For an invertible matrix $A$, we have the identity \begin{align} \dfrac{\partial \det A}{\partial A} = \det A (A^{-1})^T \end{align} where the $T$ denotes the transpose operation.

How does this formula change when considering the Kronecker product of $A$ with some other (invertible) matrix $B$? For example, how does one compute the following:

\begin{align} \dfrac{\partial \det \left[A\otimes B\right]}{\partial A} \end{align}


1 Answer 1


I find this question very interesting. I could not find any reference to this problem in any book. So, here is my derivation for square matrices and I would like to hear other people to know if it is correct.

$A$ being ${n\times n}$ matrix and $B$ being ${p\times p}$ matrix. We know that $$\det[A\otimes B]=(\det{A})^p(\det{B})^n$$ (Notice how the power switches between $n$ and $p$)

Then the derivative is $$\frac{\partial(\det{A})^p(\det{B})^n}{\partial A}=(\det{B})^np(\det A)^{p-1}\det A(A^{-1})^T=p(\det B)^n(\det A)^p(A^{-1})^T$$

To put it "nicer":

$$\frac{\partial\det[A\otimes B]}{\partial A}=p\det[A\otimes B](A^{-1})^T$$

  • $\begingroup$ Thank you! I have confirmed that this is the correct answer. I had initially derived this result including the determinant of the Kronecker product, but I had made a mistake computing the determinant of my matrix leading me to think this answer was incorrect. $\endgroup$
    – Matze
    Apr 24, 2014 at 4:12
  • $\begingroup$ @Matt. I would be actually interested in some application where you use this result. Are you working on some "bigger" problem? $\endgroup$
    – pisoir
    Apr 24, 2014 at 6:53

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