Let $X$ be an $(n \times n)$ matrix. Let $V$ be the $(n \times n-k)$ be the matrix of eigenvectors of $X$ which correspond to non-zero eigenvalues of $X$. Let $E$ be the $(n-k \times n-k)$ diagonal matrix of eigenvalues of $X$. I wish to prove that:


I can prove it when there are no zero-valued eigenvalues but not the general case. Would anyone be able to help?


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  • $\begingroup$ Hint: can you compute $V'XV$ ? $\endgroup$ – user617446 May 22 at 13:28
  • $\begingroup$ Ah yep thank you! Got it! $\endgroup$ – JDoe2 May 22 at 14:19


Ah worked out the answer!

$XV=VE \hspace{1cm}$ (by the eigendecomposition)

$ \implies V'XV=V'VE$

$ \implies V'XV=E \hspace{1cm}$ (as $V'V=I$)

$ \implies VV'XVV' = VEV'$


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