SVD decomposition of a square matrix of complex numbers

Question

Le $M$ be any matrix in $C^{n \times n}$. Consider the matrix $MM^*$. This matrix is Hermitian ($(MM^*)^* = MM^*$), and positive semi-definite ($\forall v^*, v^*MM^* v = (v^* M) (M^* v) = (v^*M) (v^*M)^* = |v^*M)|^2 \geq 0$). Thus, its eigenvalues have to be real and non-negative numbers (for an Hermitian $H$, $\lambda v^* v = v^* H v = v^* H^* v = \overline{\lambda} v^* v$). Thus, the Spectral Theorem can be applied to $MM^*$. Therefore, $MM^* = {UDU}^*$, where U is an unitary matrix, and $D$ is a diagonal matrix composed of the eigenvalues of $MM^*$. Let $\Sigma$ be the diagonal matrix composed of the square roots of the eigenvalues of $MM^*$ (i.e., the singular values of $M$). Then, ${D} = \Sigma \Sigma = \Sigma \Sigma^*$, since the Hermitian transpose does not have any effect on a diagonal matrix composed of real numbers. Thus, \begin{eqnarray*} MM^* & = & {UDU}^* = U \Sigma \Sigma^* U^*= (U \Sigma I) (I^* \Sigma^* U^*) = (U \Sigma I) (U \Sigma I)^*. \end{eqnarray*} This implies that $M = U \Sigma I$. As well as $M = U \Sigma V$, for any unitary matrix. Which is impossible! What did I miss?

Answer 1

From $$ \begin{bmatrix}0&1\\1&0\end{bmatrix} \begin{bmatrix}0&1\\1&0\end{bmatrix}^*=I\,I^* $$ you cannot conclude that $$ \begin{bmatrix}0&1\\1&0\end{bmatrix}=I. $$