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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?

Rodrigo de Azevedo
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Soheil Feizi
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  • What do you mean by "closed form"? – Igor Rivin Feb 13 '18 at 00:06
  • 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 ${\rm Tr},(1/G+1/G^t)$. – Carlo Beenakker Feb 13 '18 at 03:22
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    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. – Peter Michor Feb 13 '18 at 11:05

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