Solution (source code)

= Solution

For $t\ne0$, factor the <matrix> as
$$
A^{-1}-tI=A^{-1}(I-tA)=A^{-1}\bigl[-t(A-t^{-1}I)\bigr].
$$
The <determinant> of a scalar multiple of an $n\times n$ <matrix> gains the scalar's $n$th power. Thus
$$
\boxed{\chi_{A^{-1}}(t)=(\det A)^{-1}(-1)^nt^n\chi_A(t^{-1}).}
$$
The right side extends to a <polynomial> at $t=0$: multiplying $\chi_A(t^{-1})$ by $t^n$ removes all negative powers. Its value there is $(\det A)^{-1}$, since the leading coefficient of $\chi_A(s)=\det(A-sI)$ is $(-1)^n$. This equals $\chi_{A^{-1}}(0)$, so the identity holds at zero with that interpretation.

For the relation between <eigenvalues> and the <characteristic polynomial>,
$$
\lambda\text{ is an eigenvalue of }A
\iff \ker(A-\lambda I)\ne\{0\}
\iff A-\lambda I\text{ is singular}
\iff \chi_A(\lambda)=0.
$$
The first equivalence is the definition of an <eigenvector>; the others use the <kernel of a linear map> and the criterion for an <invertible matrix> to have nonzero <determinant>. Since $\chi_A$ has degree $n$, the <fundamental theorem of algebra> gives $n$ roots over $\mathbb C$ counted with multiplicity. The multiplicity of a root is its <algebraic multiplicity>, and the <eigenspace> is exactly $\ker(A-\lambda I)$; its dimension is the <geometric multiplicity>, which need not equal the <algebraic multiplicity>.

Finally, factor $p(s)=s^4-\mu$ over $\mathbb C$ as $p(s)=\prod_{j=1}^4(s-\lambda_j)$, including repeated roots when $\mu=0$. Substituting $A$ into this <polynomial> is valid because the resulting factors commute. If $\mu$ is an <eigenvalue> of $A^4$, then
$$
0=\det(A^4-\mu I)=\prod_{j=1}^4\det(A-\lambda_jI).
$$
At least one factor must be zero, so its $\lambda_j$ is an <eigenvalue> of $A$ and satisfies $\lambda_j^4=\mu$. Thus the reverse direction of this instance of the <spectral mapping theorem> is
$$
\boxed{\mu\text{ an eigenvalue of }A^4
\ \Longrightarrow\ \exists\lambda\text{ an eigenvalue of }A\text{ with }\lambda^4=\mu.}
$$
This argument also covers $\mu=0$, when the conclusion is that $A$ is singular and has <eigenvalue> zero.