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I was studying a Moore-Penrose pseudo inverse matrix and found out below theorem in https://www.kybernetika.cz/content/1979/5/341/paper.pdf.

If $V$ is $n \times n $ symmetrical matrix and if $X$ is an arbitrary $n \times q$ real matrix, then \begin{align} (V+XX^T)^+ = V^+ - V^+X(I+X^TV^+X)^{-1}X^TV^+ + (X_\perp^+)^TX_\perp^+, \\ where \;\; X_\perp = (I-VV^+)X. \end{align}

What if $X$ is an arbitrary $n \times 1$ real vector? Actually, I already calculate those term with R and it turns out when $X$ is a vector, that theorem went wrong. But I am not sure about my result.

Does theorem holds when X is an arbitrary $n\times 1$ vector?

or does not?

If not, how could I expand $(V+XX^T)^+$ when $V :n \times n $ symmetrical matrix and $X : n\times 1$ vector?

I also attach my R studio result imageR studio code

someeed
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  • @user1551 I also thought same as you, but as you can see in my R studio result image, they don't match. – someeed Dec 29 '21 at 12:02
  • Oh, no, the cross terms are missing in (10). Formula (8) is correct, though. – user1551 Dec 29 '21 at 13:48
  • @user1551 what do the cross terms are missing in (10) mean?? – someeed Dec 30 '21 at 00:08
  • Let $P=VV^+$. It is the orthogonal projection onto the range of $V$. We can decompose $XX^T$ into the sum $PXX^TP+PXX^T(I-P)+(I-P)XX^T+(I-P)XX^T(I-P)$ (where $(I-P)XX^T(I-P)$ can also be written as $X_\perp X_\perp^T$). Apparently, the authors have mistaken $XX^T$ as $PXX^TP+(I-P)XX^T(I-P)$. – user1551 Dec 30 '21 at 13:24
  • @user1551 It takes time though, I understand what you mean. But I'm now confused that how to expand correctly $(V+XX^T)^+$. Could you explain about that? – someeed Dec 31 '21 at 00:57
  • This isn't easy. You may take a look at the papers mentioned by J.M. to see if there are any useful formulae. – user1551 Dec 31 '21 at 08:01

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