I have to prove that $A$ characteristic's polynomial $P(x)$ and $A^{-1}$ characteristic's polynomial $Q(x)$ are related by: $$ Q(\lambda) = \frac{\lambda^{n} P(\lambda ^{-1})}{P(0)} $$
I've tried by considering the matrix $A(A^{-1} − λI)$, but I've obteined no reults...
Any help?
Hint:
Suppose $\;c\neq0\;$ is a root of a polynomial with free coefficient$\,\neq0\;$ :
$$p(x)=\sum_{k=0}^na_kx^k\implies 0=p(c)=a_0+a_1c+\ldots+ a_nc^n\implies $$
$$a_0c^{-n}=-\left(a_1c^{1-n}+a_2c^{2-n}+\ldots+a_n\right)\implies c^{-n}=\frac{-\left(a_1c^{1-n}+a_2c^{2-n}+\ldots+a_n\right)}{a_0}$$
since $\;p(x)\;$ is irreducible and $\;a_0\neq0\;$ since $\;c\neq0\;$ . The above means that $\;c^{-1}\;$ is a root of
$$\frac{a_n}{a_o}+\frac{a_{n-1}}{a_0}x+\ldots+\frac{a_2}{a_0}x^{n-2}+\frac{a_1}{a_0}x^{n-1}+x^n=\frac{a_n+a_{n-2}x+\ldots+a_1x^{n-1}+x^n}{a_0}=$$
$$=\frac{x^n\,p\left(x^{-1}\right)}{p(0)}\;\ldots\ldots$$
Note that $P(0)=det(-A)=(-1)^n det(A)$. Then we compute for $\lambda\neq 0$ $$Q(\lambda)=det(\lambda I-A^{-1}) = det (A^{-1}) det(\lambda A-I)=\lambda^n det(A^{-1}) det (A-\lambda^{-1}I)= (-\lambda)^n det(A)^{-1} P(\lambda^{-1}I-A) =\lambda^n P(0)^{-1} P(\lambda^{-1}) .$$
