5.G Exercises‣ Chapter 5 Jordan normal form (2024)

Exercise 5.62:

$A=\begin{bmatrix}5&-3&-6\\4&-2&-6\\2&-1&-3\end{bmatrix}$
$B=\begin{bmatrix}4&0&-1\\-4&2&2\\2&0&1\end{bmatrix}$
$C=\begin{bmatrix}2&3&-1\\-1&-1&1\\1&1&-1\end{bmatrix}$
$D=\begin{bmatrix}2&2&-1\\-1&-1&1\\-1&-2&2\end{bmatrix}$
$E=\begin{bmatrix}7&1&2&2\\1&4&-1&-1\\-2&1&5&-1\\1&1&2&8\end{bmatrix}$
For $E$ you may assume that $c_{E}(x)=(x-6)^{4}$ .

You may assume that all eigenvalues of these matrices are integers (they were constructed this way for ease of computation, but this will not be true in general). For each of these matrices :

i.
Find the characteristic polynomial and minimal polynomial.
ii.
For each eigenvalue, find a basis for every generalized eigenspace $V_{\lambda}^{(i)}$ .
iii.
Find the JNF.
iv.
Find a Jordan basis.

Solution.[5.62] $A=\left[\begin{smallmatrix}5&-3&-6\\4&-2&-6\\2&-1&-3\end{smallmatrix}\right]$ . We compute its characteristic polynomial as follows:

c_{A}(x)=\det\begin{bmatrix}5-x&-3&-6\\4&-2-x&-6\\2&-1&-3-x\end{bmatrix}=\cdots=-x^{3}+x=-(x^{2}-1)x=-(x-1)(x+1)(x-0).

So there are three different eigenvalues. Therefore there is only one possibility for the minimal polynomial (which is a factor of $c_{A}(x)$ , is monic, and also has the same roots as $c_{A}(x)$ ):

m_{A}(x)=x(x-1)(x+1).

Next we determine all of the generalized eigenspaces. First we find the generalized eigenspaces of index 1 for each eigenvalue:

	$\displaystyle V_{0}^{(1)}$	$\displaystyle=\ker\begin{bmatrix}5&-3&-6\\4&-2&-6\\2&-1&-3\end{bmatrix}=\ker\begin{bmatrix}1&0&-3\\0&1&-3\\0&0&0\end{bmatrix}=\operatorname{span}_{\operatorname{\mathbb{C}}}\{\begin{%bmatrix}3\\3\\1\end{bmatrix}\},$
	$\displaystyle V_{1}^{(1)}$	$\displaystyle=\ker\begin{bmatrix}4&-3&-6\\4&-3&-6\\2&-1&-4\end{bmatrix}=\ker\begin{bmatrix}1&0&-3\\0&1&-2\\0&0&0\end{bmatrix}=\operatorname{span}_{\operatorname{\mathbb{C}}}\{\begin{%bmatrix}3\\2\\1\end{bmatrix}\},$
	$\displaystyle V_{-1}^{(1)}$	$\displaystyle=\ker\begin{bmatrix}6&-3&-6\\4&-1&-6\\2&-1&-2\end{bmatrix}=\ker\begin{bmatrix}1&0&-2\\0&1&-2\\0&0&0\end{bmatrix}=\operatorname{span}_{\operatorname{\mathbb{C}}}\{\begin{%bmatrix}2\\2\\1\end{bmatrix}\}.$

These are all 1 dimensional. One could check that $V_{\lambda}^{(2)}$ are also 1-dimensional, by computing $\ker(A-\lambda I_{3})^{2}$ . But alternatively, the existence of the Jordan normal form of a matrix shows that any collection of generalized eigenvectors for different eigenvalues must be linearly independent. Therefore the dimensions of $V_{0}^{(n)}$ , $V_{1}^{(n)}$ , and $V_{-1}^{(n)}$ must together add up to less than or equal to 3. But that means they are each of dimension 1.

