Show that each eigenvalue of $A$ is a zero of $f$. With $f=a_0+a_1X+....+a_{n-1}X^{n-1}+X^n \in R[X]$

Question

Let $A$ be a real square matrix and let $I$ be the identity matrix with the same size. Let $n$>0 be the smallest natural number for which $a_0,a_1,...a_{n-1}$$\in$ $R$ exist such that $A^n+a_{n-1}A^{n-1}+...+a_1A+a_0I=0$ It's not evident that A satisfies this condition, but you don't need to show this.

The first question was to show that $A$ is invertable if and only if $a_0 \neq 0$, this i managed to do, but I don't know how to tackle the second question, which was...

Consider the polynomial $f=a_0+a_1X+....+a_{n-1}X^{n-1}+X^n \in R[X]$. Show that each eigenvalue of $A$ is a zero of $f$.

Answer 1

Let $c$ be the eigenvalue associated to the eigenvector $c$, $A(x)=cx$ implies that $(a_0+a_1A+...+a_{n-1}A^{n-1}+A^n)(x)=(a_0+a_1c+..+c^n)(x)=f(c)(x)=0$, $f(c)=0$ since $x\neq 0$

Answer 2

Suppose $Ax = \lambda x$ for some $x \ne 0$. Then, for each $i \in \mathbf N$, $$ A^i x = \lambda^i x $$ by induction. Hence $$ 0 = f(A)x = \sum_{i=0}^n a_iA^i x = \sum_{i=0}^n a_i\lambda^i x = f(\lambda)x $$ As $x \ne 0$, we must have $f(\lambda) = 0$.