Prove that $A^{-1}=A^n$ if $A^n+A^{n-1}+...+A+E=O,n\in \mathbb{N}$ and $A$ is regular matrix

Question

I took the example of $2\times 2$ matrix $$ \begin{bmatrix} x & y \\ u & v \\ \end{bmatrix}$$ which gives matrix equation: $$A^2+A+E=O$$ After addition, I get system of equations which I don't know how to solve: \begin{matrix} x^2+uy+x+1=0 \\ xy+yv+y=0\\ ux+vy+u=0\\ uy+v^2+v+1=0\end{matrix} Is there some other approach to prove that $A^{-1}=A^n$ rather than this?

Answer 1

Note: I guess that your $E$ is the matrix identity that I note $I$.

Multiply $A^n +A^{n-1}+ \dots + I= 0$ by $I-A$.

You get $I-A^{n+1}=0$. Therefore the minimal polynomial of your matrix divides the polynomial $P(X)=X^{n+1}-1=0$. As $0$ is not a root of $P$, $A$ is invertible. Now you can speak of $A^{-1}$! Multiplying $I-A^{n+1}=0$ by $A^{-1}$ you get the expected result.

PS: a simple way to see that $A$ is invertible is to notice that if $A.x=0$ then $(I-A^{n+1}).x=x-A^{n+1}.x=x=0$. Hence $\ker A=0$ and $A$ is invertible.

Answer 2

you have $$A^2 + A + I = 0 \to -A^2 - A = I \to A(-A-I) = I $$ therefore $$A^{-1} = -A - I. $$