User talk:Ashigabou/Matrix calculus

Inverse of matrix

I think there is something fishy in the computation for the derivative of the inverse of a matrix. The thing to keep in mind if you use the Frechet definition is that a derivative is a map from M(n,n) to M(n,n), like you do when you compute the derivative of A^TA.

I think the computation should go as follows:

{\begin{matrix}AA^{-1}&=&I\\{\frac {\partial AA^{-1}}{\partial A}}(X)&=&{\frac {\partial I}{\partial A}}(X)\\{\frac {\partial A}{\partial A}}(X)\,A^{-1}+A{\frac {\partial A^{-1}}{\partial A}}(X)&=&0\\A{\frac {\partial A^{-1}}{\partial A}}(X)&=&-{\frac {\partial A}{\partial A}}(X)\,A^{-1}\\A{\frac {\partial A^{-1}}{\partial A}}(X)&=&-XA^{-1}\\{\frac {\partial A^{-1}}{\partial A}}(X)&=&-A^{-1}XA^{-1}\\\end{matrix}}

This also agrees with the formula at matrix inverse. I hope that makes sense. -- Jitse Niesen (talk) 17:22, 3 February 2006 (UTC)[reply]

You are of course right (I didn't put this page on the main article for this formula, as I was not satisfied with my way of doing things...). To be rigorous, one should be able to prove the differentiability of a^-1 to use the product rule. I am not sure how to do it, though Ashigabou 05:10, 4 February 2006 (UTC)[reply]