# A question on eigenvalues

Let $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$, $A_{5}$ be linearly independent Hermitian matrices in the the space of $6$ by $6$ Hermitian matrices as a vector space over $\mathbb{R}$. Does there always exist a non-zero matrix $A_{0}\in\mbox{span}_{\mathbb{R}}\left(A_{1},A_{2},A_{3},A_{4},A_{5}\right)$ such that at least four of the eigenvalues of $A_{0}$ are $0$?

• umm. Why do you mention Lie groups in the tags? Nov 22 '13 at 2:57
• BTW, the definition of uniform measure on the Grassmannian can be found here, en.wikipedia.org/wiki/Grassmannian#Associated_measure.
– Ayna
Nov 22 '13 at 17:23
• I think you should retain the original formulation of the problem and add the new one as an update---otherwise, things look confusing to read---to give you and idea of what I mean, I'm editing the question to reflect what I mean. If you disagree with the edit, feel free to rollback Nov 22 '13 at 17:23

In the paper "On matrices whose real linear combinations are nonsingular", by J. F. Adams, P. D. Lax, R. S. Phillips, (Proc. AMS, 16 (1965), the authors show that $2b+1$ is the largest number of Hermitian matrices that one could pick so as to ensure that all of their nontrivial linear combinations are nonsingular.
Here, $b$ is implicitly defined as follows. For $n\times n$ matrices, where $n=(2a+1)2^b$, $b=c+4d$, where $a,b,c,d$ are integers with $0\le c < 4$. In our case $n=4$, so we use $a=0$, $b=2$, which completes our conclusion.
• I wonder what happens if instead of real linear combinations one considers complex linear combinations (or real combinations but only for symmetric matrices). Then, of course, there will be always singular nontrivial linear combinations and the question is whether there is always a nontrivial linear combination with nullity $\ge 2$. Nov 22 '13 at 6:11
• @Ayna - In that case, you're asking whether there is always a matrix with rank $\leq 2$ in their span. Once again, the answer is no. Proposition 4 in arXiv:1109.5478 gives a way of constructing 16 different Hermitian 6x6 matrices with the property that any linear combination of them has rank $\geq 3$. Do you have a more general question that you would like to ask so that we don't keep considering this problem case-by-case? Dec 15 '13 at 15:54
• @Misha - If we consider complex linear combinations, then the problem is much simpler than the real linear combination case. It was shown in arXiv:0706.0705 that the largest subspace of $d\times d$ complex matrices such that every matrix in their complex span has rank $\geq k$ has dimension $(d-k+1)^2$ (and the same bound is attained if we restrict to Hermitian matrices). Dec 15 '13 at 16:13