Let $V$ be an $n \times n$ symmetric, positive definite matrix (of rank $n$). Let $X$ be an $n \times p$ matrix of rank $p$.
Define $A^- = (A^\top A)^{-1} A^\top$ as the pseudo inverse of $A$ when $A$ is of full column rank. Note that $V^- = V^{-1}$ because $V$ is invertible.
I'd like to prove that
$$ (VX)^- = X^- V^{-1} $$
but the only theorem I know about the pseudo-inverses of products requires that both of the matrices be of the same rank AND that the second matrix has full row rank. (To wit: If $B$ is an $m \times r$ matrix of rank $r$ and $C$ is an $r \times m$
matrix of rank $r$, then $(BC)^- = C^-B^-$.)
There is likely something obvious I'm missing. Any clues?
Best Answer
I am assuming that by a "pseudoinverse" you mean Moore–Penrose pseudoinverse $A^+$ of a matrix $A$. Let us check the defining properties of the Moore-Penrose pseudoinverse against $X^+ V^{-1}$:
So, the above is O.K. if and only if item 3 is O.K., i.e.,
$$((VX) (X^+ V^{-1}))^* = V^{-2} (VX)(X^+ V^{-1}) V^2.$$
However, this is not generally true. For example (by Pedro Milet in comments),
$$V = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} 1 \\ 0 \end{bmatrix}.$$
Then
$$(VX)^+ = \frac{1}{5} \begin{bmatrix} 2 & 1 \end{bmatrix} \ne \begin{bmatrix} 1 & -1 \end{bmatrix} = X^+ V^{-1}.$$
Notice, however, that it would work if $V$ was unitary, instead of positive definite.