Suppose I have a differentiable function $\phi: \mathbb{R}^{p\times p} \mapsto \mathbb{R}$ defined as $\phi(\exp(tA))$ where $t$ is a positive scalar and $A$ is a $p\times p$ real matrix. How can I find gradient $\nabla \phi$ with respect to $A$?
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2$\begingroup$ As far as I know the exponental of a matrix id defined only for square martices. $\endgroup$– Emilio NovatiCommented Aug 24, 2015 at 19:38
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$\begingroup$ @Taha can we assume that $p=q$? $\endgroup$– Ben GrossmannCommented Aug 24, 2015 at 19:38
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$\begingroup$ I fixed the typo. Now p = q. $\endgroup$– TahaCommented Aug 24, 2015 at 19:59
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Suppose that we have $\psi$ defined by $$ \psi(A) = \phi(\exp(tA)) $$ The chain rule tells us that $$ D_A\psi = [D\phi](\exp(tA)) [D_A \exp(tA)] $$ The derivative of the exponential map may be taken as given here, for example.
answered Aug 24, 2015 at 19:42
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$\begingroup$ Thank you for the answer. Is it possible to point out a simpler article for derivative of just matrix exponential? I do not have the proper background in group theory and it takes some time to fully understand that article. $\endgroup$– TahaCommented Aug 25, 2015 at 1:04
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$\begingroup$ Thanks. Do you mean the answer will be far from being used in a practical gradient descent optimization algorithm? $\endgroup$– TahaCommented Aug 25, 2015 at 1:25
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$\begingroup$ @Taha actually, this might work well numerically. One expression is a rapidly converging infinite sum, and the other is an integral. It might be easy to get a good approximation... do you want me to clarify that infinite sum? $\endgroup$ Commented Aug 25, 2015 at 13:24
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$\begingroup$ Thank you. I also found a couple of papers on numerical computation. After I read them, I will supplement your answer with explicit formulas for future reference of people like me who are just looking for explicit solutions :) $\endgroup$– TahaCommented Aug 25, 2015 at 15:34