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

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.

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?

• I do not understand the first line in the expression for $var(v).$ – Justpassingby Dec 7 '15 at 0:39
• It is variance of $v$ but fair enough, this bit is statistical. – aimi Dec 7 '15 at 0:43
• But where does it come from? It does not seem to follow the same rule as your earlier computation of $var(u).$ – Justpassingby Dec 7 '15 at 0:45
• Hi, we let $var(v) = G$. Then since $u=Mv$ and it is normal rv, it follows that $var(u) = MGM^\top$. – aimi Dec 7 '15 at 1:21
• Might be worth noting that the rest of the expression after the first $G$ on the lhs gives the orthogonal projection onto the row space of $M$ relative to the scalar product defined by $G$. – amd Dec 7 '15 at 1:28

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).

• Thanks, elegant contradictory proof. I just added my logic of how I got to this but I'm not sure where it fell short. Will you be enlighten me where it might have gone wrong? – aimi Dec 7 '15 at 0:28
• I would like to understand where the equality $\text{var}(v)=GM^\top(MGM^\top)^{-1}MGM^\top(MGM^\top)^{-1}MG$ comes from. It gives you the variance of a possible $v$ that solves $u=Mv$ but there is no guarantee that it is the variance of the original $v$ that you started with. – Justpassingby Dec 9 '15 at 7:22
• Actually you are correct. That is probably where my logic failed. Sorry must have been staring at the problem too long to not realise that obvious -.- – aimi Dec 9 '15 at 8:56

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.

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.