Minimal polynomial of $T(X)=A^{-1}XA$

A possible way to show that those are the eigenvalues is to exibit the matrix of T.

It is useful to first note that the function $T(X)=AXA^{-1}$ is equivalent to $S(X)=DAD^{-1}$, in fact $$T(X)= Q D Q^{-1} X Q D^{-1} Q^{-1} = \beta_{Q} \circ S \circ \beta_{Q^{-1}}(X) $$ where $\beta_Q(X) = Q X Q^{-1}$ is an invertible linear map with determinant one.

We can now study $S$.

Let $\mathcal{E}$ be the canonical basis of $V$, space of the $n \times n$ matrices, ordered by columns.

The matrices of $L_{D^{-1}}(X) = D^{-1}X$ and $R_{D}(X)=XD$ in the basis $\mathcal{E}$ are easily shown by direct computation to be diagonal. Denoting with $l_k$ the $k$-th element on the diagonal of the first matrix and $r_k$ the $k$-th element on the diagonal of the second we have $l_{ni+j} = \frac{1}{\lambda _j}$ and $r_{ni+j}= \lambda_i$ for each $i,j \in \{1,..,n\}$ .

The matrix of $S$ is their product, wich is diagonal with entries of the form $s _{ni+j} = \frac{\lambda_i}{\lambda_j}$. The charateristic polinomial is then as follows: $$ p_S(x)= \prod_{\substack{1 \leq i \leq n \\ 1 \leq j \leq n }} (x - \frac{\lambda_i}{\lambda_j})$$ The minimal is the same with repeating factors removed.