Determinant of a linear map $\phi: \text{Mat}(n \times n, \mathbb{F}) \rightarrow \text{Mat}(n \times n, \mathbb{F})$ I have some difficulties applying the Kronecker-Delta to the elementary matrices, such that I can compute the determinant correctly.
My problem is, that I have a linear map $\phi: \text{Mat}(n \times n, \mathbb{F}) \rightarrow \text{Mat}(n \times n, \mathbb{F})$ given by $M \mapsto X \cdot M \cdot Y$ and I want to compute its determinant.
I tried to use the following identity for an elementary matrix $(E_{ij})_{kl} = \delta_{ij}\delta_{kl} = \delta_{il}$. As a second step I tried to plug in all elements of a basis from $\text{Mat}(n \times n, \mathbb{F})$ into the linear map, but this is somehow to complicated I guess. I would appreciate any suggestions. Succinct, the question is to compute $\det(\phi)$.
 A: There are a few different approaches. I think that the most straightforward is to note that this map can be written as the composition $\phi = \phi_1 \circ \phi_2$, where $\phi_1:M \mapsto XM$ and $\phi_2:M \mapsto MY$, so that $\det(\phi) = \det(\phi_1)\det(\phi_2)$.
First, we compute the determinant of $\phi_1$. To do so, we can first find the matrix of $\phi_1$ relative to the basis $\{E_{ij}:1 \leq i,j \leq n\}$, where the pairs $(i,j)$ are taken in lexicographical order. That is, we proceed in the order $(1,1),(1,2),\dots,(1,n),(2,1),\dots$. You should find that the matrix of $\phi_1$ relative to this basis is given by the block-diagonal matrix
$$
[\phi_1]_{\mathcal B} = \pmatrix{X \\ & \ddots \\ && X}.
$$
It follows that $\det(\phi_1) = \det(X)^n$.
We could apply a similar line of reasoning to find that $\det(\phi_2) = \det(Y)^n$. If we were to use the same basis, we would find that
$$
[\phi_2]_{\mathcal B} = \pmatrix{y_{11} I & \cdots & y_{n1} I\\
\vdots & \ddots & \vdots \\ y_{1n} I & \cdots & y_{nn} I},
$$
where $I$ is the $n \times n$ identity. From there, the fact that these blocks commute would allow us to compute the determinant $\det(\phi_2) = \det(Y^\top)^n = \det(Y)^n$ directly.
Alternatively, we could reuse our previous calculation by taking advantage of the following similarity. If we take $\tau$ to denote the transpose map and $\psi$ to denote the map $M \mapsto Y^\top M$, then we find that $\phi_2 = \tau \circ \psi \circ \tau^{-1}$. It follows that $\det(\phi_2) = \det(\psi)$, and from our previous calculation we know that $\det(\psi) = \det(Y^\top)^n$.

In fact, all this amounts to proving certain relationships between vectorization and the Kronecker product. If we use $\operatorname{vec}$ to denote the row-major vectorization operator (rather than the column-major operator considered in the first link), then we find that
$$
\operatorname{vec}(XMY) = (X \otimes Y^\top) \operatorname{vec}(M).
$$
Thus, $\operatorname{vec}^{-1} \circ \phi \circ \operatorname{vec}$ is the operator over $\Bbb F^{n^2}$ corresponding to $X \otimes Y^\top$. The properties of the Kronecker product are such that $\det(X \otimes Y^\top) = \det(X)^n \det(Y^\top)^n = [\det(X)\det(Y)]^n$.
