(i)}=4$ for $i\geq 3$

So by Theorem 6.48, the number of Jordan blocks is $\dim V_{-3}^{(1)}=2$ .

Applying the Theorem 6.48 to $i=2$ , we see there are $\dim V_{-3}^{(2)}-\dim V_{-3}^{(1)}=2$ Jordan blocks of size $\geq 2$ . Since the sum of the sizes of all the blocks must add up to 4, the two blocks must both have size 2. In other words, the JNF of $A$ is

J_{-3}^{(2)}\oplus J_{-3}^{(2)}=\left[\begin{smallmatrix}-3&1&0&0\\ 0&-3&0&0\\ 0&0&-3&1\\ 0&0&0&-3\end{smallmatrix}\right].

Exercise 6.50:

Assume $A\in\operatorname{M}_{{5}}({{\mathbb{C}}})$ has a single eigenvalue, $\lambda=3$ . Furthermore, assume $\ker(A-3I_{5})$ has dimension 2, and $\ker(A-3I_{5})^{2}$ has dimension 3.

•

What is the JNF of $A$ ?
•

What can you say about the dimensions of $\ker(A-3I_{5})^{3}$ and $\ker(A-3I_{5})^{4}$ ?

[End of Exercise]

Example 6.51.

Find a Jordan basis for $A=\begin{bmatrix}-3&0&0&0\\ 0&-3&1&0\\ 0&0&-3&0\\ 1&0&-3&-3\end{bmatrix}$ .

(Solution:) Since the Jordan blocks correspond to some Jordan chain, and we already found that the JNF is $J_{-3}^{(2)}\oplus J_{-3}^{(2)}$ , we know that a Jordan basis must consist of two Jordan chains of length 2, each for the eigenvalue $\lambda=-3$ . To make a Jordan chain of length 2, we need to find elements in $V_{-3}^{(2)}\backslash V_{-3}^{(1)}$ . According to our calculation above,

V_{-3}^{(2)}=\ker(A+3I_{4})^{2}=\ker\vec{0}={\mathbb{C}}^{4}.

Also,

V_{-3}^{(1)}=\ker(A+3I_{4})=\{\vec{v}\;|\;(A+3I_{4})\vec{v}=\vec{0}\}=\left\{% \left[\begin{smallmatrix}0\\ y\\ 0\\ w\end{smallmatrix}\right]\;|\;y,w\in{\mathbb{C}}\right\}=\operatorname{span}_{% {\mathbb{C}}}\left\{\left[\begin{smallmatrix}0\\ 1\\ 0\\ 0\end{smallmatrix}\right],\left[\begin{smallmatrix}0\\ 0\\ 0\\ 1\end{smallmatrix}\right]\right\}.

So any vector $\vec{x_{2}}$ in ${\mathbb{C}}^{4}$ which is not in $V_{-3}^{(1)}$ will create a Jordan chain, by taking $\vec{x_{1}}:=(A+3I_{4})\vec{x_{2}}$ . But we want two such chains, $\vec{x_{1}},\vec{x_{2}},$ and $\vec{y_{1}},\vec{y_{2}}$ which together form a basis of ${\mathbb{C}}^{4}$ . So we must ensure that the resulting 4 vectors are linearly independent; this isn’t guaranteed. For example, if we choose $\vec{x_{2}}=(1,1,0,0)$ and $\vec{y_{2}}=(1,0,0,0)$ then this would imply $\vec{x_{1}}=(0,0,0,1)$ and $\vec{y_{1}}=(0,0,0,1)$ ; so they wouldn’t form a Jordan basis.

But let’s take the two Jordan chains as follows:

$\vec{x_{1}}=\begin{bmatrix}0\\ 0\\ 0\\ 1\end{bmatrix}$
$\vec{x_{2}}=\begin{bmatrix}1\\ 0\\ 0\\ 0\end{bmatrix}$
$\vec{y_{1}}=\begin{bmatrix}0\\ 1\\ 0\\ -3\end{bmatrix}$
$\vec{y_{2}}=\begin{bmatrix}0\\ 0\\ 1\\ 0\end{bmatrix}$

To verify we haven’t made a mistake, let’s write the change of basis matrix:

P=\left[\begin{smallmatrix}0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&-3&0\end{smallmatrix}\right]

and compute

P^{-1}AP=\left[\begin{smallmatrix}-3&1&0&0\\ 0&-3&0&0\\ 0&0&-3&1\\ 0&0&0&-3\end{smallmatrix}\right].

This is, indeed, the Jordan normal form, as we found above.

Exercise 6.52:

Find the JNF, and Jordan basis for $A=\begin{bmatrix}-1&-3&-1&0\\ 0&2&1&0\\ 0&0&2&0\\ 0&3&1&-1\end{bmatrix}$ .

[End of Exercise]

Summary of the above method of finding the Jordan normal form:

•

Find the characteristic polynomial, and solve for the eigenvalues $\lambda$
•

For each eigenvalue, find the dimensions of each generalized eigenspace: $V_{\lambda}^{(1)},V_{\lambda}^{(2)},\cdots$ .
•

For each eigenvalue, use Theorem 6.48 to compute the number of Jordan blocks, and their sizes
•

The sum of all the Jordan blocks, found above, is the JNF

Summary of the above method of finding a Jordan basis, once you know the JNF:

•

For each eigenvalue, find Jordan chains whose lengths correspond to the Jordan blocks, being careful to ensure distinct Jordan chains are linearly independent
•

If done correctly, those Jordan chains together should form a Jordan basis
•

Check that $P^{-1}AP$ is the JNF, using the change of basis matrix $P$ .