## Derivative in Matrix Calculus

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 2012 at 20:34 You can restrict to $m ## 1 Answer 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 2012 at 23:22