Gradient of a $\log\det$ function

Question

Given symmetric $d \times d$ matrices ${\bf G}, {\bf F}_1, \dots, {\bf F}_n$, let ${\bf F} : {\Bbb R}^n \to {\Bbb R}^{d \times d}$ be defined by

$$ {\bf F} ({\bf x}) := {\bf G} + \sum_{i=1}^n x_i {\bf F}_i $$

What is the gradient of the scalar field ${\bf x} \mapsto \log\det({\bf F}({\bf x}))$?

I tried to use the chain rule to find the gradient but I find the first part, derivative of $\log\det({\bf F}({\bf x}))$ is ${\bf F}({\bf x})^{-1}$ which shape is $n^2$, but the second part, ${\bf F} ({\bf x})$ is a vector to matrix function where its derivative is shape $n \times n^2$ where the final derivative shape is $n \times n^2$ but the expected shape of gradient should be same shape with $x$, I was wondering which step goes wrong?

The linked posts will help you understand how to calculate the matrix-wise gradient of $\log\det X,;$ after which the component-wise gradients of your function are easy to calculate $$\large{ \frac{\partial\log\det F}{\partial x_k} ;=; \operatorname{trace}!\big(F_k,F^{-1}\big) }$$ — greg, Dec 12 '23 at 15:50
I have read the related question you post; I now understand how to compute it thank you for helping me! — david shao, Dec 12 '23 at 16:49

score 0 · Answer 1 · answered Dec 12 '23 at 13:04

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The differential is the linear form on $R^I$ defined by $$(h_i)_{i\in I}\mapsto \sum_{i\in I}\mathrm{Trace}(F^{-1}(x)F_i)h_i$$

answered Dec 12 '23 at 13:04

Gérard Letac

4,357

Is that the directional derivative in the direction of $h$? – Rodrigo de Azevedo Dec 12 '23 at 21:05

Gradient of a $\log\det$ function

1 Answers1