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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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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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