Why can't singular values be complex numbers?

Question

Reading both this paper (p. 4) and the Wikipedia article regarding singular value decomposition, they state that the diagonal matrix ${\boldsymbol {\Sigma }}$ in $$\mathbf {M} =\mathbf {U} {\boldsymbol {\Sigma }}\mathbf {V} ^{*}$$ has non-negative real entries in a context where the general field is assumed to be $\mathbb{C}$. My question is: why can't singular values be complex?

Answer 1

The singular values in $\mathbf{\Sigma}$ are by definition the nonnegative square roots eigenvalues of the matrix $\mathbf{M}^{\dagger}\mathbf{M}$. This matrix is Hermitian and nonnegative-definite, and hence has nonnegative real eigenvalues:

Let $A$ be Hermitian and nonnegative-definite. Then the eigenvalues are real and nonnegative.

Proof: Let $x$ be an eigenvector of $A$ with eigenvalue $\lambda$. Then $$ x^{\dagger} A x = \lambda x^{\dagger} x = \lambda \lVert x \rVert^2. $$ Taking the conjugate, $$ (x^{\dagger} A x)^{*} = x^{\dagger} A^{\dagger} x = x^{\dagger}Ax, $$ and hence $x^{\dagger} A x - (x^{\dagger}Ax)^*$ vanishes, which can only occur if this number is real. Since $\lVert x \rVert^2$ is real, $\lambda$ is the quotient of two real numbers and so is real.

The definition of nonnegative-definite is that $x^{\dagger} A x \geq 0$ for all $x$, so $\lambda \lVert x \rVert^2 \geq 0$, and $\lVert x \rVert^2$ is nonnegative, so $\lambda$ must also be nonnegative. $\square$