Let $n\in \mathbb N.$ Let $W$ be a non-trivial subspace of $n\times n$ symmetric matrices such that for every $x\in \mathbb R^n\setminus \{0\}$ there exists $a_x\in \mathbb R^n\setminus \{0\}$ such that $x\otimes a_x+a_x\otimes x\in W.$ Then prove that $\dim W\geq n.$
Notation: For $a,b\in \mathbb R^n$, $a\otimes b+b\otimes a=ab^T+ba^T$ is an $n\times n$ symmetric matrix.
We can prove this till $n\leq 3$ by contradiction.
For $n=1,$ it's simple because dimension of $1\times 1$ symmetric matrices is $1$.
For $n=2$, dimension of $2\times 2$ symmetric matrices is $3$. So, if $\dim W\leq n-1=1$ then for the canonical basis $\{e^1,e^2\}$ of $\mathbb R^2$ there exist $p^1,p^2\in \mathbb R^2\setminus \{0\}$ such that $e^1\otimes p^1+p^1\otimes e^1, e^2\otimes p^2+p^2\otimes e^2\in W$ and $\{e^1\otimes p^1+p^1\otimes e^1, e^2\otimes p^2+p^2\otimes e^2\}$ is liearly dependent. From there after some calculations, we got that $W=span \{e^1\otimes e^2+e^2\otimes e^1\}.$ After that I took the element $e^1+e^2$ and for this element there exists $p\in \mathbb R^2\setminus \{0\}$ such that $(e^1+e^2)\otimes p+p\otimes (e^1+e^2)\in W$. As $\dim W\leq 1,$ the set $\{(e^1+e^2)\otimes p+p\otimes (e^1+e^2), e^1\otimes e^2+e^2\otimes e^1\}$ is linearly dependent. From here also we got a contradiction that $p=0.$ Hence $\dim W\geq 2.$
For $n=3,$ similarly I considered three elements $e^1\otimes p^1+p^1\otimes e^1, e^2\otimes p^2+p^2\otimes e^2, e^3\otimes p^3+p^3\otimes e^3$ and taking the set $\{e^1\otimes p^1+p^1\otimes e^1, e^2\otimes p^2+p^2\otimes e^2, e^3\otimes p^3+p^3\otimes e^3\}$ as linearly dependent, I again got contradiction (It reduces into several cases, subcases and much more calculations than for $n=2$).
For $n\geq 4$, I was trying in a similar way but the calculations were getting much more difficult and I can not find any other way to prove for the general case.
Any help is appreciated. Thank you.
