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6.F Jordan normal form and the minimal polynomial

The method from the previous section is, in general, a lot of work. For some matrices, and in particular matrices of size $2\times 2$ or $3\times 3$ , there is a faster way. In these cases, if you know the characteristic polynomial and the minimal polynomial, then the JNF can be deduced directly from them. If $P^{-1}AP$ is the JNF of $A$ , then

c_{P^{-1}AP}(x)=c_{A}(x),

and

m_{P^{-1}AP}(x)=m_{A}(x).

This is one of the good things about similar matrices. So to understand the different situations, we will simply write down all possible JNF’s for $2\times 2$ and $3\times 3$ matrices.

Theorem 6.53.

Let $A\in\operatorname{M}_{{2}}({{\mathbb{C}}})$ . The only possible JNF’s for $A$ are as follows; we also list the corresponding characteristic and minimal polynomials. Here we assume that $a,b\in{\mathbb{C}}$ , but that $a\neq b$ .

$J_{a}^{(1)}\oplus J_{a}^{(1)}=\begin{bmatrix}a&0\\ 0&a\end{bmatrix}$ , $c_{A}(x)=(a-x)^{2}$ , $m_{A}(x)=(x-a)$ .
$J_{a}^{(2)}=\begin{bmatrix}a&1\\ 0&a\end{bmatrix}$ , $c_{A}(x)=(a-x)^{2}$ , $m_{A}(x)=(x-a)^{2}$ .
$J_{a}^{(1)}\oplus J_{b}^{(1)}=\begin{bmatrix}a&0\\ 0&b\end{bmatrix}$ , $c_{A}(x)=(a-x)(b-x)$ , $m_{A}(x)=(x-a)(x-b)$ .

So there are only three possibilities in the $2\times 2$ case. Let’s also say how to find a Jordan basis in each of these cases:

$J_{a}^{(1)}\oplus J_{a}^{(1)}$ : Here $J_{a}^{(1)}\oplus J_{a}^{(1)}$ is a scalar multiple of the identity. So if $P^{-1}AP=aI_{2}$ , then by multiplying that equation by $P$ on the left and $P^{-1}$ on the right, we see that $A=aPI_{2}P^{-1}=aI_{2}$ . So any basis of ${\mathbb{C}}^{2}$ is a basis of eigenvectors, and in particular, it is a Jordan basis.

$J_{a}^{(2)}$ : If we take any vector $\vec{x_{2}}$ in the generalized eigenspace $V_{a}^{(2)}$ which is not in $V_{a}^{(1)}$ , then set $\vec{x_{1}}:=(A-aI_{2})\vec{x_{2}}$ , so that $\vec{x_{1}},\vec{x_{2}}$ is a Jordan chain of length 2. So it must be a Jordan basis.

$J_{a}^{(1)}\oplus J_{b}^{(1)}$ , where $a\neq b$ : If $\vec{x}$ is an eigenvector of $a$ and $\vec{y}$ is an eigenvector of $b$ , then $\vec{x},\vec{y}$ forms a Jordan basis of ${\mathbb{C}}^{2}$ (consisting of two Jordan chains of length 1).

Example 6.54.

Find the JNF and Jordan basis for $A=\begin{bmatrix}3&5\\ 1&-1\end{bmatrix}$ .

(Solution:) The characteristic polynomial of $A$ is

c_{A}(x)=\det\begin{bmatrix}3-x&5\\ 1&-1-x\end{bmatrix}=x^{2}-2x-8=(x-4)(x+2).

This means $A$ has distinct eigenvalues $4$ and $-2$ . So by Theorem 6.53 the JNF is $J_{4}^{(1)}\oplus J_{-2}^{(1)}$ . A Jordan basis consists of an eigenvector for $4$ , such as $(5,1)$ , and an eigenvector of $-2$ , such as $(-1,1)$ . So a Jordan basis is $(5,1),(-1,1)$ .

To check that we haven’t made a mistake, we could use the change of basis matrix:

P=\begin{bmatrix}5&-1\\ 1&1\end{bmatrix},

then

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

Example 6.55.

Find the JNF and Jordan basis for $A=\begin{bmatrix}2&-1\\ 1&4\end{bmatrix}$ .

