0
$\begingroup$

Suppose $G$ and $A$ are full rank matrices. Is there a closed-form solution for

$$\nabla_G \mbox{Tr} (A \log GG^\top)$$

when $A$ is a PSD matrix?

$\endgroup$
3
  • $\begingroup$ What do you mean by "closed form"? $\endgroup$
    – Igor Rivin
    Commented Feb 13, 2018 at 0:06
  • $\begingroup$ unless $A=I$ this is a complicated calculation because you need to take the derivative of the logarithm of a matrix, see for example this MO question; for $A=I$ the answer is just ${\rm Tr}\,(1/G+1/G^t)$. $\endgroup$ Commented Feb 13, 2018 at 3:22
  • 1
    $\begingroup$ Write the series for $\log$ and differentiate each term: Differentiating a power needs $D_{A,X} A^n = XA^{n-1} + AXA^{n-2} +\dots+A^{n-1}X$ (for the directional derivative with respect to the variable $A$ in direction $X$), since the matrices do not commute. Passing to the gradient needs an inner product on the space of matrices. $\text{Tr}(XY^\top)$ is a suitable one. $\endgroup$ Commented Feb 13, 2018 at 11:05

0

You must log in to answer this question.

Browse other questions tagged .