Question

A matrix subspace $S\subset M_n(C)$ is called "good", if there is two linear independent elements of $S$, says $E_1,E_2$ which are simultaneously singular valued decomposable, i.e., $E_1=UD_1V$ and $E_2=UD_2V$ with $D_1$, $D_2$ diagonal and $U,V$ unitary.

Now the question becomes: if all three-dimensional subspace $S\subset M_3(C)$ is good?

This problem is try to undertand how hard could simultaneously singular valued decomposation be, and how powerful could linear combination of matrix be. $n=3$ is the simplest case.

Answer 1

wlog one can assume that $||E_1||=1$.

Let $X$ be the span of the matrix units $E_{11},E_{21},E_{31}$. Then for every $2$ linearly independent operators $E_1, E_2$ in this space there exists a unitary matrix $U$ such that $UE_1=E_{11}$, $UE_2=\lambda E_{21}+E_{11}$. But $E_{11}$ and $\lambda E_{21}+E_{11}$ are not simultaneously singular valued decomposable.