Question:
Provided that $A$ is a $n×n$ real matrix with $A^2＝0$, is it true that $\operatorname{rank}(A+A^\text{T})＝2\operatorname{rank}(A)$ ?
I believe it’s true.
I’ve tried matrix operations, Jordan standard form and dividing $A$ into column vectors, but all failed.
Now I’m aware that linear mapping might be a good choice, because $Ax$ and $A^\text{T}x$ are perpendicular, so we can prove that $\ker(A+A^\text{T})=\ker(A)\cap\ker(A^\text{T})$.
Then we need to prove that $\dim(\ker(A+A^\text{T}))=n-2 \operatorname{rank}(A)$, but I don’t know how.
Best Answer
You had the good idea with your observation about orthogonality. $$ \dim \operatorname{Ker}(A+A^\text{T}) = \dim\big(\operatorname{Ker}(A) \cap \operatorname{Ker}(A^\text{T})\big) = \dim \operatorname{Ker} (A) + \dim \operatorname{Ker}(A^\text{T}) - \dim \big( \operatorname{Ker} (A)+ \operatorname{Ker} (A^\text{T})\big)$$
Obviously $\dim \operatorname{Ker} (A)=\dim \operatorname{Ker} (A^\text{T}) = n - \operatorname{rank}(A)$. Then we always have that $\operatorname{Im}(A)$ is the orthogonal complement of $\operatorname{Ker}(A^\text{T})$, and since $A^2=0$, $\operatorname{Im}(A)\subset \operatorname{Ker}(A)$. Hence $\operatorname{Ker}(A)+\operatorname{Ker}(A^\text{T})=\mathbb{R}^n$, so $\dim \operatorname{Ker}(A+A^\text{T})=2(n-\operatorname{rank}(A))-n=n-2\operatorname{rank}(A)$.
As you observed already, this allows us to conclude that $\operatorname{rank}(A+A^\text{T})=2\operatorname{rank}(A)$.