So a Jordan basis is given by 3 distinct Jordan chains of length 1 (i.e. eigenvectors), and we can take:

$\vec{x}=\begin{bmatrix}3\\3\\1\end{bmatrix}$ ,
$\vec{y}=\begin{bmatrix}3\\2\\1\end{bmatrix}$ ,
$\vec{z}=\begin{bmatrix}2\\2\\1\end{bmatrix}$ .

There together form a Jordan basis. Now forming the transition matrix and performing the final matrix multiplication:

$P=\begin{bmatrix}3&3&2\\3&2&2\\1&1&1\end{bmatrix}$ ,
$P^{-1}=\begin{bmatrix}0&1&-2\\1&-1&0\\-1&0&3\end{bmatrix}$ ,
$P^{-1}AP=\begin{bmatrix}0&0&0\\0&1&0\\0&0&-1\end{bmatrix}$ .

This is the Jordan normal form of $A$ . Notice that had we chose to write our basis in a different order, then the Jordan blocks of our final matrix would have been in a different order.

$B=\left[\begin{smallmatrix}4&0&-1\\-4&2&2\\2&0&1\end{smallmatrix}\right]$ . We find the characteristic polynomial of $B$ as follows.

c_{B}(x)=\begin{bmatrix}4-x&0&-1\\-4&2-x&2\\2&0&1-x\end{bmatrix}=(2-x)(6-5x+x^{2})=(2-x)(2-x)(3-x).

So there are two possible polynomials which are simultaneously (i) a factor of $c_{B}(x)$ , (ii) monic, and (iii) have the same roots as $c_{B}(x)$ . So $m_{B}(x)$ is one of these two:

$(x-2)(x-3)$ ,
$(x-2)^{2}(x-3)$ .

First we test the one of smallest degree to see whether or not $m_{B}(B)=\vec{0}$ . We compute:

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

Therefore the minimal polynomial of $m_{B}(x)=(x-2)(x-3)$ .

To find the generalized eigenspaces for each eigenvalue, let’s start with the generalized eigenspaces of index 1:

	$\displaystyle V_{2}^{(1)}$	$\displaystyle=\ker\begin{bmatrix}2&0&-1\\-4&0&2\\2&0&-1\end{bmatrix}=\ker\begin{bmatrix}2&0&-1\\0&0&0\\0&0&0\end{bmatrix}=\operatorname{span}_{\operatorname{\mathbb{C}}}\{\begin{%bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}1\\0\\2\end{bmatrix}\},$
	$\displaystyle V_{3}^{(1)}$	$\displaystyle=\ker\begin{bmatrix}1&0&-1\\-4&-1&2\\2&0&-2\end{bmatrix}=\ker\begin{bmatrix}1&0&-1\\0&1&2\\0&0&0\end{bmatrix}=\operatorname{span}_{\operatorname{\mathbb{C}}}\{\begin{%bmatrix}1\\-2\\1\end{bmatrix}\}.$

Since these dimensions add up to 3, there is a basis of eigenvectors, and therefore the generalized eigenspaces of higher index are no bigger than the usual eigenspaces. In other words, $V_{2}^{(n)}=V_{2}^{(1)}$ and $V_{3}^{(n)}=V_{3}^{(1)}$ for all $n\geq 1$ .

So the Jordan normal form of $A$ is a diagonal matrix, $J_{2}^{(1)}\oplus J_{2}^{(1)}\oplus J_{3}^{(1)}$ , and a Jordan basis is made up of three Jordan chains of length 1, which we could take as follows:

$\vec{x}=\begin{bmatrix}0\\1\\0\end{bmatrix}$ ,
$\vec{y}=\begin{bmatrix}1\\0\\2\end{bmatrix}$ ,
$\vec{z}=\begin{bmatrix}1\\-2\\1\end{bmatrix}$ .

