Representing the inverse of a matrix by a polynomial

Question

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.

Answer 1

The characteristic polynomial is $\chi_A(s) = \det (sI-a)$ and the Cayley Hamilton theorem tells us that $\chi_A(A) = 0$. Since $A$ is invertible, we have $c_0=\chi_A(0) \neq 0$ (the parameter $0$ means the scalar zero) and so we have $\chi_A(A) = 0 = A^n+c_{n-1}A^{n-1}+\cdots+ c_1A +c_0 I = 0$. Now multiply through by $A^{-1}$ to get $A^{n-1}+c_{n-2}A^{n-1}+\cdots+ c_1 I +c_0 A^{-1} = 0$, from which we get $A^{-1} = - {1 \over c_0} (A^{n-1}+c_{n-2}A^{n-1}+\cdots+ c_1 I)$.

Answer 2

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.