Let $M_i$, $i=1,\dots,(n-1)^2+1$, $M_1 = 1_{n\times n}$ be a set of linearly-independent Hermitian $n\times n$ matrices. Show that there exists a rank-1 matrix $P$, which is a linear combination of $M_i$s: $$ P = \sum_{i=1}^{(n-1)^2+1} a_i M_i $$ Motivaion: The positive answer to this question will resolve an earlier stated problem Traceless Hermitian matrices with simultaneously vanishing Rayleigh quotients.
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2$\begingroup$ For what it's worth, the answer is known to be "yes" if you remove the "Hermitian" requirement (i.e., any $((n-1)^2+1)$-dimensional subspace of $n \times n$ complex matrices contains a rank-1 matrix). Unfortunately, that doesn't solve this problem since it relies crucially on the fact that it's a complex vector space (whereas the $n \times n$ Hermitian matrices is a real vector space, which makes things trickier). I don't know of a way around this obstacle yet. $\endgroup$– Nathaniel JohnstonCommented Jul 24 at 18:18
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1$\begingroup$ Thanks! Could you provide me a reference to the complex case? $\endgroup$– Michał JanCommented Jul 24 at 18:20
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2$\begingroup$ Sure! This is my go-to reference for this (there may be an earlier/more canonical reference that is not in quantum information theory language; not sure): arXiv:0706.0705. Theorem 11 with r=2 is the result I described. $\endgroup$– Nathaniel JohnstonCommented Jul 24 at 18:21
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2$\begingroup$ @Nathaniel Johnston: I wrote it at mathoverflow.net/questions/474920/…. $\endgroup$– Narutaka OZAWACommented Jul 25 at 4:03
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2$\begingroup$ Thank you @NarutakaOZAWA, this answers my question here, counterexample being $P = \left( \begin{array}{cccc} a_{1,1} & 0 & a_{1,3} & a_{1,4} \\ 0 & a_{1,1} & a_{2,3} & a_{2,4} \\ a_{3,1} & a_{3,2} & a_{3,3} & 0 \\ a_{4,1} & a_{4,2} & 0 & a_{3,3} \\ \end{array} \right)$. $\operatorname{rank}P = 1$ does not have solutions in Hermitian matrices. $\endgroup$– Michał JanCommented Jul 25 at 8:20
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