So we can now form the transition matrix, and perform the final verification:

$P=\begin{bmatrix}0&1&1\\1&0&-2\\0&2&1\end{bmatrix}$ ,
$P^{-1}=\begin{bmatrix}4&1&-2\\-1&0&1\\2&0&-1\end{bmatrix}$ ,
$P^{-1}AP=\begin{bmatrix}2&0&0\\0&2&0\\0&0&3\end{bmatrix}=J_{2}^{(1)}\oplus J_{2}^{(1)}\oplus J_{3}^{(1)}$ .
See Also
Jordan's Math Work—Free Games and Resources — Mashup Math 5.D Jordan chains and Jordan bases‣ Chapter 5 Jordan normal form

$C=\left[\begin{smallmatrix}2&3&-1\\-1&-1&1\\1&1&-1\end{smallmatrix}\right]$ . JNF is $\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$ .

$D=\left[\begin{smallmatrix}2&2&-1\\-1&-1&1\\-1&-2&2\end{smallmatrix}\right]$ . JNF is $\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}$ .

$E=\left[\begin{smallmatrix}7&1&2&2\\1&4&-1&-1\\-2&1&5&-1\\1&1&2&8\end{smallmatrix}\right]$ . JNF is $\begin{bmatrix}6&1&0&0\\0&6&1&0\\0&0&6&0\\0&0&0&6\end{bmatrix}$ .

Exercise 5.63:

Assume a matrix $A$ has the following characteristic polynomial. Find all possible JNF’s for $A$ (up to reordering of the Jordan blocks).

i.
$(x-1)^{2}(x+2)^{2}$ ,
ii.
$(x-1)^{3}(x+2)$ .

Solution.[5.63] (i) Since the characteristic polynomial does not change under a similarity transformation, the Jordan norm form of $A$ must have diagonal entries consisting of two 1’s and two $-2$ ’s. The only choice is whether, for each eigenvalue, there is a single Jordan block of size 2, or two blocks of size 1. So there are four possibilities:

$J_{1}^{(1)}\oplus J_{1}^{(1)}\oplus J_{-2}^{(1)}\oplus J_{-2}^{(1)}$ ,
$J_{1}^{(2)}\oplus J_{-2}^{(1)}\oplus J_{-2}^{(1)}$ ,
$J_{1}^{(1)}\oplus J_{1}^{(1)}\oplus J_{-2}^{(2)}$ ,
$J_{1}^{(2)}\oplus J_{-2}^{(2)}$ .

These may also be written explicitly as matrices as follows:

$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&-2&0\\0&0&0&-2\end{bmatrix}$ ,
$\begin{bmatrix}1&1&0&0\\0&1&0&0\\0&0&-2&0\\0&0&0&-2\end{bmatrix}$ ,
$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&-2&1\\0&0&0&-2\end{bmatrix}$ ,
$\begin{bmatrix}1&1&0&0\\0&1&0&0\\0&0&-2&1\\0&0&0&-2\end{bmatrix}$ .

It is equally correct to have written the Jordan blocks in a different order.

[5.63] (ii) Since the characteristic polynomial doesn’t change under similarity, the JNF must have three 1’s on the diagonal, and one $-2$ . So the only possible choices for Jordan normal forms as:

$J_{1}^{(1)}\oplus J_{1}^{(1)}\oplus J_{1}^{(1)}\oplus J_{-2}^{(1)}$ ,
$J_{1}^{(2)}\oplus J_{1}^{(1)}\oplus J_{-2}^{(1)}$ ,
$J_{1}^{(3)}\oplus J_{-2}^{(1)}$ .

So, up to rearranging the Jordan blocks, there are only three possible JNF’s. If we were to write these as matrices, they would look as follows:

$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-2\end{bmatrix}$ ,
$\begin{bmatrix}1&1&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-2\end{bmatrix}$ ,
$\begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&-2\end{bmatrix}$ .