(Solution:) The characteristic polynomial of $A$ is $c_{A}(x)=(x-3)^{2}$ . So $3$ is the only eigenvalue. Since $A$ is not a scalar multiple of the identity, it doesn’t satisfy the matrix equation $A-3I_{2}=\vec{0}$ . In other words, the minimal polynomial is not $x-3$ . The only other option for the minimal polynomial is $m_{A}(x)=(x-3)^{2}$ . Therefore, the JNF of $A$ is $J_{3}^{(2)}$ .

To find a Jordan basis we just need to take any non-zero vector which is not an eigenvector (because $V_{3}^{(2)}={\mathbb{C}}^{2}$ , and $V_{3}^{(1)}$ is the eigenspace). Since $\vec{x_{2}}=(1,0)$ is not an eigenvector, it will do. Define $\vec{x_{1}}:=(A-3I_{2})\vec{x_{2}}=(-1,1)$ . Therefore a Jordan basis is $(-1,1),(1,0)$ .

To verify that this is a Jordan basis, create the change of basis matrix, and multiply:

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

Exercise 6.56:

Find the JNF of the following matrices, by first finding the characteristic and minimal polynomials, and then applying Theorem 6.53:

i.

$\begin{bmatrix}-6&9\\ -1&0\end{bmatrix}$ ,
ii.

$\begin{bmatrix}-10&4\\ -25&10\end{bmatrix}$ ,
iii.

$\begin{bmatrix}2&4\\ -1&-2\end{bmatrix}$ .

[End of Exercise]

Next we consider the $3\times 3$ matrix case. Again, the JNF can be completely described, just from looking at the characteristic polynomial together with the minimal polynomial.

Theorem 6.57.

Let $A\in\operatorname{M}_{{3}}({{\mathbb{C}}})$ . The only possible JNF’s are listed as follows; we also list the corresponding characteristic and minimal polynomials. We assume that $a, b, c$ are all different from each other.

$J_{a}^{(1)}\oplus J_{a}^{(1)}\oplus J_{a}^{(1)}=\left[\begin{smallmatrix}a&% \cdot&\cdot\\ \cdot&a&\cdot\\ \cdot&\cdot&a\end{smallmatrix}\right]$ , $c_{A}(x)=(a-x)^{3}$ , $m_{A}(x)=x-a$ .
$J_{a}^{(2)}\oplus J_{a}^{(1)}=\left[\begin{smallmatrix}a&1&\cdot\\ \cdot&a&\cdot\\ \cdot&\cdot&a\end{smallmatrix}\right]$ , $c_{A}(x)=(a-x)^{3}$ , $m_{A}(x)=(x-a)^{2}$ .
$J_{a}^{(3)}=\left[\begin{smallmatrix}a&1&\cdot\\ \cdot&a&1\\ \cdot&\cdot&a\end{smallmatrix}\right]$ , $c_{A}(x)=(a-x)^{3}$ , $m_{A}(x)=(x-a)^{3}$ .
$J_{a}^{(1)}\oplus J_{a}^{(1)}\oplus J_{b}^{(1)}=\left[\begin{smallmatrix}a&% \cdot&\cdot\\ \cdot&a&\cdot\\ \cdot&\cdot&b\end{smallmatrix}\right]$ , $c_{A}(x)=(a-x)^{2}(b-x)$ , $m_{A}(x)=(x-a)(x-b)$ .
$J_{a}^{(2)}\oplus J_{b}^{(1)}=\left[\begin{smallmatrix}a&1&\cdot\\ \cdot&a&\cdot\\ \cdot&\cdot&b\end{smallmatrix}\right]$ , $c_{A}(x)=(a-x)^{2}(b-x)$ , $m_{A}(x)=(x-a)^{2}(x-b)$ .
$J_{a}^{(1)}\oplus J_{b}^{(1)}\oplus J_{c}^{(1)}=\left[\begin{smallmatrix}a&% \cdot&\cdot\\ \cdot&b&\cdot\\ \cdot&\cdot&c\end{smallmatrix}\right]$ , $\begin{array}[]{l}c_{A}(x)=(a-x)(b-x)(c-x),\\ m_{A}(x)=(x-a)(x-b)(x-c).\end{array}$ .

Here “ $\cdot$ ” means 0.

