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.


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



Derivative in Matrix Calculus
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
$$\det(X-XA^T(AXA^T)^{-1}AX)\equiv0.$$