Exercise 5.64:

Let $D:\mathcal{P}_{3}(\operatorname{\mathbb{C}})\to\mathcal{P}_{3}(\operatorname{%\mathbb{C}})$ be the differentiation map.

i.
Choose a basis $\operatorname{\mathcal{B}}$ of $\mathcal{P}_{3}(\operatorname{\mathbb{C}})$ , and write down the matrix ${}_{\operatorname{\mathcal{B}}}[D]_{\operatorname{\mathcal{B}}}$ .
ii.
What is the minimal polynomial of $D$ ?
iii.
Find the JNF of $D$ .

Solution.[5.64] First we will write down the matrix of $D$ in the standard basis $\{1,x,x^{2},x^{3}\}$ of $\mathcal{P}_{3}(\operatorname{\mathbb{C}})$ . To do this, we compute $D(\vec{b_{i}})$ for each basis vector:

$D(1)=0$ ,
$D(x)=1$ ,
$D(x^{2})=2x$ ,
$D(x^{3})=3x^{2}$ .

Then we find the coordinate matrices of these vectors, and the matrix of $D$ is constructed by taking those coordinate matrices as its columns:

A:=[D]=\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{bmatrix}.

The characteristic polynomial of this matrix is $c_{A}(x)=x^{4}$ , and so the choices for the minimal polynomial are:

$x$ ,
$x^{2}$ ,
$x^{3}$ ,
$x^{4}$ .

To see which it is, we need to evaluate each of these polynomials are the matrix $A$ . We evalute as follows:

	$\displaystyle A$	$\displaystyle=\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{bmatrix}$
	$\displaystyle A^{2}$	$\displaystyle=\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{bmatrix}\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{bmatrix}=\begin{bmatrix}0&0&2&0\\0&0&0&6\\0&0&0&0\\0&0&0&0\end{bmatrix},$
	$\displaystyle A^{3}$	$\displaystyle=\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{bmatrix}\begin{bmatrix}0&0&2&0\\0&0&0&6\\0&0&0&0\\0&0&0&0\end{bmatrix}=\begin{bmatrix}0&0&0&6\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix},$
	$\displaystyle A^{4}$	$\displaystyle=\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}.$

So the minimal polynomial of $A$ is $m_{A}(x)=x^{4}$ . Unlike $3\times 3$ matrices, for $4\times 4$ matrices, to find the Jordan normal form it is not enough to consider the characteristic and minimal polynomials. For matrices of size 4 and higher, it is possible to have matrices with different JNF’s (even up to reordering of blocks), but with the same characteristic and minimal polynomials. So to find the JNF we need to consider the generalized eigenspaces.

Since the only eigenvalue is $\lambda=0$ , we could consider the generalized eigenspaces for that eigenvalue. In fact, all we will need is the dimension of the first one:

V_{0}^{(1)}=\ker\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{bmatrix}=\operatorname{span}_{\operatorname{\mathbb{C}}}\{\begin{%bmatrix}1\\0\\0\\0\end{bmatrix}\}.

So $\dim V_{0}^{(1)}=1$ . By Theorem 5.47 the number of Jordan blocks is 1, so the JNF must be $J_{0}^{(4)}$ .

Exercise 5.65:

Assume $A,B\in\operatorname{M}_{{n}}({\operatorname{\mathbb{C}}})$ are similar matrices, and let $\lambda\in\operatorname{\mathbb{C}}$ be an eigenvalue (by Exercise 3.55, $A$ and $B$ have the same eigenvalues). Prove that $\dim V_{\lambda}^{(i)}$ is the same for $A$ as it is for $B$ .

Solution.[5.65] Assume that $B=P^{-1}AP$ for some invertible matrix $P$ . Firstly we need to understand how similarity changes the kernel of a matrix. If $C$ is a square matrix of size $n$ , then the following statements are equivalent to each other:

$\vec{x}\in\ker P^{-1}CP$ ,
$P^{-1}CP\vec{x}=\vec{0}$ ,
$C(P\vec{x})=\vec{0}$
$P\vec{x}\in\ker C$
$\vec{x}\in P^{-1}\ker C$ .