Using analogous arguments to the $2\times 2$ case, we can find a Jordan basis in each case. Instead of writing out how this is done for each case, we will consider a few examples.

Example 6.58.

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

(Solution:) The characteristic polynomial is $c_{A}(x)=-x(1-x)^{2}$ , so the eigenvalues are 0 and 1. Since the minimal polynomial is a factor of $c_{A}$ , and shares the same roots (by Theorem 6.10), it must be either $x(x-1)$ or $x(x-1)^{2}$ . One can check, by matrix multiplication, that $A(A-I_{3})=\vec{0}$ . Therefore, the minimal polynomial is $m_{A}(x)=x(x-1)$ . So by Theorem 6.57, the JNF of $A$ is $J_{1}^{(1)}\oplus J_{1}^{(1)}\oplus J_{0}^{(1)}$ .

To find a Jordan basis, we need three Jordan chains of length 1. In other words, we need three linearly independent eigenvectors. We can choose the eigenvectors $(1,0,1),(0,1,0)$ for the eigenvalue 1, and the eigenvector $(1,-1,0)$ for the eigenvalue 0. These together form a Jordan basis.

To verify that we haven’t made a mistake, we can use the change of basis matrix:

$P=\begin{bmatrix}1&0&1\\ 0&1&-1\\ 1&0&0\end{bmatrix}$
$P^{-1}AP=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&0\end{bmatrix}=J_{1}^{(1)}\oplus J_{1}^{(1)}\oplus J_{0}^{(1)}$

Example 6.59.

Find the JNF and Jordan basis for $A=\begin{bmatrix}3&2&1\\ 0&3&1\\ -1&-4&-1\end{bmatrix}$ .

(Solution:) We have already found the Jordan normal form (Example 6.42), and Jordan basis for $A$ in (Example 6.30). But let’s do it again, this time using the minimal polynomial method. We compute the characteristic polynomial by expanding the determinant and find that $c_{A}(x)=-x^{3}+5x^{2}-8x+4$ . In general, cubic polynomials are difficult to factor by hand; in this case we are lucky that $x=1$ is a root, because $c_{A}(1)=0$ ; so $(1-x)$ is a factor. Therefore

c_{A}(x)=(2-x)^{2}(1-x).

So there are two eigenvalues, 1 and 2. Since the minimal polynomial must also have 1 and 2 as roots, and $c_{A}$ is divisible by $m_{A}$ , the only choices are

m_{A}(x)=(x-1)(x-2),

m_{A}(x)=(x-1)(x-2)^{2}.

To check whether or not the first one is the minimal polynomial, we perform the matrix multiplication:

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

Therefore, $(x-1)(x-2)$ is not the minimal polynomial, so it must be

m_{A}(x)=(x-1)(x-2)^{2}.

By Theorem 6.57, the JNF must be $J_{2}^{(2)}\oplus J_{1}^{(1)}$ .

So a Jordan basis consists of a Jordan chain of length 2, for the eigenvalue $\lambda=2$ , and a Jordan chain of length 1 (i.e. an eigenvector), for the eigenvalue $\lambda=1$ . See Example 6.42 for the details.

Exercise 6.60:

Find the JNF of the following matrices, by first finding the characteristic and minimal polynomials, and then applying Theorem 6.57:

i.

$\begin{bmatrix}-3&-10&-10\\ 0&-3&0\\ 0&5&2\end{bmatrix}$ ,
ii.

$\begin{bmatrix}5&4&3\\ -1&0&-3\\ 1&-2&1\end{bmatrix}$
iii.

$\begin{bmatrix}4&0&-1\\ 0&5&0\\ 1&0&6\end{bmatrix}$ .

[End of Exercise]

Example 6.61.

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

(Solution:) Find we find the characteristic polynomial:

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

So there is only one eigenvalue, $\lambda=-1$ . Therefore, the minimal polynomial is one of the following three polynomials:

m_{A}(x)=(x+1),

m_{A}(x)=(x+1)^{2},

m_{A}(x)=(x+1)^{3}.

Since $A$ is not equal to $-I_{3}$ , it is not the first one. To test whether the middle polynomial is the correct one:

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

This proves that the minimal polynomial is

m_{A}(x)=(x+1)^{2}.

