Linear Algebra - Polynomial Functions with Matices

Question

Show that if a square matrix $A$ satisfies the equation $p(A)=0$, where $p(x) = 2+a_1x+a_2x^2+...+a_kx^k$ where $a_1,a_2,...,a_k$ are constant scalars, then $A$ must be invertible. Find $A^{-1}$.

So I am aware of the fact that $A(-A-2I)=I$ but I don't see a way with the polynomial to get that form to then show that $A$ is invertible. The best I was able to do was that $p(A) = 2I+a_1A+a_2A^2+...+a_kA^k$ and $0=2I+a_1A+a_2A^2+...+a_kA^k$ since $p(A)=0$ but even from here I am not sure how to really approach the problem.

Answer 1

Let $Ax=0$. Then $0=p(A)x=2x+0+\cdots+0$ so $x=0$. So $A$ is injective, hence invertible.

Answer 2

Let $u \in ker(A)$, then $0=p(A)u=2u+a_1Au+...+a_kA^ku=2u$, hence $u=0$. This shows that $A$ is invertble.

Can you proceed ?

Answer 3

You have a polynomial with non-zero constant term satisfied by $A$. What you have to do is write reformulate $p(A)=0$ as $2I=-(a_1A+a_2A^2+\cdots +a_kA^k )$. This means $I = -\frac12(a_1+a_2A+a_3A^2+\cdots +a_kA^{k-1}) A$. This shows that $B=-\frac12(a_1+a_2A+a_3A^2+\cdots +a_kA^{k-1})$ is the inverse of $A$.