Ch.17 Nilpotent Matrices and Jordan Canonical Form

Strings

Consider the shift map defined by
$\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}0\\x\end{pmatrix}$
On the basis, this map's action gives a string
$\vec{e}_1\mapsto\vec{e}_2\mapsto\vec{0}$

Example 17.1

A linear function $\hat{n}:\mathbb{C}^4\to\mathbb{C}^4$ whose action on $\mathcal{E}_4$ is given by the string
$\vec{e}_1\mapsto\vec{e}_2\mapsto\vec{e}_3\mapsto\vec{e}_4\mapsto\vec{0}$
has $\mathscr{R}(\hat{n})\cap\mathscr{N}(\hat{n})=[\{\vec{e}_4\}]$ , $\mathscr{R}(\hat{n}^2)\cap\mathscr{N}(\hat{n}^2)=[\{\vec{e}_3,\vec{e}_4\}]$ , and $\mathscr{R}(\hat{n}^3)\cap\mathscr{N}(\hat{n}^3)=[\{\vec{e}_2,\vec{e}_3,\vec{e}_4\}]$
The matrix representation is
$\hat{N}=\text{Rep}_{\mathcal{E}_4,\mathcal{E}_4}(\hat{n})=\begin{pmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}$
None of $\hat{n},\hat{n}^2,\hat{n}^3$ are the zero map, but $\hat{n}^4$ is

Example 17.2

Transformations can also act on two strings. The transformation $t$ acting on the basis $B=\langle\vec{\beta}_1,...,\vec{\beta}_5\rangle$ by
$\begin{array}{l}\vec{\beta}_1\mapsto\vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{0}\\\vec{\beta}_4\mapsto\vec{\beta}_5\mapsto\vec{0}\end{array}$
has, for example, $\vec{\beta}_3$ in the intersection of the range space and null space.
The matrix representation is
$\text{Rep}_{B,B}(t)=\left(\begin{array}{ccc|cc}0&0&0&0&0\\1&0&0&0&0\\0&1&0&0&0\\\hline0&0&0&0&0\\0&0&0&1&0\end{array}\right)$

A nilpotent transformation is one with a power that is the zero map, and a nilpotent matrix is similar. In either case, the least such power is the index of nilpotency
Example 16.1 has an index of nilpotency of 4
Example 16.2 has an index of nilpotency of 3
The derivative map $d/dx:\mathcal{P}_2\to\mathcal{P}_2$ has index of nilpotency of 3 since taking the derivative of any quadratic three times makes it $0$

The subdiagonal in a matrix is the array formed by the entries just below the diagonal (shown in red)
$\begin{pmatrix}0&0&0\\\textcolor{#EE4266}{1}&0&0\\0&\textcolor{#EE4266}{1}&0\end{pmatrix}$

Since the zero matrix is only similar to itself, any matrix similar to a nilpotent matrix is also nilpotent.

Example 17.3

With the matrix $\hat{N}=\begin{pmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}$ and the four-vector basis
$D=\langle\begin{pmatrix}1\\0\\1\\0\end{pmatrix},\begin{pmatrix}0\\2\\1\\0\end{pmatrix},\begin{pmatrix}1\\1\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\rangle$
a change of basis operation produces this representation with respect to $D,D$
$\begin{pmatrix}1&0&1&0\\0&2&1&0\\1&1&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}\begin{pmatrix}1&0&1&0\\0&2&1&0\\1&1&1&0\\0&0&0&1\end{pmatrix}^{-1}=\begin{pmatrix}-1&0&1&0\\-3&-2&5&0\\-2&-1&3&0\\2&1&-2&0\end{pmatrix}$
This new matrix is nilpotent with index of nilpotency 4

Let $t$ be a nilpotent transformation on $V$ . A $t$ -string generated by $\vec{v}\in V$ is a sequence $\langle\vec{v},t(\vec{v}),...,t^{k-1}(\vec{v})\rangle$ such that $t^k(\vec{v})=\vec{0}$ . A $t$ -string basis is a basis that is a concatenation of $t$ -strings (which can only happen if they are disjoint)