So, according the Theorem 6.57, the JNF of $A$ is $J_{-1}^{(2)}\oplus J_{-1}^{(1)}$ .

To find a Jordan basis, we need to find two Jordan chains for the eigenvalue $-1$ , of length 1 and 2, which together are linearly independent (and hence form a basis of ${\mathbb{C}}^{3}$ ). To form a Jordan chain of length 2, we need a vector in the generalized eigenspace $V_{-1}^{(2)}\backslash V_{-1}^{(1)}$ , which is not an eigenvector. Since $(A+I_{3})^{2}=\vec{0}$ , the kernel of this matrix is all of ${\mathbb{C}}^{3}$ . Also, $\ker(A+I_{3})=\{(-2y-z,y,z)\;|\;y,z\in{\mathbb{C}}\}$ . Therefore

$V_{-1}^{(1)}=\operatorname{span}\{(-2,1,0),(-1,0,1)\}$
$V_{-1}^{(2)}={\mathbb{C}}^{3}$ .

Let’s define $\vec{x_{2}}=(1,0,0)$ , because this vector is in $V_{-1}^{(2)}$ but not in $V_{-1}^{(1)}$ . Therefore the following vectors define a Jordan chain of length 2:

$\vec{x_{1}}:=(A+I_{3})\vec{x_{2}}=(1,-1,1)$
$\vec{x_{2}}=(1,0,0)$

To complete the Jordan basis, we just need another Jordan chain of length 1, which is linearly independent to the one above. Pretty much any other eigenvector will do, such as $\vec{y_{1}}=(-1,0,1)$ . We can verify that we haven’t made a mistake, using the change of basis matrix whose columns are the basis vectors:

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

Therefore, a Jordan basis is $\vec{x_{1}},\vec{x_{2}},\vec{y_{1}}$ , where these vectors are the columns of $P$ .

Example 6.62.

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

(Solution:) First, we compute its characteristic polynomial, and find the eigenvalues.

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

As in the previous example, we have only one eigenvalue, $\lambda=-1$ . So the only possibilities for the minimal polynomial are:

$m_{A}(x)=(x+1)$ ,
$m_{A}(x)=(x+1)^{2}$ ,
$m_{A}(x)=(x+1)^{3}$ .

Since $A$ is not $-I_{3}$ , we can rule out the first one. To determine whether or not the middle one is the minimal polynomial:

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

This rules out the middle polynomial. Therefore, the minimal polynomial is

m_{A}(x)=(x+1)^{3}.

So, by Theorem 6.57, the JNF must be $J_{-1}^{(3)}$ .

To find a Jordan basis, we just need to find a single Jordan chain of length 3, since there is only one Jordan block, and it is size 3. We need a vector in $V_{-1}^{(3)}\backslash V_{-1}^{(2)}$ . According to the above calculation, we can express $V_{-1}^{(2)}=\ker(A+I_{3})^{2}$ as the span of two vectors as follows:

$V_{-1}^{(2)}=\operatorname{span}\{(1,0,0),(0,1,-1)\}$ ,
$V_{-1}^{(3)}={\mathbb{C}}^{3}$ .

So define $\vec{x_{3}}=(0,1,0)$ , which is clearly not in $V_{-1}^{(2)}$ . This vector defines the rest of the Jordan chain:

$\vec{x_{1}}:=(A+I_{3})^{2}\vec{x_{3}}=(1,-1,1)$
$\vec{x_{2}}:=(A+I_{3})\vec{x_{3}}=(1,0,0)$
$\vec{x_{3}}=(0,1,0)$

So this is our Jordan basis. To verify that we haven’t made a mistake, let’s form the change of basis matrix whose columns are the basis elements:

$P=\begin{bmatrix}1&1&0\\ -1&0&1\\ 1&0&0\end{bmatrix}$
$P^{-1}=\begin{bmatrix}0&0&1\\ 1&0&-1\\ 0&1&1\end{bmatrix}$
$P^{-1}AP=\begin{bmatrix}-1&1&0\\ 0&-1&1\\ 0&0&-1\end{bmatrix}$

Since this matrix equation holds, we have found a Jordan basis, $\vec{x_{1}},\vec{x_{2}},\vec{x_{3}}$ .