Home page for accesible maths 6 Jordan normal form 6.A The Cayley-Hamilton theorem 6.C Generalized eigenspaces

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6.B Minimal polynomials

For a square matrix $A\in\operatorname{M}_{{n}}({F})$ , we already know how to associate a certain polynomial: its characteristic polynomial. In this section we define and describe a procedure to find another polynomial, called the minimal polynomial. The characteristic polynomial can be used to find eigenvalues. It turns out that the minimal polynomial will also tell you the eigenvalues, but additionally, it tells you whether or not the matrix is diagonalizable (see Exercise 6.71).

A polynomial is called monic if the coefficient of the term of highest degree is equal to 1. Notice that the characteristic polynomial is monic if and only if the size of the matrix is even.

Definition 6.7:

Let $A\in\operatorname{M}_{{n}}({F})$ be a square matrix. A polynomial $m\in\mathcal{P}(F)$ is called a minimal polynomial if

i.

$m(A)=\vec{0}$ , and
ii.

$m$ has the smallest possible degree among polynomials obeying (i), and
iii.

$m$ is monic.

Theorem 6.8.

Let $A\in\operatorname{M}_{{n}}({F})$ be a square matrix.

i.

There is exactly one minimal polynomial of $A$ , and we denote it by $m_{A}$ .
ii.

If $p\in\mathcal{P}(F)$ is any polynomial that obeys $p(A)=\vec{0}$ , then $p$ is divisible by the minimal polynomial $m_{A}$ .

Proof.

For part (i) see Exercise 6.69.

To prove part (ii), assume $p\in\mathcal{P}(F)$ is a polynomial such that $p(A)=\vec{0}$ . By using polynomial long division, we can write $p=q\cdot m_{A}+r$ , where $q,r\in\mathcal{P}(F)$ , and $r$ has degree less than $m_{A}$ . But this equation implies that $p(A)=q(A)m_{A}(A)+r(A)$ , which implies $r(A)=\vec{0}$ . If $r$ is not the zero polynomial, then it contradicts the minimality of $m_{A}$ . Therefore, $r=0$ , and $p=q\cdot m_{A}$ . In other words, $p$ is divisible by the minimal polynomial. ∎

Example 6.9.

i.

Find the minimal polynomial of $A=\begin{bmatrix}5&0\\ 0&5\end{bmatrix}$ .

Solution: By Theorem 6.8(ii), the minimal polynomial is a factor of the characteristic polynomial $c_{A}(x)=(x-5)^{2}$ . The only monic factors are: $1$ , $x-5$ and $(x-5)^{2}$ . Of these, only the second two obey $p(A)=\vec{0}$ . Since $x-5$ is of smaller degree than $(x-5)^{2}$ , we must have $m_{A}(x)=x-5$ .
ii.

Find the minimal polynomial of $A=\begin{bmatrix}5&1\\ 0&5\end{bmatrix}$ .

Solution: Again, by Theorem 6.8(ii), the minimal polynomial is a factor of $c_{A}(x)=(x-5)^{2}$ . But now, the only monic factor which obeys $p(A)=\vec{0}$ is $(x-5)^{2}$ . Therefore, $m_{A}(x)=(x-5)^{2}$ .

The method in the previous example started with factoring the characteristic polynomial. The Fundamenthal Theorem of Algebra says that (for $F={\mathbb{C}}$ ) we can always factor a polynomial into a product of degree 1 factors. From that we can deduce all possible monic factors, by combining the various degree 1 factors in all possible ways. For example $x^{3}+x=x(x+i)(x-i)$ , there are 8 monic factors:

1,\;x,\;x+i,\;x-i,\;x(x+i),\;x(x-i),\;(x+i)(x-i),\;x(x+i)(x-i).

As the degree increases, the number of monic factors increases very quickly, so it would be easier to find the minimal polynomial if we could immediately reject many of the entries in the list. This is the purpose of the following Theorem.

Theorem 6.10.

Let $A$ be a square matrix.

i.

If $\lambda$ is an eigenvalue of $A$ , then $m_{A}(\lambda)=0$ .
ii.

The polynomials $c_{A}(x)$ and $m_{A}(x)$ have the same roots.

Proof.

