TITLE: A question on eigenvalues
QUESTION [8 upvotes]: 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 $0$?

REPLY [11 votes]: Unless I'm mistaken, the answer to your question is negative.
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.