Since $P$ is a bijective linear map, this proves that the dimension of $\ker P^{-1}CP$ is the same as the dimension of $\ker C$ . In particular,

\dim\ker(B-\lambda I_{n})^{i}=\dim\ker(P^{-1}AP-\lambda I_{n})^{i}=\dim\ker P^%{-1}(A-\lambda I_{n})^{i}P=\dim\ker(A-\lambda I_{n})^{i}.

Therefore, the generalized eigenspaces of index $i$ have equal dimensions for similar matrices $A$ and $B$ .

Exercise 5.66:

Consider the function $T:\mathcal{P}_{2}(\operatorname{\mathbb{C}})\to\mathcal{P}_{2}(\operatorname{%\mathbb{C}})$ defined by

T(f)(x)=f(x+1).

For example, $T(x^{2}+7)=(x+1)^{2}+7=x^{2}+2x+8$ .

•
Choose a basis $\operatorname{\mathcal{B}}$ of $\mathcal{P}_{2}(\operatorname{\mathbb{C}})$ , and find the matrix ${}_{\operatorname{\mathcal{B}}}[T]_{\operatorname{\mathcal{B}}}$ .
•
Find the JNF of that matrix
•
Find a Jordan basis for $T$ ; this should form a basis of $\mathcal{P}_{2}(\operatorname{\mathbb{C}})$ .

Solution.[5.66] Let $\operatorname{\mathcal{B}}=\{1,x,x^{2}\}$ , which is the easiest basis of $\mathcal{P}_{2}(\operatorname{\mathbb{C}})$ . To find the matrix of $T$ , we need to compute where $T$ transforms the elements of the basis:

$T(1)=1$ ,
$T(x)=x+1$ ,
$T(x^{2})=(x+1)^{2}=x^{2}+2x+1$ .

So the columns of the matrix ${{}_{\operatorname{\mathcal{B}}}[}T]_{\operatorname{\mathcal{B}}}$ are the coordinates of the vectors $T(\vec{b_{i}})$ with respect to the basis $\operatorname{\mathcal{B}}$ . Therefore

A:={{}_{\operatorname{\mathcal{B}}}[}T]_{\operatorname{\mathcal{B}}}=\begin{%bmatrix}1&1&1\\0&1&2\\0&0&1\end{bmatrix}.

The characteristic polynomial of this matrix is $c_{A}(x)=(1-x)^{3}$ , since it is upper-triangular. This shows that the only eigenvalue is 1. Since it is a $3\times 3$ matrix, to find the JNF, it is enough to compute the minimal polynomial of $A$ (recall that this technique doesn’t work for matrices of size 4 or larger). The possibilities for the minimal polynomial are:

$(x-1)$ ,
$(x-1)^{2}$ ,
$(x-1)^{3}$ .

To check which one is correct, we evaluate the matrix $A$ at each of these possibilities, and the one of smallest degree which evaluates $A$ to zero is the minimal polynomial. We compute:

	$\displaystyle A-I_{3}$	$\displaystyle=\begin{bmatrix}0&1&1\\0&0&2\\0&0&0\end{bmatrix},$
	$\displaystyle(A-I_{3})^{2}$	$\displaystyle=\begin{bmatrix}0&1&1\\0&0&2\\0&0&0\end{bmatrix}\begin{bmatrix}0&1&1\\0&0&2\\0&0&0\end{bmatrix}=\begin{bmatrix}0&0&2\\0&0&0\\0&0&0\end{bmatrix},$
	$\displaystyle(A-I_{3})^{3}$	$\displaystyle=\begin{bmatrix}0&1&1\\0&0&2\\0&0&0\end{bmatrix}\begin{bmatrix}0&0&2\\0&0&0\\0&0&0\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}.$

Therefore the minimal polynomial is $(x-1)^{3}$ , and hence the JNF is $J_{1}^{(3)}$ .

