I saw a question that asks in the first part to prove that if $A, B \in M_n(\mathbb C)$ such that $AB=BA$ and for any 2 polynomials $f$ and $g$, we have $f(A)g(B)=g(B)f(A)$. Thats not difficult.
In the second part the question wa to prove that $A^{−1}=p(A)$ for some polynomial with complex coefficients. So how can I use part 1 to prove part 2? Or are they not related to each other ?
I think in general they use the fact that the characteristic polynomial $c_A(t) = \det(t I – A)$ of any square matrix $A$ satisfies $c_A(A) = 0$.
But I also didn't know how to prove from here.
Best Answer
Here's a quick and dirty way, not using determinants.
Because $M_n$ itself has dimension $n^2$, we can find constants $c_1,\dots,c_{n^2+1}$, not all zero, such that $c_1 A + \dots + c_{n^2+1} A^{n^2+1}=0$.
Suppose now $A$ is invertible. Let $j_0$ the smallest index satisfying $c_{j_0}\ne 0$. Then $j_0<n^2+1$ because $A$ is invertible.
Multiply the linear combination by $\frac{1}{c_{j_0}}A^{-(j_0+1)}$ to obtain
$$ \boxed{ A^{-1}} + \frac{c_{j_0+1}}{c_{j_0}} A + \dots + \frac{c_{n^2+1}}{c_{j_0}}A^{n^2+1-j_0} =0,$$
completing the proof.