Example 17.4

This he linear map $t:\mathbb{C}^3\to\mathbb{C}^3$
$\begin{pmatrix}x\\y\\z\end{pmatrix}\xmapsto{t}\begin{pmatrix}y\\z\\0\end{pmatrix}$
is nilpotent with index 3. A $t$ -string is
$\langle\begin{pmatrix}0\\0\\1\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}1\\0\\0\end{pmatrix}\rangle$
and is also a $t$ -string basis for the space $\mathbb{C}^3$

For Example 16.2, the two strings can be concatenated to form a basis $\langle\vec{\beta}_1,\vec{\beta}_2,\vec{\beta}_3,\vec{\beta}_4,\vec{\beta}_5\rangle$ for the domain $\mathbb{C}^5$ of $t$

If a space has a $t$ -string basis then the index of nilpotency of $t$ equals the length of the longest string in that basis.

Any nilpotent transformation $t$ is associated with a $t$ -string basis. While it is not unique, the number and length of the strings is determined by $t$
As a result, every matrix is similar to one that is all zeroes except for blocks of subdiagonal ones; it is represented by such matrix for some basis
If there are two such matrices, then they're just in a different order -> call the one with blocks in longest to shortest the canonical form

Example 17.5

The matrix
$M=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}$
has index of nilpotency of 2. This is a dimension two matrix, the space the transformation is applied to is dimension 2. The null space of one application of this transformation has dimension 1, and the null space of two applications has dimension 2. Thus, the action of this transformation on a string basis is $\vec{\beta}_1\to\vec{\beta}_2\to\vec{0}$ , making the canonical form
$N=\begin{pmatrix}0&0\\1&0\end{pmatrix}$
So, choosing $\vec{\beta}_2\in\mathscr{N}(m)$ and $\vec{\beta}_1$ such that $m(\vec{\beta}_1)=\vec{\beta}_2$ , for example
$\vec{\beta}_2=\begin{pmatrix}1\\1\end{pmatrix}\space\space\vec{\beta}_1=\begin{pmatrix}1\\0\end{pmatrix}$
shows that the canonical form is $\text{Rep}_{B,B}(m)=PMP^{-1}$ , where
$P^{-1}=\text{Rep}_{B,B}(\text{id})=\begin{pmatrix}1&1\\0&1\end{pmatrix}\space\space P=(P^{-1})^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$
We can check that indeed
$\begin{pmatrix}1&-1\\0&1\end{pmatrix}\begin{pmatrix}1&-1\\1&-1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}$

Example 17.6

Consider the matrix
$N=\begin{pmatrix}0&0&0&0&0\\1&0&0&0&0\\-1&1&1&-1&1\\0&1&0&0&0\\1&0&-1&1&-1\end{pmatrix}$
From this chart

power	$N^p$	$\mathscr{N}(N^p)$
1	$\begin{pmatrix}0&0&0&0&0\\1&0&0&0&0\\-1&1&1&-1&1\\0&1&0&0&0\\1&0&-1&1&-1\end{pmatrix}$	$\left\{\begin{pmatrix}0\\0\\u-v\\u\\v\end{pmatrix}\space\middle\|\space u,v\in\mathbb{C}\right\}$
2	$\begin{pmatrix}0&0&0&0&0\\0&0&0&0&0\\1&0&0&0&0\\1&0&0&0&0\\0&0&0&0&0\end{pmatrix}$	$\left\{\begin{pmatrix}0\\y\\z\\u\\v\end{pmatrix}\space\middle\|\space y,z,u,v\in\mathbb{C}\right\}$
3	zero matrix	$\mathbb{C}^5$

