Does a singular value decomposition always exist for complex matrix? 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.

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.
