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I am trying to find out how to get the error associated with with this equation G = MD where I compute the pseudo-inverse of D and left multiply by G to get M.


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The Moore-Penrose pseudo-inverse of $D$ is a matrix $D^+$ such that $D D^+ D = D$. If the "error" is $E = M - G D^+ $, this says $E D = 0$, i.e. $E$ annihilates the column space of $D$. On the other hand, $E v = M v$ for $v \in \text{Ker} D^*$, i.e. vectors orthogonal to the column space of $M$.

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Wouldn't E = M − GD+ always give you an E of all zeros because M = GD+? – user14041 Aug 1 '11 at 0:06
Sorry, I somehow got things turned around. You want to take $M = G D^+$ as your "approximate solution" to the equation $G = M D$. The error is then $E = G - M D = G - G D^+ D$. Now $D^+ D$ is the orthogonal projection on $\text{Ran}D^* = (\text{Ker} D)^\perp$. So $E v = 0$ for vectors $v$ orthogonal to $\text{Ker}(D)$, while $E v = G v$ for vectors $v \in \text{Ker} D$. – Robert Israel Aug 1 '11 at 19:30

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