we see that after one application the null space has dimension 2, so 2 basis vectors map directly to zero. The nullity after 2 applications is 4, so two more basis vectors map to zero after 2 iterations. The nullity after 3 applications is 5, so the last remaining vector maps to zero after 3 iterations.
$\begin{array}{l}\vec{\beta}_1\mapsto\vec{\beta}_2\mapsto\vec{\beta}_3\mapsto\vec{0}\\\vec{\beta}_4\mapsto\vec{\beta}_5\mapsto\vec{0}\end{array}$
To produce such a basis, first pick two vectors from $\mathscr{N}(n)$ that form a linearly independent set
$\vec{\beta}_3=\begin{pmatrix}0\\0\\1\\1\\0\end{pmatrix}\space\space\vec{\beta}_5=\begin{pmatrix}0\\0\\0\\1\\1\end{pmatrix}$
Then, we find two vectors $\vec{\beta}_2,\vec{\beta}_4\in\mathscr{N}(n^2)$ such that $n(\vec{\beta}_2)=\vec{\beta}_3$ and $n(\vec{\beta}_4)=\vec{\beta}_5$
$\vec{\beta}_2=\begin{pmatrix}0\\1\\0\\0\\0\end{pmatrix}\space\space\vec{\beta}_4=\begin{pmatrix}0\\1\\0\\1\\0\end{pmatrix}$
Finally, find vector $\vec{\beta}_1\in\mathscr{N}(n^3)$ such that $n(\vec{\beta}_1)=\vec{\beta}_2$

Jordan Canonical Form

We have given canonical forms for nilpotent and diagonalizable matrices. We now look for canonical forms for more general matrices.
We will show that any linear transformation $t$ can be represented by a map $\text{Rep}_{B,B}(t)$ under basis $B$ such that it is the sum of a diagonal matrix and a nilpotent matrix.

First: a linear transformation on a nontrivial vector space is nilpotent iff its only eigenvalue is zero.

Proof

If the transformation $t$ on a nontrivial vector space is nilpotent, then some $t^n$ is $0$ . Thus, $t$ is a root of the polynomial $p(x)=x^n$ . Since $p(t)=0$ , we must have $p(\lambda)=0$ for every eigenvalue $\lambda$ of $t$ . Since $p(\lambda)=\lambda^n$ , the only eigenvalue of $t$ is zero.
Conversely, if the only eigenvalue of an $n$ -dimensional vector space $t$ is $0$ , then its characteristic polynomial is $(-1)^nx^n$ . By the Cayley-Hamilton Theorem, $t$ satisfies its characteristic polynomial so $t^n=0$ , thus $t$ is nilpotent.

The "nontrivial vector space" is important because a trivial vector space would have no eigenvalues since it has no associated nonzero eigenvectors.

The transformation $t-\lambda\text{ id}_V$ is nilpotent iff $t$ 's only eignevalue is $\lambda$ .

Proof

The transformation is nilpotent iff $t-\lambda\text{id}_V$ 's only eigenvalue is $0$ . But that holds iff $t(\vec{v})=\lambda\vec{v}$ , which is true iff $(t-\lambda\text{ id}_V)(\vec{v})=0\cdot\vec{v}$

If the matrices $T-\lambda I$ and $N$ are similar, then $T$ and $N+\lambda I$ are also similar via the same change of basis matrices.

Proof

We have $N=P(T-\lambda I)P^{-1}=PTP^{-1}-P(\lambda I)P^{-1}=PTP^{-1}-\lambda PIP^{-1}=PTP^{-1}-\lambda$ . Rearranging gives $N+\lambda I=PTP^{-1}$

Since nilpotent matrices, which have a single eigenvalue of $0$ are similar to the canonical form matrix consisting of all zeroes except blocks of subdiagonal ones, this can be extended to any matrix with a single eigenvalue of $\lambda$ , in which case all the entries are $\lambda$ .

