# Determinant of Jacobian of matrix multiplication

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 $$f_A \circ f_B = f_{AB}$$.

• What does the Jacobian matrix have to do with anything? Since $f_A$ is linear, wouldn't its Jacobian just be itself? Aug 16 at 21:25
• But to represent $A$ as a matrix in the usual way you have to do the trick that Thomas Peru has suggested in an answer.
– T_M
Aug 16 at 21:52

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!): $$\det(B)=\det(A)^n.$$
• Instead of $n$ separate column vectors you mean one large column vector. Aug 16 at 22:26
$$A$$ 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.