Proving matrix equation: $GM^\top(MGM^\top)^{-1}MG=G$

Question

I want to show $$GM^\top(MGM^\top)^{-1}MG=G$$ where $MGM^\top$ is invertible, $G$ is a symmetric square matrix. $M$ does not have to be a square matrix. I am not 100% sure this is true (90% sure it is) so a contradictory proof will be great as well.

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Thanks Justingpassby for the contradictory proof. I came into above from the following but I'm not sure where it is wrong?

Let $u = Mv$, $\text{var}(v)=G$, $u$ and $v$ be both Normal random variables with mean 0 so $\text{var}(u)=MGM^\top$. Then $v=GM^\top(MGM^\top)^{-1}u$ is one possible solution of $u = Mv$ since $$ Mv=MGM^\top(MGM^\top)^{-1}u = u.$$ Now $$\begin{array}\text{var}(v) &=& GM^\top(MGM^\top)^{-1}MGM^\top(MGM^\top)^{-1}MG\\ &=& GM^\top(MGM^\top)^{-1}MG \end{array}$$ Therefore $GM^\top(MGM^\top)^{-1}MG=G$. But as shown this it not true for some non-square $M$ so there must be a hole in my logic somewhere above. Any idea anyone?

Answer 1

Let $G$ be the unit 3 by 3 matrix and $M$ any nonzero 1 by 3 row. Then the LHS has rank 1 (and determinant 0).

Answer 2

If $L$ is the left side, $$M L M^T = (MGM^T)(MGM^T)^{-1}(MGM^T) = MGM^T$$ If $M$ is a square matrix, $M$ must be invertible for $MGM^T$ to be invertible, so $L = G$. If $M$ is not square, then as Justpassingby noted $L$ does not have full rank, whereas $G$ could have full rank.

Answer 3

Formally,

$$(MGM^\top)^{-1}= M^{\top(-1)}G^{-1}M^{-1}$$

So

$$\begin{align} GM^\top(MGM^\top)^{-1}MG &= GM^\top M^{\top(-1)}G^{-1}M^{-1} MG \\ & = G \end{align}$$

since everything else cancels.

This assumes that both $G$ and $M$ are invertible.