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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.

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Your $\in$ should be $\subset$, and adding some sort of motivation (e.g. why consider $n=3$ for instance) would probably encourage people to solve your question. – Thierry Zell Nov 7 '10 at 17:15
What is your supervisor's reaction or advice to this question? – Yemon Choi Nov 8 '10 at 0:26
@Yemon Choi :Beautiful, but not so important. – gondolf Nov 8 '10 at 3:03
up vote 5 down vote accepted

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.

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@Kate Juschenko:There is a problem of your proof: if there exists a unitary $U$ such that $UE_1=E_{11}, UE_2=E_{21}$, then $$0=tr(E_{11}^*E_{21})=tr(E_1^*U*UE_2)=tr(E_1^*E_2)$$, which leads to $E_1,E_2$ are orthogonal. – gondolf Nov 8 '10 at 1:58
sure, the second should be corrected to: $E_{11}+\lambda E_{21}$, where $\lambda\in \mathbb{C}$. – Kate Juschenko Nov 8 '10 at 5:28
Yes, you are correct! Thanks! – gondolf Nov 8 '10 at 6:06

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