# Intuition for the differences between characteristic and minimal polynomial

I understand the definitions of the characteristic and minimal polynomials, but I don't quite see an easy way to explicitly come up with examples of matrices for which the two polynomials satisfy some properties.

For example, consider the following three exercises from Sheldon Axler's Linear Algebra Done Right, chapter 8C:

1. Give an example of an operator on $\mathbb{C}^4$ whose characteristic polynomial equals $(z-1)(z-5)^3$ and whose minimal polynomial equals $(z-1)(z-5)^2$.

2. Give an example of an operator on $\mathbb{C}^4$ whose characteristic and minimal polynomials both equal $z(z-1)^2(z-3)$.

3. Give an example of an operator on $\mathbb{C}^4$ whose characteristic polynomial equals $z(z-1)^2(z-3)$ and whose minimal polynomial equals $z(z-1)(z-3)$.

For Exercise 6, it is obvious that we just take a diagonal matrix with diagonal $0,1,1,3$. However, I don't quite see how to come up with examples of operators in questions 4 and 5 - I don't have any intuition between how the structure of an operator/matrix relates to its characteristic and minimal polynomials.

Can someone please explain the reasoning one can use to find answers to the questions?

From a quick search I understand that Jordan forms of matrices could be useful in finding explicit forms of matrices satisfying these properties, but these exercises are actually before the introduction of Jordan forms in the book so I would appreciate not using them in the answers. Thank you.

• What are the characteristic and minimal polynomial of the identity matrix $n \times n$ ? – reuns Jul 22 '17 at 22:28
• @reuns The minimal polynomial of a diagonal matrix will have no repeating zeros, and it's the minimal polynomials with repeated zeros where I'm struggling to come up with intuitive examples. – Ali Jul 22 '17 at 22:36
• What are those polynomials exactly ? – reuns Jul 22 '17 at 22:37
• @reuns $(z-n)^n$ and $(z-n)$, respectively. – Ali Jul 22 '17 at 22:43
• Not exactly. What is the obvious (non-zero) polynomial $P(x)=\sum_{k=0}^d a_k x^k$ such that $P(I_n) = \sum_{k=0}^d a_k (I_n)^k= 0$ ? – reuns Jul 22 '17 at 22:47

• You already know that the elements on the main diagonal of an upper triangular matrix are the eigenvalues of the matrix and the determinant of this matrix is the product of these eigenvalues. For example $$A = \begin{pmatrix} 2 & \star & \star \\ 0 & 3 & \star \\ 0 & 0 & 4 \end{pmatrix} \rightarrow det(A) = 2\cdot3\cdot4$$
• You already know that the powers of upper triangular matrices with only zeros on the main diagonal will "finally vanish" (this property is called "nilpotent"). For example $$A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \rightarrow A^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \rightarrow A^3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
• You may already have noted that "diagonally" blocked matrices multiply "blockwise". For example: $$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} \mbox{ and } B = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \rightarrow C = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix} = \begin{pmatrix} 2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 3 \\ \end{pmatrix}$$ $$C^2 = \begin{pmatrix} A^2 & 0 \\ 0 & B^2 \end{pmatrix} = \begin{pmatrix} 4 & 4 & 0 & 0 & 0\\ 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 9 & 0 & 0 \\ 0 & 0 & 0 & 9 & 0 \\ 0 & 0 & 0 & 0 & 9 \end{pmatrix}$$
Now, you can construct your operators easily, that is, "blockwise". Let's take exercise 4 from above (let $I_3$ denote the identity matrix of dimension $3$): $$(z-1)(z-5)^3 \mbox{ with minimal polynom } (z-1)(z-5)^2$$ So, a possible operator may look like $$A =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 5 & \star & \star \\ 0 & 0 & 5 & \star \\ 0 & 0 & 0 & 5 \end{pmatrix}$$ Let's call the bottom right matrix $B$. Since the minimal polynomial of $B$ is supposed to be $(z-5)^2$, you know that $B-5I_3$ needs to "vanish" when squared: $$B-5I_3 =\begin{pmatrix} 0 & \star & \star \\ 0 & 0 & \star \\ 0 & 0 & 0 \end{pmatrix} \mbox{ and } (B-5I_3)^2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
From the above mentioned prerequisites you see that $$B-5I_3 =\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \mbox{ or } (B-5I_3) = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \mbox{ will do.}$$ Altogether:
$$A =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 5 & 0 & 1 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \end{pmatrix} \mbox{ or } A =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 5 \end{pmatrix} \mbox{ or }A =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 5 & 1 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \end{pmatrix}$$