TITLE: Sums of a set of symmetric matrices
QUESTION [5 upvotes]: Say we have a set of symmetric $n \times n$ matrices $M_i$ for $1 \leq i \leq k$, elements in $\mathbb{R}$.  Suppose that for every $\boldsymbol{\lambda} = (\lambda_1, \dots , \lambda_k) \in \mathbb{R}^k$ we have that the kernel of 
\begin{equation*}
M_{\boldsymbol{\lambda}} = \sum_i \lambda_i M_i 
\end{equation*}
is nontrivial.  Does it follow that there exists some nonzero $n$ vector $\textbf{v}$ with $M_i \textbf{v} = 0$ for all $i$?

REPLY [4 votes]: Still no.  Counterexample:
$M = \lambda M_1 + \mu M_2 = \pmatrix{0 & \mu & \mu \cr \mu & \lambda & 0 \cr \mu & 0 & -\lambda}$.
Obviously $\det(M)\equiv 0$ for all $\lambda$ and $\mu$.  However, the only $(x,y,z)$ that satisfies
$M\pmatrix{x\cr y\cr z} = \pmatrix{0 & \mu & \mu \cr \mu & \lambda & 0 \cr \mu & 0 & -\lambda}\pmatrix{x\cr y\cr z} = \pmatrix{\mu(y+z)\cr \mu x + \lambda y\cr \mu x - \lambda z} = \mathbf{0}\quad \forall \lambda, \mu$
is the zero vector.