To find a Jordan basis, we know we need to find a single Jordan chain of length 3. To do this, we need to find an element $\vec{x_{3}}\in V_{1}^{(3)}\backslash V_{1}^{(2)}$ . According to our matrix multiplications above, we can deduce that

$V_{1}^{(3)}=\operatorname{\mathbb{C}}^{3}$ ,
$V_{1}^{(2)}=\operatorname{span}_{\operatorname{\mathbb{C}}}\{\begin{bmatrix}1%\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix}\}$ .

So we define

$\vec{x_{3}}:=\begin{bmatrix}0\\0\\1\end{bmatrix}$ ,
$\vec{x_{2}}:=(A-I_{3})\vec{x_{3}}=\begin{bmatrix}1\\2\\0\end{bmatrix}$ ,
$\vec{x_{1}}:=(A-I_{3})\vec{x_{2}}=\begin{bmatrix}2\\0\\0\end{bmatrix}$ .

Then the sequence $\vec{x_{1}},\vec{x_{2}},\vec{x_{3}}$ forms a Jordan chain of length 3. In particular, it forms a Jordan basis.

As a final check, we could construct the transition matrix, and make the following verification:

$P=\begin{bmatrix}2&1&0\\0&2&0\\0&0&1\end{bmatrix}$ ,
$P^{-1}=\begin{bmatrix}1/2&-1/4&0\\0&1/2&0\\0&0&1\end{bmatrix}$ ,
$P^{-1}AP=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}=J_{1}^{(3)}$ .

Exercise 5.67:

A student is asked to prove Theorem 5.9(i). He writes the following:

[Start of student’s work]

Take two monic polynomials $p_{1},p_{2}\in\mathcal{P}(F)$ of minimal degree such that $p_{1}(A)=p_{2}(A)=\vec{0}$ , and assume $p_{1}\neq p_{2}$ . Since $p_{1}$ and $p_{2}$ are both of the same degree $r$ , and are monic, the polynomial $p_{1}-p_{2}$ is monic of degree $r-1$ . But notice that

(p_{1}-p_{2})(A)=p_{1}(A)-p_{2}(A)=\vec{0}-\vec{0}=\vec{0}.

So $p_{1}-p_{2}$ contradicts the minimality of $r$ . Therefore, our assumption $p_{1}\neq p_{2}$ must have been false. This proves $p_{1}=p_{2}$ , and in other words, the polynomial is unique.

[End of student’s work]

This student has made a logical mistake. What is it, and how could it be fixed?

Exercise 5.68:

A student is asked to prove the Cayley-Hamilton theorem (which is not an easy thing to do, and is omitted from this module). He writes the following:

[Start of student’s work]

Substitute $\lambda$ with $A$ in the characteristic polynomial. Then

c_{A}(A)=\det(A-A\cdot I_{n})=\det(\vec{0})=0.

Therefore, $c_{A}(A)=0$ for any square matrix $A$ , as required.

[End of student’s work]

Identify the student’s mistake. You do not have to give a correct proof.

Exercise 5.69:

Prove that a matrix is diagonalizable if and only if

m_{A}(x)=(x-a_{1})(x-a_{2})\cdots(x-a_{r}),

where $a_{i}\neq a_{j}$ for $i\neq j$ .

Solution.[5.69] There are two statements that need to be proved, because it says “if and only if”. For either direction, we can define distinct eigenvalues $a_{i}$ . By putting $A$ into Jordan normal form, we see that the characteristic polynomial of $A$ must be

c_{A}(x)=(a_{1}-x)^{m_{1}}(a_{2}-x)^{m_{2}}\cdots(a_{r}-x)^{m_{r}}.

Here $m_{i}$ is called the multiplicity of the eigenvalue $a_{i}$ .

