For a non-square matrix $X$ of size $n \times p$ ($n>p$) and another non-square matrix $A$ of size $p \times n$, what is the derivative of $\det(X^TA)$ w.r.t. $X$? i.e.,

$\frac{\partial\det(X^TA)}{\partial X}$

It'd be a matrix of size $n \times p$ while I got stuck in some intermediate steps of using chain rules.

Update: I think I got the solution. Thanks to AlexR and Bob.

Reference: The Matrix Cookbook; Section 2.1.2; Equation (45)

$\frac{\partial\det(AXB)}{\partial X} = \det(AXB){(X^{-1})}^T = \det(AXB){(X^T)}^{-1}$

This is, actually, a special case when A and B are non-singular matrices.

The more general form, according to matrix reference manual, is

(URL: http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/calculus.html#deriv_det)

$\frac{\partial\det(A^TXB)}{\partial X} = \frac{\partial\det(B^TX^TA)}{\partial X} = A\operatorname{adj}(A^TXB)^TB^T = A(\det(A^TXB)(A^TXB)^{-1})^TB^T = \det(A^TXB)A(B^TX^TA)^{-1}B^T$

When $B = B^T= I$

$\frac{\partial\det(X^TA)}{\partial X} = A\operatorname{adj}(A^TX)^T = A(\det(A^TX)(A^TX)^{-1})^T = \det(X^TA)A(X^TA)^{-1}$

  • $\begingroup$ Did you find $\partial_X \det X$ already? If not, you can compute $\partial_X \log \det X$ by using $\det X = \prod_{i=1}^p \lambda_i$ where $X\in\mathbb R^{p\times p}$. Then use the generalisation of the logarithmic derivative trick to obtain what you're looking for. $\endgroup$ – AlexR May 28 '15 at 22:02
  • $\begingroup$ @AlexR. Thanks for your suggestions first. However, $\det(X)$ is not meaningful given that $X$ is not a square matrix. $\endgroup$ – Will Yongxin Yang May 28 '15 at 22:29
  • $\begingroup$ Note I write $\log \det : \mathbb R^{p\times p} \to \mathbb R$ wich is meaningful. $\endgroup$ – AlexR May 28 '15 at 22:30
  • $\begingroup$ Hi AlexR. Sorry I don't really get the idea behind $\log\det$. Can you name a reference? Cheers! $\endgroup$ – Will Yongxin Yang May 28 '15 at 22:36
  • $\begingroup$ It's used in Convex Optimisation. The matrix cookbook is a better source if you don't need to prove it. $\endgroup$ – AlexR May 28 '15 at 22:40

We know that $\frac{\partial\det(XA)}{\partial X}=\det(XA)(X^T)^{-1} $ (see The Matrix Cookbook, for example). Thus,

$\frac{\partial\det(X^TA)}{\partial X}=\det(XA)(X)^{-1}$.

  • $\begingroup$ Hi Bob. Thanks for your answer first. I have noticed that a similar form in The Matrix Cookbook but I don't think it's that straightforward as $X^{-1}$ suggests that X is a square matrix. $\endgroup$ – Will Yongxin Yang May 28 '15 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.