Show matrix A is invertible if and only if $a_0 \neq 0$ with $A^n+a_{n-1}A^{n-1}+....+a_1A+a_0I=O$.

Question

Let $A$ be a real square matrix en let $I$ be the identity matrix with the same size. Let $n>0$ be the smallest natural number for which there exist $a_0,a_1,...a_{n-1} \in {R}$ such that $A^n+a_{n-1}A^{n-1}+....+a_1A+a_0I=O$.

Show that A is invertible if and only if $a_0\neq0 $.

So i need to proof

1. $A$ invertible $\rightarrow$$a_0 \neq 0$
2. $a_0 \neq0 \rightarrow$ $A$ invertible

I don't really know where to start with 1, and with 2 I tried to prove it with a contradiction but I don't think it's right either.

Answer 1

Hint:

$$A^n+a_{n-1}A^{n-1}+\ldots+a_1A+a_0I=0\implies a_0I=-\left(A^n+a_{n-1}A^{n-1}+\ldots a_1A\right)\implies$$

$$a_0I=A\;\overbrace{\left(-A^{n-1}-a_{n-1}A^{n-2}-\ldots-a_1I\right)}^{=B}$$

Now, observe that $\;B\;$ is an excellent candidate to almost be $\;A^{-1}\;$ , but this will happen iff ...

Answer 2

Let $A$ be an invertible matrix. If $a_0 = 0$ then $$A^n + a_{n-1}A^{n-1} + \ldots + a_1A = O$$ but then $$A^{n-1} + a_{n-2}A^{n-1} + \ldots + a_1I = A^{-1}(A^n + a_{n-1}A^{n-1} + \ldots + a_1A) = O$$

but that contradicts the assumption that $n$ was minimal. Therefore $a_0$ must be nonzero.

Conversely, if $a_0\neq 0$ then

$$\dfrac{-1}{a_0}(A^{n-1} + a_{n-2}A^{n-1} + \ldots + a_1I)A = I$$

so $$\dfrac{-1}{a_0}(A^{n-1} + a_{n-2}A^{n-1} + \ldots + a_1I) = A^{-1}$$

Answer 3

Forward part $A$ is invertible implies $a_0\neq 0$. Lets look at the contrapositive. If $a_0=0$ then $$A(A^{n-1}+a_{n-1}A^{n-2}+...+a_1I)=0$$ Now if A is invertible then $A^{n-1}+a_{n-1}A^{n-2}+...+a_1I=0$ which contradicts the fact that n is the smallest number which annihilates A.
Backward implication if $a\neq 0$ then we can find the inverse of $A$ as $\frac{-1}{a_0}(A^{n-1}+a_{n-1}A^{n-2}+...+a_1I)$