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.
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.