Complex SO(n) and Degenerations Let's say you have a matrix $A\in\Bbb C^{n\times n}$ with $A^T A = 0$, like for example this one:
$$
A = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}
$$
I ask the following:

Question: Is there a sequence of matrices $(S_k)_{k\in\Bbb N}$ with $S_k^T\cdot S_k = I$ and $\lim\limits_{k\to\infty} S_k A = 0$ ?

For the matrix above, you can pick $\theta_k:=k\cdot(1-i)$ and
$$
S_k = \begin{pmatrix} \cos(\theta_k) & -\sin(\theta_k) \\ \sin(\theta_k) & \cos(\theta_k) \end{pmatrix}.
$$
This is not entirely obvious, but a straightforward computation. I say that because a computer algebra system did it.
PS: I have somewhat involved reasons to ask this question, and laying them out would basically just yield many paragraphs full of buzzwords, and I do not think it would make the question any more clear. I do believe that the answer is affirmative, though.
Edit: I decided to add at least a few buzzwords. Let $O_n=\{ S\in\mathbb C^{n\times n} \mid S^T S = I \}$ be the orthogonal group. It is a reductive group acting on the space $W=\Bbb C^{n\times n}$ by left multiplication. I am asking whether the Nullcone of this action is equal to $N=\{ A \mid A^T A = 0 \}$. Note that $N$ is closed and certainly contains the Nullcone, because if a sequence $(S_k)$ with the above property exists, then $A^TA=A^TS_k^TSA=(S_kA)^T(S_kA)\to 0$ as $k\to\infty$, which means $A^TA=0$.

The answer is affirmative.
I should have done some literature research. In fact, this is the First Fundamental Theorem for $\operatorname{O}_n$ and $\operatorname{SO}_n$, as can be found in the book Classical Invariant Theory - A Primer by Kraft & Procesi, page 117. 
The proof is quite clever and not too hard, but let me just describe how you get my version from the statement of the theorem in the book:
Take $p=n$, so we are looking at the invariants on $(\Bbb C^n)^n = \Bbb C^{n\times n}$. You can check that the action they describe is left-multiplication by matrices. According to the theorem, the invariants for this action are the maps 
\begin{align*}
f_{ij}:\Bbb C^{n\times n} &\longrightarrow \Bbb C
\\ A=(A_{k,\ell}) & \longmapsto \sum_{k=1}^n A_{ki} A_{kj}
\end{align*}
i.e. you take the inner product of the $i$-th and $j$-th column of $A$. Now, $A$ is in the Nullcone if and only if $f_{ij}(A)=0$ for all $1\le i\le j\le n$ and this precisely means $A^TA=0$.
