Does a singular value decomposition always exist for complex matrix?

Question

I know for real matrix A, SVD always exists, but I am wondering for any complex matrix, will SVD still exist for any scenarios? Thanks.

Answer 1

Yes.

One way to see this is by thinking of SVD as $X = U^\ddagger V$ where:

• $X\in \mathbb{C}^{N\times P}$, where $N > P$, without loss of generality
• $U \in \mathbb{C}^{N\times N}$ and $U^\ddagger U = I$
• $V \in \mathbb{C}^{N\times P}$ and $V^\ddagger V = \Lambda$ , where $\Lambda$ is $P \times P$ diagonal and real matrix

That is, given $x$ to we want to solve for $u$ and $v$, where (in einstein notation) the following equations hold $$ x_{i,j} = u_{i,k}.v_{k,j} \\ u_{i,k}.u{j,k} = \delta_{i,j} \\ v_{i,k}.v_{j,k} = \delta_{i,j} .\lambda_{i}, $$ where $\lambda_{i} \in \mathbb{R}$, and as always, $\delta_{i,j} = 1$ if $i=j$ and $0$ otherwise.

So now we have a system with $2\times(N^2 + NP)$ unknowns, subject to $2NP$ + $N^2$ + $P^2$ constraints (the 2 factor since they are complex values), and $P < N$ as assumed.

So because the number of unknowns is less than the number of linear constraints the problem is solvable.