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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?

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    You forgot to put nonnegative in bold too. There's no such thing as a nonnegative complex number that isn't real. – Matt Samuel Apr 01 '17 at 23:04

1 Answers1

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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. $\square$

Chappers
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    Why do you use the pseudoinverse? Is that just another notation for the conjugate transpose? –  Apr 01 '17 at 23:11
  • $\mathbf{M}^{\dagger}$ is standard notation for the conjugate transpose/adjoint (i.e. $(\mathbf{M}^T)^*$. – Chappers Apr 01 '17 at 23:12
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    Apparently in quantum mechanics $M^\dagger$ is standard notation for the conjugate transpose of $M$, but I think in math $M^*$ is standard notation for the conjugate transpose of $M$. – littleO Apr 01 '17 at 23:26
  • I think in my university's undergraduate mathematics course, the adjoint of a complex matrix is denoted exclusively by $M^{\dagger}$. Sometimes one does take the complex conjugate of a matrix without transposing it! – Chappers Apr 01 '17 at 23:37
  • Yes seems to be some confusion, I have also seen $^*$ used for conjugate transpose. – mathreadler Apr 01 '17 at 23:37
  • I never saw $M^\dagger$. Surely $M^*$ and $M^H$ are the usual notation in maths – reuns Apr 02 '17 at 00:58
  • What if a matrix for which we wat to calculate singular values is not Hermitian, i.e. just a general square matrix? Such a general square matrix can have negative and/or complex eigenvalues. Do we still define singular values as the square root of these eigenvalues in this case? Thank you. – Confounded May 30 '23 at 05:13