Find a similiar matrix with constraints I need to find a $4 \times 4$ unitary matrix $U$ such that for any $M=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$, where $a, b, c, d$ are arbitrary complex numbers,
$$
U(M \otimes M) U^{\dagger}=\left(\begin{array}{cccc}
* & 0 & 0 & 0 \\
0 & * & * & * \\
0 & * & * & * \\
0 & * & * & *
\end{array}\right)
$$
If $M$ were diagonalizable, I could find a $2\times 2$ unitary $U_2$ such that $U_2MU_2^\dagger=D$, where $D$ is a diagonal matrix. Then, taking $U=U_2\otimes U_2$ would have solved the problem. But in general $M$ is not diagonalizable. Hence I'm still lacking a general solution.
 A: Solution Outline: A matrix $U$ satisfies this condition if and only if its first column (which I will call $u$) is such that both $\operatorname{span}(u)$ and $u^\perp$ are $M \otimes M$ invariant subspaces.
Now, note that if $u$ is of the form $x \otimes y - y \otimes x$ for vectors $x,y \in \Bbb C^2$, then $(M\otimes M)u$ is another vector of this form. Thus, $\mathcal U_1 = \operatorname{span}(\{x \otimes y : x,y \in \Bbb C^2\})$ is an $M \otimes M$ invariant subspace. Show that $\mathcal U_1$ is $1$-dimensional. Similarly, show that its orthogonal complement $\mathcal U_2 = \mathcal U_1^\perp$ is an $M \otimes M$ invariant subspace. In order to show that this is the case, it may help to first show that
$$
\mathcal U_2 = \operatorname{span}(\{x \otimes x: x \in \Bbb C^2\}).
$$
You might find this all more intuitive if you think of vectors of the form $x \otimes y$ as $2 \times 2$ matrices. Note that there is a (unique) linear isomorphism $\phi:\Bbb C^4 \to \Bbb C^{2 \times 2}$ such that $\phi(x \otimes y) = xy^T$. What kind of matrix is $\phi(u)$ for $u \in \mathcal U_1$? What about for $u \in \mathcal U_2$?
Further Hint: The first column of $U$ is $u = \frac 1{\sqrt{2}}(0,1,-1,0)$. It makes no difference what the remaining columns are, so long as $U$ has orthonormal columns.
