Show that
$X′V^{-1}X(X′V^{−1}X)^{+}X′ = X′$
for any positive definite matrix V.
My attempt:
I am writing $V^{-1} = GG'$, and then substituting it in the LHS of the above equation.
LHS = $X'GG'X(X'GG'X)^{+}X'$
Assuming $X'G = B$, I get
LHS = $BB'(BB')^{+}X'$ = $BB'B'^{+}B^{+}X'$
I am stuck here and don't know how to reduce it from here to RHS.
Note that $X'=BG^{-1}$. Hence you need to show that $$BB'(BB')^+B=B$$ Let $B=P\Sigma Q'$ be the singular value decomposition where $$\Sigma=\begin{bmatrix}\Sigma_1&0\\0&0\end{bmatrix}$$, $\Sigma_1$ is $r\times r$ diagonal matrix where $r=\operatorname{rank}B$. So, $$BB'=P\begin{bmatrix}\Sigma_1^2&0\\0&0\end{bmatrix} P' ~~\text{and}~~ (BB')^+=P\begin{bmatrix}\Sigma_1^{-2}&0\\0&0\end{bmatrix} P'$$ Therefore, $$BB'(BB')^+B=P\begin{bmatrix}I_r&0\\0&0\end{bmatrix} \begin{bmatrix}\Sigma_1&0\\0&0\end{bmatrix}Q'=B$$