Want to show that suppose column space $C(X) = C(V^{-1}X)$ then $X(X^{T}X)^{-1}X^T = X(X^{T}V^{-1}X)^{-1}X^TV^{-1}$ where V is a diagonal matrix and X$(n\times p)$ is full rank.
My try:
since $C(X) = C(V^{-1}X)$, and $(X^{T}X)^{-1}$, $(X^{T}V^{-1}X)^{-1}$ are invertible
then $C(X(X^{T}X)^{-1}) = C(V^{-1}X(X^{T}V^{-1}X)^{-1})$
But I got stuck here. I think I am close to the final form - just need a transpose and multiply X on the left - but it is not feasible.
However, even if I prove $C(X(X^{T}X)^{-1}X^T) = C(X(X^{T}V^{-1}X)^{-1}X^TV^{-1})$ it doesn't imply that $X(X^{T}V^{-1}X)^{-1}X^TV^{-1} = X(X^{T}X)^{-1}X^T$. Should I find another method?
Here's an approach that works in the case that $V$ has non-negative entries. This argument could potentially work in the case that $V$ has positive and negative entries, but would require complex numbers. Notably, for a complex matrix $M$, $M^TM$ might fail to be invertible when $M$ has full column rank.
The reason this works is that like $X^T(X^TX)^{-1}X$, $X(X^{T}V^{-1}X)^{-1}X^TV^{-1}$ is also an orthogonal projection matrix. Note that two projection matrices will be the same if their column spaces are the same.
Let $D$ denote the diagonal matrix with positive diagonal entries for which $D^2 = V$. Because there is a polynomial $p$ for which $D^{-1} = p(V^{-1})$, we can deduce that $C(D^{-1}X) = C(X)$ from the fact that $C(V^{-1}X) = C(X)$. It follows that the projection onto $C(X)$ is equal to the projection $P$ onto $C(D^{-1}X)$, which is given by $$ P = [D^{-1}X]([D^{-1}X]^T[D^{-1}X])^{-1}[D^{-1}X]^T =\\ D^{-1} X(X^TD^{-T}D^{-1}X)^{-1}X^TD^{-T} =\\ D^{-1} X(X^TV^{-1}X)^{-1}X^TD^{-1}. $$ Now, from the fact that $C(D(D^{-1}X)) = C(D^{-1}X)$, we can deduce that $DP = PD$. Thus, we can write $$ P = PDD^{-1} = DPD^{-1} =\\ DD^{-1} X(X^TV^{-1}X)^{-1}X^TD^{-1}D^{-1} =\\ X(X^TV^{-1}X)^{-1}X^TV^{-1}. $$ So indeed, the projection onto $C(X)$ is equal to $ X(X^TV^{-1}X)^{-1}X^TV^{-1}$.