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Following a problem, it was not explained why the $\Sigma$ matrix is the square root of $V$'s Eigen values rather than the values themselves.

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  • $\begingroup$ What were you taught about SVD? Weren't you taught that the singular values of $C$ would be the square roots of the eigenvalues of $C^TC$? $\endgroup$ – Minus One-Twelfth May 22 at 2:06
  • $\begingroup$ It is not clear what the sentence "$\Sigma$ is the square root of $V$'s eigenvector". It certainly doesn't sound like a correct sentence, whatever it's supposed to mean. The diagonal entries of $\Sigma$ are the roots of the eigenvalues of $C^TC$, and the columns of $V$ are the corresponding eigenvectors. $\endgroup$ – Omnomnomnom May 22 at 2:07
  • $\begingroup$ I see, sorry I am learning this so it is new to me. I understand now that this is just the definition of the $\Sigma$ matrix. Thanks. $\endgroup$ – Jacob B May 22 at 2:08
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Theorem Let $A$ be a matrix with SVD $A= U\Sigma V^*$. The nonzero singular values of $A$ are the square roots of the nonzero eigenvalues of $AA^*$, $A^*A$. If $A=A^*$, then the singular values are the absolute values of the eigenvalues of $A$.

Pf. Observe that $$ AA^*= (U \Sigma V^*)(V \Sigma^* U^*)= U(\Sigma \Sigma^*) U^* $$ is an eigenvalue decomposition of $AA^*$ so that $\Sigma\Sigma^*$ is similar to $AA^*$. Therefore, $AA^*$ and $\Sigma\Sigma^*$ have the same eigenvalues, namely $\sigma_1^2,\ldots,\sigma_r^2$ (recalling that the $\sigma_i$ are real), with $n-r$ additional zero eigenvalues if $n>r$. The case of $A^*A$ follows mutatis mutandis.

Notice the square root becomes necessary because, although $\Sigma$ contains the singular values, what you are ultimately is working with is $\Sigma \Sigma^*$. If you are working over the reals, then this is $\Sigma\Sigma^T$ so all the diagonal entries are of the form $\sigma_i^2$, where $\sigma_i$ is a singular value.

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    $\begingroup$ This is a great explanation as to WHY thank you so much for taking the time to explain this. $\endgroup$ – Jacob B May 22 at 2:29

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