Example 17.7

Find the canonical form of the matrix
$T=\begin{pmatrix}2&-1\\1&4\end{pmatrix}$

The characteristic polynomial of $T$ is $(x-3)^2$ , so $3$ is the only eigenvalue. Thus, $T-3I$ has a unique eigenvalue of $0$ , making it nilpotent. To find the canonial form, we find the string basis for $T-3I$

power $p$	$(T-3I)^p$	$\mathscr{N}((T-3I)^p)$
$1$	$\begin{pmatrix}-1&-1\\1&1\end{pmatrix}$	$\{\begin{pmatrix}-y\\y\end{pmatrix}\|\space y\in\mathbb{C}\}$
$2$	$\begin{pmatrix}0&0\\0&0\end{pmatrix}$	$\{\begin{pmatrix}x\\y\end{pmatrix}\|\space x,y\in\mathbb{C}\}$

So the nullspace of $t-3\text{id}$ has dimension 1 and $(t-3\text{id})^2$ has dimension 2, making the string basis $B$ $\vec{\beta}_1\mapsto\vec{\beta}_2\mapsto\vec{0}$ , giving the canonical form
$\text{Rep}_{B,B}(t-3\text{id})=N=\begin{pmatrix}0&0\\1&0\end{pmatrix}$
To find such a basis $B$ , we look for a $\vec{\beta}_2$ that is in the nullspace of $t-3\text{id}$ and a $\vec{\beta}_1$ that is in the nullspace of $(t-3\text{id})^2$ . Note that they must form a linearly independent set.
One such basis is
$\vec{\beta}_1=\begin{pmatrix}1\\1\end{pmatrix}\space\space\vec{\beta}_2=\begin{pmatrix}1\\-1\end{pmatrix}$
So we can conclude $T$ is similar to
$N+3I=\begin{pmatrix}3&0\\1&3\end{pmatrix}$

A Jordan block is a square matrix, or square block of entries within a matrix, that is all zero except every diagonal entry is some $\lambda\in\mathbb{C}$ and every subdiagonal entry is $1$ .
The strings in the associated basis are Jordan strings or Jordan chains.

The previous examples show that for single-eigenvalue matrices, the Jordan block matrices are the canonical representations of the similarity class. We can make this matrix form unique by arranging the basis elements so that the block of subdiagonal ones go from longest to shortest from left to righ.

Example 17.8

There are three similarity classes for $3\times 3$ matrices with unique eigenvalue $2$ :
$\begin{array}{ccc}\begin{pmatrix}2&0&0\\0&2&0\\0&0&2\end{pmatrix}&\begin{pmatrix}2&0&0\\1&2&0\\0&0&2\end{pmatrix}&\begin{pmatrix}2&0&0\\1&2&0\\0&1&2\end{pmatrix}\end{array}$
Of these, the matrix
$\begin{pmatrix}2&0&0\\0&2&0\\0&1&2\end{pmatrix}$
belongs to the middle similarity class due to the convention mentioned above.

Any transformation $t:\mathbb{C}^n\to\mathbb{C}^n$ can be represented in Jordan form, where each $J_\lambda$ is a Jordan block
$\begin{pmatrix}J_{\lambda_1}&&-\text{zeroes}-&\\&J_{\lambda_2}&&\\\\&&\ddots&\\\\&&-\text{zeroes}-&J_{\lambda_n}\end{pmatrix}$

Example 17.9

This matrix with eigenvalues $3$ , $2$ , and $1$ ,
$\begin{pmatrix}3&0&0&0&0&0\\1&3&0&0&0&0\\0&0&3&0&0&0\\0&0&0&2&0&0\\0&0&0&1&2&0\\0&0&0&0&0&-1\end{pmatrix}$
has four Jordan blocks: two associated with the eigenvalue $3$ , and one each with eigenvalues $2$ and $-1$ . The associated string representations are
$\begin{array}{r}\vec{\beta}_1\xmapsto{t-3\text{id}}\vec{\beta}_2\xmapsto{t-3\text{id}}\vec{0}\\ \vec{\beta}_3\xmapsto{t-3\text{id}}\vec{0}\\\vec{\beta}_4\xmapsto{t-2\text{id}}\vec{\beta}_5\xmapsto{t-2\text{id}}\vec{0}\\ \vec{\beta}_6\xmapsto{t+1\text{id}}\vec{0}\end{array}$