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There is $A$ which is a matrix: $$\begin{bmatrix}2 & 4 \\ 1 & -4 \\ -2 & 2\end{bmatrix}.$$ While I have easily worked out the singular value decomposition of this matrix, but I am not sure how to go about trying to present the pseudo-inverse of $A$ (i.e. $A^+$) in SVD form. What I have found out is that: $$ A^+ = V \cdot \Sigma^{-1} \cdot U^\top $$ But trying this out has caused a problem since the matrix $\Sigma$ is not a square matrix so the inverse is not possible. So I am not quite sure if I am following the right route or not ...

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Just work out the pseudo inverse of $S$. You can do this by working out the inverse of the square portion of $S$ and then padding the matrix with zeroes to obtain the appropriate size.

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  • $\begingroup$ Not quite sure what exactly you are inferring, but is it something like [1/6 0; 0 1/3; 0 0]... $\endgroup$ – aspiringProgrammer May 25 '14 at 14:51
  • $\begingroup$ So the pseudo inverse has diagonal elements six and three and you just keep adding columns with zeroes to make the dimensions of the pseudo inverse the same as the dimensions of the transpose of SS$. $\endgroup$ – JPi May 25 '14 at 16:09
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You first transpose $S$ then replace the non-zero entries with their reciprocals to get $S^{-1}$.

Alternatively, you can use the thin SVD where $S$ is diagonal, so $S^{-1}$ just reciprocates the non-zero entries of the diagonal of $S$.

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