If $\det(A-\lambda I)=\det(A^{-1}-\lambda I)$, then characteristic polynomial is coefficient symmetry

Question

In one class, professor mentioned the following without proof,

Suppose

1. $A\in \mathbb{R}^{2N\times 2N}$.
2. $\det(A-\lambda I)=\det(A^{-1}-\lambda I)=p(\lambda)$

then if $$p(\lambda) = a_{2N}\lambda^{2N}+a_{2N-1}\lambda^{2N-1}+ \cdots +a_1\lambda^1 +a_0,$$ we have $a_{2N}=a_0, a_{2N-1}=a_1, \cdots.$

I refer to the following discussion Show that $\det(A-\lambda I)=\det(B-\lambda I)$

However, I still have no idea how to see it. Is there any good method without expanding the "$\det$" to show it?

Thanks in advanced.

Answer 1

Hint: $\lambda$ is an eigen value of $A$ iff $\frac 1 {\lambda}$ is an eigen value. Hence $\lambda$ is an root of $p(\lambda)$ value of $A$ iff $\frac 1 {\lambda}$ is a root of $p(\lambda)$.

Answer 2

$det(A-\lambda I)=det(A^{-1}-\lambda I)$ suppose $\lambda=0$, you deduce that $det(A)=det(A^{-1})=1$.

Write $P(\lambda)=det(A-\lambda I)$

$det(A-\lambda I)=det(\lambda A)det({1\over\lambda}I-A^{-1})=\lambda^{2n}P({1\over \lambda})$

Comparing the coefficients of $\lambda^{2n}P({1\over \lambda})$ and $P(\lambda)$ you obtain the result.

Answer 3

For $A$ of order $4$,

$p(\lambda)=(\lambda-a)(\lambda-1/a)(\lambda-b)(\lambda-1/b)$ gives a symmetrical polynomial of degree $4$. Generalize this for all $n$.