# In SVD why is $\Sigma$ the square root of $V$'s Eigen values?

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.

• 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 $C^TC$? – Minus One-Twelfth May 22 at 2:06
• 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. – Omnomnomnom May 22 at 2:07
• 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. – Jacob B May 22 at 2:08

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.