Assume $A$ is diagonalizable, so there is an invertible matrix $P$ such that $P^{-1}AP=D$ , where $D=\operatorname{diag}(a_{1},\cdots,a_{r})$ ; it may be the case that each $a_{i}$ occurs multiple times in the diagonal matrix $D$ . The minimal polynomials of $A$ and $D$ are the same, since they are similar matrices. Now I claim that $p(x)=(x-a_{1})(x-a_{2})\cdots(x-a_{r})$ is the minimal polynomial. Since the minimal polynomial must share the same roots as the characteristic polynomial, this is the smallest possibility that might be the minimal polynomial. We just need to check that it sends the matrix $D$ to the zero matrix. Indeed, $p(D)=(D-a_{1}I_{n})(D-a_{2}I_{n})\cdots(D-a_{r}I_{n})$ is the product of diagonal matrices, and the resulting matrix is the zero matrix. This proves that $p(x)$ is the minimal polynomial, as required.

Next, assume that $A$ is not diagonalizable. Then we can put $A$ is Jordan normal form, $J$ , and since it is not diagonalizable, there must exist at least one Jordan block of size $\geq 2$ . To prove that $p(x)$ is not the minimal polynomial, it is enough to check that $p(J)$ is not the zero matrix; in other words there is a vector $\vec{v}$ such that $p(J)\vec{v}\neq\vec{0}$ . Since we have assumed $A$ is not diagonalizable, we can choose a $\vec{v}\in V_{\lambda}^{(2)}\backslash V_{\lambda}^{(1)}$ , for some eigenvalue $\lambda$ . Since we are free to reorder the eigenvalues, there is no loss of generality in assuming $\lambda=a_{r}$ , the last eigenvalue. Then

\vec{w}:=(J-a_{r}I_{n})\vec{v}\in V_{a_{r}}^{(1)}\backslash\{\vec{0}\}.

Since we have assumed $a_{1},\cdots,a_{r}$ are the distinct eigenvalues, for any $i\neq r$ , we use that $\vec{w}$ is an eigenvector to see that:

(J-a_{i}I_{n})\vec{w}=(a_{r}-a_{i})\vec{w}\neq\vec{0}.

In particular, this proves $(J-a_{1}I_{n})\cdots(J-a_{r-1}I_{n})\vec{w}=p(J)\vec{v}\neq\vec{0}$ . Therefore $p$ is not the minimal polynomial of $J$ , and hence it is also not the minimal polynomial of $A$ . This finishes the proof.

Exercise 5.70 (Bonus):

Let $A\in\operatorname{M}_{{n}}({\operatorname{\mathbb{C}}})$ , and consider the set of all matrices which are similar to $A$ . This is called the orbit of $A$ under the conjugation action. [ Aside: The words ‘‘orbit’’, ‘‘conjugation’’, and ‘‘action’’ will all be defined in MATH321.]

How many different orbits are there, among nilpotent $5\times 5$ complex matrices? Recall the definition of “nilpotent” from Exercise 3.62.

Learning objectives for Chapter 5:

Pass Level: You should be able to…

$\square$
Verify the Cayley-Hamilton theorem for specific matrices (e.g. Exercise 5.4).
$\square$
For a given (mostly factored) polynomial, produce a list of all possible monic factors which share the same roots (e.g. Exercise 5.14).
$\square$
Given a matrix and its factored characteristic polynomial, find its minimal polynomial (e.g. Exercise 5.16).
$\square$
Be able to express any generalized eigenspace of a matrix as the kernel of another matrix (e.g. Definition 5.18).
$\square$
Given the generalized eigenspaces of a matrix, deduce its JNF using Theorem 5.47 (e.g. Exercise 5.49).
$\square$
Correctly answer, with full justification, at least 50% of the true / false questions relevant to this Chapter.

First class level: You should be able to…

$\square$
Explain, in your own words, the main ideas used in the proofs of Theorem 5.11, Theorem 5.22, and Theorem 5.26
$\square$
Identify, in your own words, the most important concepts and results of this Chapter.
$\square$
Write a complete solution, without referring to any notes, to at least 80% of the exercises in this Chapter, and in particular the proof questions.
$\square$
Correctly answer, with full justification, all of the true / false questions relevant to this Chapter.