Proving the inverse of a matrix as a polynomial of matrices Suppose $A$ is invertible $n\times n$ matrix. Show that $A^{-1}$ can be written as a polynomial of degree at most $n-1$, i.e.
$$A^{-1} = c_{n-1}A^{n-1} + \dots + \ c_{1}A + c_{0}I$$
Are there any tips on how may this proof be approached? 
 A: If $m(\lambda)$ is the minimal polynomial for $A$, then
$$
       m(A) = A^{k}+\alpha_{k-1}A^{k-1}+\cdots+\alpha_{1}A+\alpha_0I = 0.
$$
If $\alpha_0 = 0$ it follows that
$$
              A(A^{k-1}+\alpha_{k-1}A^{k-2}+\cdots+\alpha_1 I)=0,
$$
which means that $A$ is not invertible because
$$
               A^{k-1}+\alpha_{k-1}A^{k-2}+\cdots+\alpha_1 I \ne 0.
$$
(This is due to the minimality of $m$.) Therefore, $\alpha_0 \ne 0$, which gives
$$
         -\frac{1}{\alpha_0}A(A^{k-1}+\alpha_{k-1}A^{k-2}+\cdots+\alpha_1 I)=I \\
         \implies A^{-1} = -\frac{1}{\alpha_0}(A^{k-1}+\alpha_{k-1}A^{k-2}+\cdots+\alpha_1 I).
$$
A: Here is a different take. The set of polynomials in $A$ forms a finite dimensional vector space $K[A]$. Since $A$ is invertible, the map $K[A] \to K[A]$ given by $X \mapsto AX$ is injective and so is surjective. Therefore $I=AP(A)$.
To get a polynomial of degree less than $n$, write $P=QC+R$, where $C$ is the characteristic polynomial of $A$ with $R$ of degree less than $n$. Then $P(A)=R(A)$ because $C(A)=0$.
A: Hints


*

*The characteristic polynomial $c_A(t) = \det(t I - A)$ of any square matrix $A$ satisfies $c_A(A) = 0$.

*The constant term of the characteristic polynomial is $c_A(0) = -\det A$.

