Eigenvalues of non-diagonalizable representations of a group acting on $V^\otimes V$, $\text{Sym}^k V$, $\bigwedge^k V$, etc. (Fulton & Harris) In Section 2.1 of Fulton & Harris, it states that if the action of a group element $g$ has eigenvalues $\{\lambda_i\}$ when acting on a vector space $V$, and eigenvalues $\{\mu_j\}$ when acting on a vector space $W$, then $g$ has eigenvalues $\{\lambda_i \mu_j\}$ on $V \otimes W$ (where the representation on $V \otimes W$ is defined via $g\cdot (v\otimes w) = (g\cdot v) \otimes (g \cdot w)$. It also states that the eigenvalues of $g$ on the exterior power $\bigwedge^2 V$ are $\{\lambda_i \lambda_j : i < j\}$.
This is all clear to me when the representations of $g$ on $V$ and $W$ are both diagonalizable, so we can just work with a basis of eigenvectors for each vector space. But what about when a representation (of some element) isn't necessarily diagonalizable (e.g., for the natural representation of the general linear group, consider any nondiagonalizable matrix)? What's a clean way to prove that these are indeed the eigenvalues (complete with multiplicities) for $V \otimes W$, $\bigwedge^2 V$, and other tensor/exterior/symmetric powers? I tried looking at the characteristic polynomial, but did not get far.
 A: Let $A \colon V \to V$ be a linear operator on the finite-dimensional $k$-vector space $V$. After going to a suitable field extension $K/k$ (the algebraic closure of $k$ will surely do), the matrix $A$ is triangularisable, i.e. there exists a basis $(v_1, \ldots, v_n)$ of $V$ such that the matrix of $A$ is upper-triangular with respect to this basis:
$$ A v_j = \sum_{i \leq j} a_{ij} v_i.$$
The characteristic polynomial $\chi_A(t)$ is defined over the original field $k$, and over $K$ it splits into linear factors $\chi_A(t) = (t - a_{11}) \cdots (t - a_{nn})$.
We can do the same for another linear operator $B \colon W \to W$: after a field extension $K$ (which can be taken to be the same for that of $A$), it is upper-triangular with respect to some basis $(w_1, \ldots, w_m)$:
$$ B w_j = \sum_{i \leq j} b_{ij} w_i.$$
Now consider the linear operator $A \otimes B \colon V \otimes W \to V \otimes W$. We can calculate its action on a basis element
$$ \begin{aligned}
(A \otimes B)(v_j \otimes w_l)
&= \left(\sum_{i \leq j} a_{ij} v_i \right)\otimes \left( \sum_{k \leq l} b_{kl} w_k \right) \\
&= \sum_{\substack{i \leq j \\ k \leq l}}a_{ij} b_{kl} (v_i \otimes w_k),
\end{aligned}$$
which is again upper-triangular in a suitable ordering on pairs $(i, k)$, where we say that $(i, k) \leq (j, l)$ if $i \leq j$ and $k \leq l$. Therefore the characteristic polynomial of $A \otimes B$ is
$$ \chi_{A \otimes B}(t) = \prod_{\substack{1 \leq i \leq n \\ 1 \leq j \leq m}} (t - a_{ii} b_{jj}).$$
This proves the claim for fields. There are some lovely arguments (search around for "universal identities") which will allow you to extend this to more general commutative rings if you like. It should hopefully also now be clear how to attack the case of the exterior square $\bigwedge^2 A$ and symmetric square $S^2 A$.
