# Eigendecomposition proof

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?

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

$$XV=VE \hspace{1cm}$$ (by the eigendecomposition)
$$\implies V'XV=V'VE$$
$$\implies V'XV=E \hspace{1cm}$$ (as $$V'V=I$$)
$$\implies VV'XVV' = VEV'$$