5
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 14, 2018 at 15:02

1 Answer 1

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.