TITLE: linearly dependent symmetric matrices
QUESTION [5 upvotes]: 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.

REPLY [5 votes]: 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.