if $A \in C^{2015,2015}$ and $rank(A) < 1000$ proof that $\dim(\ker(A+A^T)) > 15$

Question

I want to solve that thesis:
if $A \in C^{2015,2015}$ and $rank(A) < 1000$ proof that $\dim(\ker(A+A^T)) > 15 $

from the fact that $$\dim(im(A)) = \dim(im(A^T))$$ and $$ \dim(\ker(A))+\dim(im(A))=n $$ it follows that $$\dim(\ker(A)) = \dim(\ker(A^T))$$ We know that $rank(A) < 1000$. Hence $\dim(\ker(A))>1015$ But there I have stuck... thanks for your time

Answer 1

You're one line from the solution:

You showed that $\dim (\ker A) > 1015$ (so $A$ sends a vector space of dimension $1015$ to zero).

Moreover since $\dim (Im A ) = \dim (Im A^T)$ you also have that the dimensions of the kernels equal and so $\dim(\ker A^T)>1015$.

Therefore by the dimensions theorem we have that the dimension of $\ker A \cap \ker A^T$ is at least $15$ (because $\dim(\ker A + \ker A^T)\leq 2015$). This completes the proof, because $\ker (A+A^T)$ contains $\ker A\cap \ker A^T$.

Answer 2

Hint: $\operatorname{dim}(\operatorname {im}(A+A^t)=\operatorname{dim}(\operatorname{im}A)+\operatorname{dim}(\operatorname{im}A^t)-\operatorname{dim}(\operatorname{im}A\cap\operatorname{im}A^t)\lt2000$.