linear algebra - Pseudo inverse of a product of two matrices with different rank

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?

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 $X^+ V^{-1}$:

1. $(VX) (X^+ V^{-1}) (VX) = VX X^+ X = VX$. Ok.


2. $(X^+ V^{-1}) (VX) (X^+ V^{-1}) = X^+ X X^+ V^{-1} = X^+ V^{-1}$. Ok.


3. $((VX) (X^+ V^{-1}))^* = V^{-*} (XX^+)^* V^* = V^{-2} (VX)(X^+ V^{-1}) V^2$. Hmmm...


4. $((X^+ V^{-1}) (VX))^* = (X^+X)^* = X^+X = (X^+ V^{-1}) (VX)$. Ok.

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.