Let $A \in \mathbb R^{10,10}$, $x_{1},x_{2},...,x_{7}\in \mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=...=Ax_{7}$. Proof that $A+A^T$ is singular (not invertible)
I am trying to do this task in two different ways and none of them works.
Firstly I wanted to show that $$\det (A+A^T) = 0 $$
but there is no nice formula for $\det A + B $ so probably it is wrong way.
From the other hand I have that idea:
Assume that exists $B$ such that
$$ B\cdot(A+A^T) = I_n$$
and now I wanted to proof that $$\ker A^T \subset \ker A$$ then I could easly fail that assumption. But unfortunately It (my sub-proof) seems to be false theorem so... have somebody any idea how to solve this task?
In fact, the conclusion should be that $A+A^T$ is not invertible. Note that the range of $A$ is spanned by $$ \langle Ax_1,Ax_2,\ldots , Ax_{10}\rangle =\langle Ax_7,Ax_8,Ax_{9}, Ax_{10}\rangle, $$ which implies that the rank of $A$ is at most $4$. By the rank theorem, $A^T$ has the same rank as $A$. This implies that $A+A^T$ has rank at most $\text{rk}(A)+\text{rk}(A^T)\le 8<10$, which says that $A+A^T$ is not invertible.