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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 $i$?

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No. As a counterexample, consider the matrices $M_1 = \pmatrix{1 & 0 \cr 0 & 0}$, $M_2 = \pmatrix{0 & 1 \cr 0 & 0}$. Any linear combination has a row of zeroes and therefore is not invertible, but the kernels of $M_1,M_2$ are disjoint (apart from 0).

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Oh sorry, right you are - I forgot the condition that the matrices must be symmetric so can be diagonalised. Does this make a difference? – I Like Trams Aug 31 '10 at 11:52

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.

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