Kernel of a linear combination of matrices Suppose I have $k$ square Hermitian real-valued matrices $B_i$, that are all singular, but that are such that the intersection of their kernels is trivial. I.e., there does not exist a vector $x \ne 0$, such that $B_i x = 0$ for all $i$. Does there necessarily exist a set of scalar weights $a_i$ so that $B = \sum_i a_i B_i$ is non-singular? 
 A: It seems that I found an answer to this question. The claim is false in general: there exist real symmetric matrices, whose kernels have a trivial intersection, but for which no linear combination is non-singular. One example of such a matrix pair is given at the bottom of the following page:
http://www.netlib.org/utk/people/JackDongarra/etemplates/node297.html
$$ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix};
   B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & -1 & 1 \\ 1 & 1 & 0 \end{bmatrix} $$
Both matrices are rank two, and the nullspace of $A$ is generated by $(0,0,1)$ while the nullspace of $B$ by $(1,-1,-1)$.  Any nontrivial linear combination of $A,B$ is again rank two.
I'm wondering now if there is a simple necessary condition for this situation to occur. The example given on the page above has the following property: $Az_b = c Bz_a$, where $z_a$ (resp. $z_b$) is in the kernel of $A$ (resp. $B$), and $c$ is some scalar. Is there a generalization of such a condition to the case of multiple matrices?
