Given two matrices $A$ (is $m \times m$) and $B$ (is $n \times n$) that are both diagonalizable, prove its Kronecker sum
$$ A \oplus B := A \otimes I_n + I_m \otimes B $$
is diagonalizable, and give its eigenvalues in terms of those of $A$ and $B$.
I tried solving the problem by first naming Diagonal matrices $D = P^{-1} A P$ and $E = Q^-1 B Q$.
We can see that $D \otimes I = (P^{-1} \otimes I)(A \otimes I)(P \otimes I)$, and similarly for $B$.
Then I tried "guessing" that the diagonalization matrix of $A \oplus B$ was $P \otimes Q$, and proving it diagonalized the Kronecker sum. I did some algebra in this, but I hit a wall. Pls halp.
You're nearly there. As user1551's comment notes, we have $(X\otimes Y)(Z\otimes W)=(XZ)\otimes(YW)$. So, we have $$ (P \otimes Q)^{-1}(A\otimes I_n + I_m \otimes B)(P \otimes Q) = \\ (P^{-1} \otimes Q^{-1})(A\otimes I_n + I_m \otimes B)(P \otimes Q) = \\ (P^{-1} \otimes Q^{-1})(A\otimes I_n)(P \otimes Q) + (P^{-1} \otimes Q^{-1})(I_m \otimes B)(P \otimes Q) = \\ (P^{-1}AP) \otimes (Q^{-1}I_nQ) + (P^{-1}I_m P) \otimes (Q^{-1}BQ) =\\ D \otimes I_n + I_m \otimes E= \\ D \oplus E. $$ Note that $D \oplus E$ is a diagonal matrix. You should find that if $\lambda_1,\dots,\lambda_m$ are the diagonal entries of $D$ and $\mu_1,\dots,\mu_n$ the diagonal entries of $E$, then the diagonal entries of $D \oplus E$ are of the form $\lambda_i + \mu_j$ for every pair $i,j$ with $1 \leq i \leq m$ and $1 \leq j \leq n$.