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

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    $\begingroup$ What does the Jacobian matrix have to do with anything? Since $f_A$ is linear, wouldn't its Jacobian just be itself? $\endgroup$ Aug 16 at 21:25
  • $\begingroup$ 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. $\endgroup$
    – T_M
    Aug 16 at 21:52

2 Answers 2


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!): $$ \det(B)=\det(A)^n. $$

  • $\begingroup$ Instead of $n$ separate column vectors you mean one large column vector. $\endgroup$ Aug 16 at 22:26
  • $\begingroup$ Well view them as seperate column vectors and then stack them on top of each other to form one large column vector - it has both aspects. $\endgroup$ Aug 16 at 22:27

$A$ 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.


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