If $AB=0$, then $A+A^T$ or $B+B^T$ is singular

Question

Define $A$ and $B$ as being square matrices of dimension $2011$. Prove that if $AB=0$, then at least one of matrices $A+A^{T}$ or $B+B^{T}$ have rank below $2011$.

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Rank of a matrix is a number of linear independent rows.

-- edit2 -- dimension instead of rank

Answer 1

If $A,B\in M_n(K)$, $K$ a field, $n$ odd, and $AB=0$, then $A+A^T$ or $B+B^T$ is singular.

If $A+A^T$ and $B+B^T$ are invertible, then their rank is $n$. But $\operatorname{rank}(A+A^T)\le 2\operatorname{rank}(A)$ and similarly $\operatorname{rank}(B+B^T)\le 2\operatorname{rank}(B)$. Set $n=2k+1$. Then $\operatorname{rank}(A)\ge k+1$ and $\operatorname{rank}(B)\ge k+1$, so $\operatorname{rank}(A)+\operatorname{rank}(B)\ge n+1$. From Sylvester Rank Inequality we have $\operatorname{rank}(A)+\operatorname{rank}(B)-n\le \operatorname{rank}(AB)=0$, a contradiction.