# Derivative for Matrix function

I have a matrix kernel function which I am trying to find the derivative to. Function is K = c * exp[-1/2 * (P(X1 - X2))' * P(X1 -X2)] where uppercase are matrices and lower case are scalars (and ' denotes transpose). I'm trying to find dK/dP. I'm pretty rusty on matrix calculus, can anyone give me a hand here?

Thanks

-
Your notation is not very clear. Is $P$ a matrix? then did you mean something like $K(P) = c \exp[-\frac12 \| P(X_1 - X_2) \|^2]$? In that case you should probably write (P(X1-X2))', the transpose should be around everything and maybe a trace before it? Is P symmetric or not? –  passerby51 Jun 13 '12 at 17:43
I was assuming this is a matrix valued function. If that's true I'd undelete my (by now corrected) answer. If @passerby51 assumption is correct, the derivative is just $$D_V K = -1/2K \langle V(X_1-X_2), P(X_1-X_2)\rangle$$ with $\langle.,.\rangle$ the scalar product on the vector space of matrices. –  user20266 Jun 13 '12 at 17:50
Sorry yeah the transpose is around the whole thing. P is not symmetric. –  tomas Jun 13 '12 at 17:50
Yeah it is a matrix valued function. I didn't quite get a chance to see your deleted answer Thomas. What was it? Thanks for the response guys. –  tomas Jun 13 '12 at 17:53
undeleted my reply. –  user20266 Jun 13 '12 at 17:53

Let $$Z(P)=(P(X_1-X_2))^TP(X_1-X_2)$$ If $K$ is viewed as depending only on $P$ and if you differentiate in direction $V$ you get $$D_V K = K*\frac{-1}{2} \left\{\frac{I-e^{-ad_{Z(P)}}}{ad_{Z(p)}} \left[ (V(X_1-X_2))^TP(X_1-X_2) + P(X_1-X_2)^TV(X_1-X_2) \right] \right\}$$
The term on the right hand side in curly parenthesis needs explanation. It is the matrix valued function $\frac{I-e^{-ad_{Z(P)}}}{ad_{Z(P))}}$ applied to the term in square brackets. This in turn means
$$\frac{I-e^{-ad_{Z}}}{ad_{Z}}[Y] = Y-\frac{[Z,Y]}{2!} + \frac{[Z,[Z,Y]]}{3!} - \ldots$$
(Sorry for posting a too simple and wrong answer first, which is true only if $Z$ and $D_VZ$ commute). (I don't like the $\frac{d}{dP}$ notation).