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Consider the following matrix derivative problem:

$\frac{\partial}{\partial\mathbb{X}}\ln\bigg(\det\Big(\mathbb{X}^T\mathbb{A}\mathbb{X}\mathbb{B}\Big)\bigg)=?$

where:

  • $\mathbb{X}$ is a $(p\times d)$ semi-orthogonal matrix
  • $\mathbb{A}$ is a $(p\times p)$ symmetric matrix
  • $\mathbb{B}$ is a $(d\times d)$ symmetric matrix

I have checked a number of resources such as matrix Cookbook, but I cannot find any property such as this.

Thanks for your time.

  • 0
    A log is not making task more complicated, so all you need is know how to evaluate derivative of det(...), this trace of some matrices. If I had have to solve such a thing, I would have tried to start from det() or even some traces and then recover full answer from that. – 2017-02-21
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    X-post: http://mathoverflow.net/q/262801/91764 – 2017-02-26

1 Answers 1

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Let $$\eqalign{ Y &= X^TAXB \cr }$$ Then write the function in terms of $Y$ and find its gradient $$\eqalign{ f &= \log\det Y \cr \cr df &= d\log\det Y = d\operatorname{tr}\log Y\cr &= Y^{-T}:dY \cr &= Y^{-T}:(dX^T\,AXB+X^TA\,dX\,B) \cr &= Y^{-T}B^TX^TA^T:dX^T + A^TXY^{-T}B^T:dX \cr &= AXBY^{-1}:dX + A^TXY^{-T}B^T:dX \cr &= \big(AXBY^{-1} + A^TXY^{-T}B^T\big):dX \cr \cr \frac{\partial f}{\partial X} &= AXBY^{-1} + A^TXY^{-T}B^T \cr \cr }$$ where I've used a colon to denote the double-dot (aka Frobenius) product in the intermediate steps.