# Why are singular values of “complex” matrices always real and non-negative?

I've already read the following related questions on math.SE:

The conclusion they seem to agree on is the following:

For $$A \in \mathbb{R}^{m\times n}$$, its singular values are real non-negative.

I do not see, however, how can this be the case for $$A \in \mathbb{C}^{m\times n}$$, even though in most answers, people say it is the same as for real matrices.

## Usual argument for singular values of $$A \in \mathbb{R}^{m\times n}$$ being real non-negative:

The SVD of $$A$$ is: $$A = U S V^T$$ We have $$B = A^T A = V S^T U^T U S V^T = V S^T S V^T$$ where $$S^T S$$ is diagonal with elements $$\sigma_i^2$$, where $$\sigma_i$$ are the singular values of $$A$$. Now, $$\sigma_i^2$$ are real-nonegative because they can be seen as the eigenvalues $$\lambda_i=\sigma_i^2$$ of the symmetric, positive-definite, matrix $$B = V \Lambda V^T$$ (i.e., $$\Lambda = S^T S$$).

Using $$\lambda_i=\sigma_i^2 \ge 0$$ to solve for $$\sigma_i$$, we find that:

• $$\sigma_i$$ is real because if it were complex, then the only way $$\sigma_i^2$$ would be real is if $$\sigma_i$$ is real or pure imaginary. However, in the latter case we get $$\sigma_i^2$$ negative, which contradicts $$\sigma_i^2 \ge 0$$.
• $$\sigma_i$$ is the square-root of $$\lambda_i$$, which can be positive or negative. By convention, however, we take the positive square-root.

Hence, $$\sigma_i \text{ are real-nonnegative themselves.}$$

## A try for a similar argument for singular values of $$A \in \mathbb{C}^{m\times n}$$ :

The SVD of $$A$$ is: $$A = U S V^H$$ We have $$B = A^H A = V S^H U^H U S V^H = V S^H S V^H$$ where $$S^H S$$ is diagonal with elements $$|\sigma_i|^2$$, where $$\sigma_i$$ are the singular values of $$A$$. Now, $$|\sigma_i|^2$$ are real-nonnegative because of the modulus square, and they can also be seen as the eigenvalues $$\lambda_i=|\sigma_i|^2$$ of the Hermitian, positive-definite, matrix $$B = V \Lambda V^H$$ (i.e., $$\Lambda = S^H S$$).

Using $$\lambda_i=|\sigma_i|^2 \ge 0$$, how can one solve for $$\sigma_i$$ and prove it is real-nonnegative?

Particularly, what prevents $$\sigma_i$$ from being complex?

## 1 Answer

The result is that a complex matrix $$A$$ has a factorization of form $$A=USV^H$$ where $$S$$ is nonnegative real diagonal and $$U$$ and $$V$$ are "unitary" in the sense that they might be rectangular but $$U^HU$$ and $$V^HV$$ are identity matrices. To produce it one takes the positive square roots of the eigenvalues of $$AA^H$$ and $$A^HA$$.

No one asserts that all factorizations of $$A$$ as a product of form $$USV^H$$ with "unitary" $$U$$ and $$V$$ and diagonal $$S$$ must have all entries of $$S$$ nonnegative real.

Take, for instance, factorizations of form $$A=UDD^{-1}SV^H$$ where $$D$$ is diagonal with complex numbers of unit magnitude on the diagonal. If $$U$$ is "unitary" so is $$UD$$, if $$S$$ is diagonal, so is $$D^{-1}S$$. So if there is a standard SVD decomposition $$A=USV^H$$ there are non-standard ones too, that have non-positive diagonal middle factors.

More concretely: suppose $$A=USV^H$$ is an SVD in the standard sense. Then $$A=(-U)(-S)V^H$$ is also true, and $$-U$$ is "unitary" since $$U$$ is. But $$-S$$ is not positive on its diagonal. The factorization $$A=(-U)(-S)V^H$$ holds, but is not a standard SVD factorization.

So the answer boils down to this: the middle factor in an SVD is defined to be nonnegative real diagonal.

• So, nothing prevents $\sigma_i$ from being complex. It's just that we've chosen among the possible choices for $\sigma_i$ one particular choice and called the relative factorization SVD. – Likely May 21 at 2:20
• Exactly. The wide range of applications of the SVD constitutes a kind of justification for the definition we've chosen. – kimchi lover May 21 at 2:58