# Is there such a thing [closed]

For any $U_{i}\in\mathcal{U}\left(4\right)$, $1\le i\le5$, are there $W\in\mathcal{U}\left(4\right)$ and nontrivial $\left(x_{1},x_{2}\right)\in\mathbb{R}^{2}$, such that $\mbox{tr}\left(U_{i}\mbox{diag}\left(1,0,0,0\right)U_{i}^{*}W\mbox{diag}\left(x_{1},x_{2},0,0\right)W^{*}\right)=0,$ $1\le i\le5$, with $\left|u_{k,1}\right|^{2}x_{1}+\left|u_{k,2}\right|^{2}x_{2}=0$ for $1\le k\le4$?

• What are $v_{k,j}$? Could you provide some context and motivation for the question? – Qfwfq Dec 22 '13 at 22:39
• How do you feel about rephrasing your question following Nathaniel Johnston's suggestion? – S. Carnahan Dec 23 '13 at 11:02
• How about a more descriptive title, too? – Nate Eldredge Dec 24 '13 at 2:13
• Welcome back. Have you nothing to say about the answer posted two weeks ago by Nathaniel Johnston? – Gerry Myerson Jan 6 '14 at 2:06

I don't know the answer, but I can at least provide some context that may convince others that this question is actually something that is studied and is indeed hard.

First, let's rewrite the question so it's easier to digest. We'll first define $Z := V{\rm diag}(x_1,x_2,0,0)V^*$ for convenience. Notice that the condition $x_1|v_{k,1}|^2 + x_2|v_{k,2}|^2 = 0$ for $1 \leq k \leq 4$ is equivalent to saying that $Z$ has zeroes on its diagonal.

Also, $U_i{\rm diag}(1,0,0,0)U_i^* = u_i u_i^*$, where $u_i$ is the first column of $U_i$, so the condition ${\rm tr}(U_i{\rm diag}(1,0,0,0)U_i^*V{\rm diag}(x_1,x_2,0,0)V^*) = 0$ for $1 \leq i \leq 7$ is equivalent to ${\rm tr}(Zu_iu_i^*) = 0$ for $1 \leq i \leq 7$.

So a rephrasal of the question is: given unit vectors $\{u_i\}_{i=1}^7$, does there exist a $4\times 4$ Hermitian matrix $Z$ with rank $2$ such that ${\rm tr}(Zu_iu_i^*) = 0$ for $1 \leq i \leq 7$ and ${\rm tr}(Ze_ie_i^*) = 0$ for $1 \leq i \leq 4$, where $\{e_i\}_{i=1}^4$ is the standard basis of $\mathbb{R}^4$?

In other words, we are given eleven Hermitian $4 \times 4$ matrices ($\{u_iu_i^*\}_{i=1}^7$ and $\{e_ie_i^*\}_{i=1}^4$), and we are asking whether or not there is always a rank-$2$ Hermitian matrix orthogonal to all of them. This is the central question studied in arXiv:1109.5478 -- it is a question about how many quantum measurements are necessary to reconstruct a pure quantum state.

Anyway, they show that the smallest set of $4 \times 4$ Hermitian matrices with the property that every orthogonal matrix has rank $\geq 3$ has $9$ elements. They also show that almost every set of $13$ or more $4 \times 4$ Hermitian matrices have the property that every orthogonal matrix has rank $\geq 3$. You have given a set that lies somewhere in the middle (it consists of $11$ matrices, but those matrices are promised to be rank $1$ and four of them are mutually orthogonal), so it's not immediately clear to me what the answer is.

• Given your reformulation, the existence of this rank-2 Hermitian is implied by the 4M-4 conjecture (specifically, taking M=4). The conjecture is already known to be true for M=2 and 3. I offer US\$100 for a proof of the conjecture on my blog: dustingmixon.wordpress.com/2013/03/19/… – Dustin G. Mixon Dec 28 '13 at 15:10