I have this expression embedded in a larger equation:

$$ A^T (A D A^T)^+ A $$

Where $A \in \mathbf{R}^{m \times n}$ is an arbitrary matrix, $D \in \mathbf{R}^{n \times n}$ is diagonal (with some elements possibly $0$), and $X^+$ is the Moore-Penrose Pseudoinverse of $X$.

Can the expression be simplified? In particular, I'd like to "distribute" the pseudoinverse in terms of the pseudoinverse of $A$ and $D$, if that's possible.

If $D = I$, for instance, it's easy to show that $ A^T (A D A^T)^+ A = V V^T$, where $A = U \Sigma V^T$, $V^T \in \mathbf{R}^{r \times n}$ is the "thin" SVD of $A$ with rank $r$.


Well obviously if you replace $B = A \sqrt{D}$ you get

$A^T (ADA^T)^+ A^T = D^{-1/2} W W^T D^{-1/2}$ with $W$ from the thin svd from of $B$ (in place of the $V$ in your question).

That is the only thing I would see. Of course that does not strictly work for you, since you have possibly zero elements in $D$.

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  • $\begingroup$ Ah, I hadn't thought about turning the problem inside out like that! So simple in hindsight. If there are zero terms in $D$, is there a way to show that the form above still holds or not? Since they're going to be zeroed out in $B$ anyway when you multiply, I'm wondering if it doesn't matter. $\endgroup$ – Jay Lemmon Jan 25 '14 at 5:53
  • $\begingroup$ Good question, I dont know. I would say it depends what the limit of the $\infty \cdot 0$ expression inside of $D^{-1/2} W$ would be. If it is always zero, then you are right. If it is some number, than we are out of luck. But before starting an mathematical investigation, you could just try it out numerically. $\endgroup$ – Andreas H. Jan 25 '14 at 17:54
  • $\begingroup$ Matlab experiments seem to indicate you can't just do something like sqrt(pinv(D)) when D has a 0 element :( Still, thanks for the help. $\endgroup$ – Jay Lemmon Jan 26 '14 at 22:29
  • $\begingroup$ too bad, thanks for the notice $\endgroup$ – Andreas H. Jan 26 '14 at 23:29

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