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In singular value decomposition (SVD), $X=USV^T$, if $X$ is $N\times D$, then $U$ is $N\times N$, $S$ is $N\times D$, and $V$ is $D\times D$.

Let's assume $N<D$.

Every column vector in $V$ is an eigenvector of $X^TX$, so we have $D$ eigenvectors in total.

Each (squared) diagonal element of $S$ is an eigenvalue of $X^TX$, so we have $\min(N, D)=N$ eigenvalues.

Here comes my question: why do we have unequal numbers of eigenvalues and eigenvectors? Shouldn't each eigenvector be associated with one eigenvalue?

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$X^TX$ is a $D\times D$ matrix, hence it has $D$ eigenvalues (taken with multiplicity). Same goes for $S^TS$.

Each element on the diagonal of $S^TS$ is an eigenvalue of $X^TX$; moreover, the corresponding eigenvectors are columns of $V$; there are $D$ such columns, so everything fits.

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    $\begingroup$ I see. So I can't compute $X^TX$'s eigenvalues as diag(S)^2. Instead, I have to do the matrix multiplication first and then take the diagonal elements: diag(S'*S). $\endgroup$ – Sibbs Gambling Apr 23 '15 at 11:32
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    $\begingroup$ @SibbsGambling exactly. $\endgroup$ – TZakrevskiy Apr 23 '15 at 11:33
  • $\begingroup$ Oh, and actually they are all 0. :-) $\endgroup$ – Sibbs Gambling Apr 23 '15 at 11:37
  • $\begingroup$ @SibbsGambling that's strange, because if all eigenvalues of a symmetric matrix are zero, then the matrix itself is zero. And if $X^TX=0$, then $X=0$. $\endgroup$ – TZakrevskiy Apr 23 '15 at 11:41
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    $\begingroup$ @SibbsGambling no problem. The last $D-N$ eigenvalues have to be zeros, because $D> N\ge rank(X) = rank(X^TX)=rank(S^TS)$. $\endgroup$ – TZakrevskiy Apr 23 '15 at 12:47

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