minimal and characteristic polynomial of this operator

Question

The following is Problem 18 from Chap8.C of Axler's Linear Algebra Done Right.

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Edited to add a transcription of the original problem(in the image)

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P18. Suppose $a_0, a_1, ...., a_{n-1} \in \mathbb{C}$, find the minimal and characteristic polynomials of the operator on $\mathbb{C}^n$ whose matrix form (w.r.t the standard basis) is

$$ A = \begin{pmatrix} 0 & & & & & -a_0\\ 1 & 0 & & & & -a_1 \\ & 1 & \ddots & & & -a_2 \\ & & \ddots & & & \vdots \\ & & & & 0 & -a_{n-2}\\ & & & & 1 & -a_{n-1} \end{pmatrix} $$

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Denote the given matrix form of the operator as $A$, then its characteristic polynomial (I think) can be obtained by induction and Row Expansion on the first row of the determinant $det(A-\lambda I)$. If I'm not mistake, the characteristic polynomial should be $$q(z) = z^n +a_{n-1}z^{n-1} + ..... + a_1 z + a_0$$

(1)How to get the minimal polynomials? The characteristic polynomial $q(z)$ should be a multiple of the minimal polynomial $p(z)$, but there's nothing to do in factorizing $q(z)$.

In Axler's book the only method to find the minimal polynomial mentioned is to find the linear dependency of $A, A^2, ..., A^n$ for the minimum integer $n$. And also here's another related method I found in another post finding the minimal polynomial without using the characteristic polynomial, where it's recommended to find the linear dependency of $v, Av, A^2 v, ...$ for some vector $v$. However either way, it's a bit overwhelming in calculations, as the entries in the last column will explode with increasing power of $A$.

(2) I heard Axler's known for his aversion in matrices and related operations (e.g. determinant). As expected, there is very few mentioning of matrix operations. I wonder if there are other ways to find the characteristic polynomial (and also the minimal polynomial) for this operator, without referring to calculating the $det(A-\lambda I)$. (I tried to find each eigenvalue of $A$ based on the fact that $(A-\lambda I)$ is not invertible/injective/surjective if $\lambda$ is an eigenvalue of $A$, but the characteristic polynomial $p(z)$ I get above may suggest the eigenvalues are not of simple forms of $a_0, a_1, a_2, ...., a_{n-1}$)

Any hint and help would be appreciated, and thanks ahead.

Answer 1

$\newcommand{\C}{\mathbb{C}}$Let $e_{0}, e_{1}, \dots , e_{n-1}$ be the standard basis of $\C^{n}$. Note that $$A e_{0} = e_{1}, A e_{1} = e_{2}, \dots, A e_{n-2} = e_{n-1}.$$

Therefore if $p \in \C[x]$ is a polynomial, the following are equivalent

• $p(A) = 0$, and
• $p(A) e_{0} = 0$.

This is because if $p(A) e_{0} = 0$, then $0 = A 0 = A p(A) e_{0} = p(A) A e_{0} = p(A) e_{1}$, etc.

If $0 \ne p \in \C(x)$ has degree less than $n$, say $p = p_{0} + p_{1} x + \dots + p_{n-1} x^{n-1}$, then \begin{align} p(A) e_{0} &= (p_{0} + p_{1} A + \dots + p_{n-1} A^{n-1}) e_{0} \\&= p_{0} e_{0} + p_{1} A e_{0} + \dots + p_{n-1} A^{n-1} e_{0} \\&= p_{0}e_{0} + p_{1} e_{1} + \dots + p_{n-1} e_{n-1} \ne 0, \end{align} as $p \ne 0$.

Now note that \begin{align} A^{n} e_{0} &= A A^{n-1} e_{0} = A e_{n-1} \\&= -a_{0} e_{0} - a_{1} e_{1} - \dots - a_{n-1} e_{n-1} \\&= -a_{0} e_{0} - a_{1} A e_{0} - \dots - a_{n-1} A^{n-1} e_{0}, \end{align} and you're done. (Actually, you have already calculated the characteristic polynomial, so you already know the latter fact.)