Solution verification: characteristic polynomial of $\mathcal{A}^{-1} $

Question

Let $\mathcal{A}$ be an invertible operator on a finite-dimensional vector space $V$ of dimension $n$. Let $g(t)= a_0 + a_1 t + a_2 t^2+ ... + a_n t^n $ be a characteristic polynomial of $\mathcal{A}^{-1}$. Then, from Cayley-Hamilton theorem, we obtain: $g(\mathcal{A}^{-1})= a_0 \mathcal{I} + a_1 \mathcal{A}^{-1} + a_2 (\mathcal{A}^{-1})^2 +... + a_n (\mathcal{A}^{-1})^n= \mathcal{O} $, where $\mathcal{I} $ and $\mathcal{O}$ are an identity and zero operator, respectively. Hence, $ (\mathcal{A}^{-1})^n = -\frac{ a_0}{a_n} \mathcal{I} - \frac{ a_1}{a_n} \mathcal{A}^{-1} - ... - \frac{ a_{n-1}}{a_n} (\mathcal{A}^{-1})^{n-1} $. If we compose both sides of last equality with operator $\mathcal{A}^{n-1}$, then we get:

$ \mathcal{A}^{-1} =-\frac{ a_0}{a_n} \mathcal{A}^{n-1} -\frac{ a_1}{a_n} \mathcal{A}^{n-2}- ... - \frac{ a_{n-1}}{a_n} \mathcal{I} $.

Is this reasoning correct, can we express $\mathcal{A}^{-1}$ in terms of $\mathcal{A}$ this way? Thanks a lot in advance.

Answer 1

Yes, your reasoning is absolutely correct.

You could obtain the same result by using the characteristic polynomial $p_A(t) = \det ( A-tI)$ of $A.$ It is even a classical method, when the constant coefficient of this $p_A$ is nonzero, to prove that $A$ is invertible and compute its inverse.

Moreover (just for fun), note that $$p_{A^{-1}}(t) = \det (A ^{-1}-tI)= \det(-tA^{-1}(A-t^{-1}I))=(-t)^n\det(A^{-1})p_A(t^{-1}).$$

Answer 2

Yes, but it is a little different: if $\;A\;$ is invertible then using CH theorem

$$0=a_0I+a_1A+\ldots+A^n\stackrel{\cdot A^{-1}}\implies 0=a_0A^{-1}+a_1I+\ldots+aA^{n-1}\implies$$

$$A^{-1}=-\frac1{a_0}\left(a_1I+a_2A+\ldots+A^{n-1}\right)$$

since of course $\;a_0=\pm\det A\neq0\;$ .

This proves $\;A^{-1}\;$ is a polynomial in $\;A\;$ of degree less than $\;n=\;$ the matrix's order.