Determinant of Jacobian of matrix multiplication

Question

Let $A \in \mathbb R^{n \times n}$. We consider the map $$f_A : \mathbb R^{n \times n} \to \mathbb R^{n \times n} ,\quad X \mapsto AX.$$ By considering easy examples of $A$ one comes up quite fast with the conjecture $\det(D f_A) = \det(A)^n$. Here $D f_A$ is the Jacobian matrix of $f_A$.

Is there an elegant proof which does not result in a long confusing computation?

Idea (edit): I just came up with an idea. Obviously, the statement has only to be proven for non-singular $A$. They are generated by elementary matrices. So it suffices to consider them since we have $f_A \circ f_B = f_{AB}$.

Answer 1

Consider $X$ to be $n$ separate column vectors and then stack them on top of each other to form one large column vector $x=(x_{1,1},x_{2,1},\ldots,x_{n,1},x_{1,2},\ldots)^\top$. The map $f_A$ can thus be represented by a block diagonal matrix $B$ of $n$ blocks, each block being $A$. The linear map $f_A$ can be viewed as $\mathbb{R}^{n^2}\rightarrow\mathbb{R}^{n^2},x\mapsto Bx$. The derivative of this linear map is just $B$. Its determinant is (block diagonal matrix!): $$ \det(B)=\det(A)^n. $$

Answer 2

$A$ acts on each column of $X$. $Df_A$ gives the best linear approximation of $f_A$ at each point. But $A$ is linear and acts on each $n$-dimensional column vector. So $det(Df_A)$ is the volume of the unit parallelepiped after applying the best linear approximation of $f_A$ to each of its vectors, giving us that $det(Df_A)=det(A)^n$ since we are scaling each column vector by a factor of $det(A)$. Completely unrigorous of course, but that is my way of thinking about it conceptually.