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Hi everyone, Given the two full rank matrices $X$ and $A$,

$X_{n\times n},~~(rank(X) = n)$

$A_{m\times n},~~(rank(A) = m \le n)$

Can I get a closed form expression for the following derivative? Thanks in advance.

$\frac{\partial det(X-XA'(AXA')^{-1}AX)}{\partial A}=?$

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Yes, of course. Apply the chain rule. –  Ryan Budney Mar 1 '12 at 20:34
    
You can restrict to $m<n$, otherwise $A$ is invertible, and the bracketed expression is $0$. –  Zack Wolske Mar 1 '12 at 20:47
    
Isn't this outside MO ? I would vote to close if I could. –  BS. Mar 1 '12 at 22:49

1 Answer 1

up vote 1 down vote accepted

Your expression is always zero, thus its derivative is zero. Proof. From the Schur complement formula, $$\det(AXA^T)\cdot\det(X-XA^T(AXA^T)^{-1}AX)=\det MXM^T,$$ where $M=\begin{pmatrix} I_n \\\\ A \end{pmatrix}$. But $MXM^T$ is a $q\times q$ matrix with $q=m+n>n$, whereas its rank is $n$. Therefore $\det(MXM^T)=0$. Because $AXA^T$ is non-singular, $\det(AXA^T)\ne0$ and there remains $$\det(X-XA^T(AXA^T)^{-1}AX)\equiv0.$$

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Thanks for your detailed and helpful response. –  Soroosh Mar 1 '12 at 23:22

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