To prove (i), assume that $\lambda$ is an eigenvalue. So there is an eigenvector $\vec{x}$ ; i.e. a vector such that $A\vec{x}=\lambda\vec{x}$ and $\vec{x}\neq\vec{0}$ . Now, by definition of the minimal polynomial, we know $m_{A}(A)$ is the zero matrix. Therefore $m_{A}(A)\vec{x}$ is the zero vector. For the rest of the proof of this part, see Exercise 6.11.

To prove (ii), use Theorem 6.8(ii), together with the Cayley-Hamilton theorem, to see the minimal polynomial is a factor of the characteristic polynomial. Therefore, any root of $m_{A}(x)$ is also a root of $c_{A}(x)$ . Conversely, if $\lambda$ is a root of $c_{A}(x)$ , that is the same as saying $\lambda$ is an eigenvalue, and so by part (i), $\lambda$ is a root of $m_{A}(x)$ . Therefore, they must have the same roots. ∎

Exercise 6.11:

Finish the proof of Theorem 6.10(i).

[Hint: Write $m_{A}(A)=m_{0}\cdot I_{n}+m_{1}A+\cdots+m_{r}A^{r}$ , and use that $A^{i}\vec{x}=\lambda^{i}\vec{x}$ .]

[End of Exercise]

In the example $x^{3}+x=x(x+i)(x-i)$ mentioned above, if we were to list all monic polynomial factors which also share the same roots, then that cuts the list of 8 down to just one:

x(x+i)(x-i).

Example 6.12.

Assume a matrix $A\in\operatorname{M}_{{4}}({{\mathbb{R}}})$ has $c_{A}(x)=(-1-x)^{3}(4-x)=0$ . List the possibilities for $m_{A}$ .

(Solution:) By Theorem 6.8, the minimal polynomial is a factor of the polynomial $(x+1)^{3}(x-4)$ . By Theorem 6.10, $m_{A}$ contains the factors $(x+1)$ and $(x-4)$ . Since we also know $m_{A}$ is monic, the only possibilities are: $(x+1)(x-4)$ , $(x+1)^{2}(x-4)$ , and $(x+1)^{3}(x-4)$ .

Exercise 6.13:

Assume a matrix $A$ has characteristic polynomial

c_{A}(x)=(x^{2}-2x+1)(3-x)(x^{2}+4x+4).

List the possible minimal polynomials associated to $A$ by finding all polynomials which obey all of the following: monic, divides $c_{A}$ , and has the same roots as $c_{A}$ .

[End of Exercise]

Example 6.14.

Find the minimal polynomial of the matrix $\begin{bmatrix}3&0&0\\ 1&3&0\\ 0&0&3\end{bmatrix}$ .

(Solution:) The characteristic polynomial is $c_{A}(x)=\det(A-xI_{3})=(3-x)^{3}.$ Since $m_{A}$ is a factor of $c_{A}$ , the only possibilities for $m_{A}$ are the monic polynomials $x-3$ , $(x-3)^{2}$ , and $(x-3)^{3}$ . We just need to check for which of these is $m_{A}(A)=\vec{0}$ ? We compute:

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

So $m_{A}(x)=(x-3)^{2}$ is the minimal polynomial.

Exercise 6.15:

Find the characteristic and minimal polynomials of the following matrices $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ .

i.

$\begin{bmatrix}3&2\\ 3&4\end{bmatrix}$ ,
ii.

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

$\begin{bmatrix}2&0&1\\ 0&6&2\\ 0&0&2\end{bmatrix}$ ,
iv.

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

Exercise 6.16:

If $B\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ is an invertible matrix, $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , and $p\in\mathcal{P}({\mathbb{C}})$ is some polynomial of degree $r\geq 1$ . Prove that $p(B^{-1}AB)=B^{-1}p(A)B$ . Deduce that similar matrices have the same minimal polynomial.

[End of Exercise]

Summary: A method for finding the minimal polynomial of a matrix $A$ is as follows:

•

Factorize the characteristic polynomial $c_{A}(x)$ into linear (degree 1) factors,
•

List all possible monic polynomial factors of $c_{A}$ ,
•

(Optional) Remove any polynomials which don’t share all the roots of $c_{A}$ ,
•

Remove all polynomials from the list for which $p(A)\neq\vec{0}$ ,
•

Of the remaining polynomials, $m_{A}$ is the one of smallest degree (there should be only one of smallest degree, by Theorem 6.8).