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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?

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  • $\begingroup$ I do not understand the first line in the expression for $var(v).$ $\endgroup$ – Justpassingby Dec 7 '15 at 0:39
  • $\begingroup$ It is variance of $v$ but fair enough, this bit is statistical. $\endgroup$ – aimi Dec 7 '15 at 0:43
  • $\begingroup$ But where does it come from? It does not seem to follow the same rule as your earlier computation of $var(u).$ $\endgroup$ – Justpassingby Dec 7 '15 at 0:45
  • $\begingroup$ Hi, we let $var(v) = G$. Then since $u=Mv$ and it is normal rv, it follows that $var(u) = MGM^\top$. $\endgroup$ – aimi Dec 7 '15 at 1:21
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    $\begingroup$ 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 $G$. $\endgroup$ – amd Dec 7 '15 at 1:28
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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).

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  • $\begingroup$ 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? $\endgroup$ – aimi Dec 7 '15 at 0:28
  • $\begingroup$ 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. $\endgroup$ – Justpassingby Dec 9 '15 at 7:22
  • $\begingroup$ 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 -.- $\endgroup$ – aimi Dec 9 '15 at 8:56
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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.

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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.

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