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Let $k$ be an infinite field. Assume that $n,m$ are two positive integers such that $n>m$. Consider symmetric matrices $A_1,\dots,A_m$ of size $n\times n$. Suppose for each $i=1,\dots,m$, every column of the matrix $A_i$ has a non-zero entry which is not on the main diagonal. Is the following implication correct?

If, for every vector $v \in k^n$ the vectors $A_1v,\dots,A_mv$ are linearly dependent, then the matrices $A_1,\dots,A_m$ are linearly dependent.

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    $\begingroup$ In the literature a tuple of matrices $(A_1,\ldots,A_m)$ is called "locally linearly dependent" if the tuple of vectors $(A_1v,\ldots,A_mv)$ is linearly dependent for every vector $v$. The case $m=2$ follows from Theorem 2.3 in Bresar and Semrl's article: On locally linearly dependent operators and derivations. The case $(A_1,\ldots,A_m)=(1,T,\ldots,T^{m-1})$ for some matrix $T$ is due to Aupetit, see Theorem 7.3 in Prasolov's book: Problems and Theorems in Linear Algebra. $\endgroup$ – Philipp Lampe Feb 14 '18 at 15:02
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The implication does not hold.

Counterexample: Let $(n,m)=(4,3)$. We consider the subspace $$M=\left\lbrace\left(\begin{matrix} a&b&b&b\\ b&c&c&c\\ b&c&c&c\\ b&c&c&c\end{matrix}\right)\colon a,b,c\in k\right\rbrace\subseteq\operatorname{Mat_{4\times 4}(k)}.$$ Note that every matrix in $M$ is symmetric by construction. We claim that there are linearly independent elements $A_1,A_2,A_3\in M$ such that every column of every matrix contains a non-diagonal element which is non-zero. This is equivalent to a choice of $3$ linearly independent vectors in $k^3$ of the form $(a,b,c)$ with $b\neq 0$, which is always possible.

Let $v\in k^4$. For every $A\in M$ the vector $Av\in k^4$ lies in the $2$-dimensional subspace $\{(x,y,y,y)\colon x,y\in k\}\subseteq k^4$. In particular, $A_1v,A_2v,A_3v$ are linearly dependent.

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