# Characteristic polynomial of an inverse

Given the characteristic polynomial $\chi_A$ of an invertible matrix $A$, I’m to find $\chi_{A^{-1}}$.

I can see that this is theoretically possible. $\chi_A$ uniquely determines the similarity class of $A$, which uniquely determines the similarity class of $A^{-1}$, which uniquely determines $\chi_{A^{-1}}$.

Calculating the coefficients of $\chi_{A^{-1}}$ explicitly and then relating them to the coefficients of $\chi_A$ seems unfeasible. I thought about calculating the factors instead, which could be easier since I at least have some idea what the linear factors $\chi_{A^{-1}}$ are (since I can see how to get the eigenvalues of $A^{-1}$ from those of $A$), but that doesn’t help me with potential higher-degree irreducibles or repeated linear factors. Also, we’re not supposed to know the eigenvalues of $A^{-1}$, at least I don’t think so, since calculating them is the next question.

Any hints?

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Hint: Look at the determinant of $(\lambda I-A)A^{-1}$.

$\chi_{A^{-1}}$ is essentially the reciprocal polynomial of $\chi_{A}$.
More precisely, if $$\chi_{A}= X^n + a_{n-1} X^{n-1} + \cdots + a_1 X + a_0$$ then
$$\chi_{A^{-1}}= X^n + \frac{1}{a_0}(a_1 X^{n-1} + \cdots + a_{n-1} X + 1)$$
This can be deduced from

$0 = (A^n + a_{n-1} A^{n-1} + \cdots + a_1 A + a_0I) A^{-n} = a_0(A^{-1})^n+\cdots + a_{n-1}A^{-1}+I$

and dividing by $a_0$ to get a monic polynomial.

Note that $a_0\ne 0$ because $a_0=(-1)^n\det A$ and $A$ is invertible.

A shorter way is to take $\mu=\frac{1}{\lambda}$ and write
$$|\lambda I – A^{-1}|\cdot|A|=|\lambda A-I|=|\lambda A-\lambda \mu I|= \lambda^n|A-\mu I| = (-\lambda)^n|\mu I – A| = (-\lambda)^n\chi_A(\mu) = (-\lambda)^n\chi_A(\frac{1}{\lambda})$$