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 ATA.

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)


 * 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)