0
$\begingroup$

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:

$VV'XVV'=VEV'$

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

$\endgroup$

migrated from stats.stackexchange.com May 22 at 11:37

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

  • $\begingroup$ 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
0
$\begingroup$

Edit:

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'$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.