finding $K$ with $KQK^∗=0$ and $KK^∗=I$ For a given nonsingular complex matrix $Q_{m\times m}$, how can we find a complex $K_{n\times m}$, $n<m$, with
\begin{align*}
KQK^∗&=0\\
KK^∗&=I?
\end{align*}
 A: Not a full answer, but a starting point:
Clearly we can assume that $Q$ is Hermitian. Take any eigenvectors of $Q$ with eigenvalue zero and add them as rows of $K$.
Now suppose $Q$ has a positive eigenvalue $\lambda_1$ with eigenvector $v_{1}$ and a negative eigenvalue $\lambda_{-1}$ with eigenvector $v_{-1}$. Take
$$w_1 = \frac{\sqrt{-\lambda_{-1}}v_1 + \sqrt{\lambda_1} v_{-1}}{\sqrt{\lambda_1-\lambda_{-1}}}.$$
We have that $\|w_1\|=1$ and $w_1^*Qw_1 = 0$, so can add $w_1$ as a row of $K$. We can repeat this process for other pairs of positive and negative eigenvalues $\lambda_2, \lambda_{-2}$, etc. of $Q$.
So if $Q$ has $x$ zero eigenvalues and at least $n-x$ positive and negative eigenvalues, the above construction can be used to find $K$. What's not clear to me is if this condition is necessary; perhaps another answer will fill this in.
Finally if $m >> n$ you won't want to diagonalize $Q$ in practice; instead you'll want to pluck out $n$ positive and negative eigenvalue/eigenvector pairs using e.g. power iteration and spectral shifting.
