Matrix inverse $A^{-1}$ as linear combination of the powers of $A$?

Question

Let $A \in \mathbb{R}^{n \times n}$ be any invertible matrix. Can $A^{-1}$ always be expressed as a linear combination of the powers of $A$, i.e. $$A^{-1}=\sum_{i=0}^\infty c_iA^i\,,$$ for an appropriate choice of coefficients $\{c_i\}_{i=0}^\infty$?

Answer 1

There is no need to use Cayley–Hamilton.

There is a nonzero polynomial $p$ such that $p(A)=0$ because $\mathbb{R}^{n \times n}$ has finite dimension, and so $I, A, A^2, \dots, A^{n^2}$ cannot be linearly independent.

Let $m$ be a polynomial of least degree such that $m(A)=0$.

Write $m(A)=0$ as $Ap(A)=-m(0)$.

If $m(0)=0$, then multiplying both sides by $A^{-1}$ gives $p(A)=0$, which cannot happen because $\deg p < \deg m$.

Therefore, $m(0)\ne0$ and $A^{-1}=-\frac{1}{m(0)}p(A)$.

Answer 2

Hint: Use Cayley Hamilton to find such an expression.