Provided that $A$ is a $n \times n$ real matrix, $A^2＝0$, is it true that $\operatorname{rank}(A+A^\text{T})＝2\times\operatorname{rank}(A)$? Question: Provided that $A$ is a $n×n$ real matrix, $A^2＝0$, is it true that $\operatorname{rank}(A+A^\text{T})＝2\times\operatorname{rank}(A)$ ?
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.
 A: 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 $\operatorname{rank}(A+A^\text{T})=2\operatorname{rank}(A)$.
A: Let $X=AA^T$ and $Y=A^TA$. Then
$$
\begin{aligned}
\operatorname{rank}(A+A^T)
&=\operatorname{rank}\left((A+A^T)^2\right)\quad\text{(as $A+A^T$ is symmetric)}\\
&=\operatorname{rank}(X+Y)\quad\text{(as $A^2=0$)}.\\
\end{aligned}
$$
Also, $A^2=0$ gives $XY=YX=0$. Hence $X$ and $Y$ are simultaneously diagonalisable and
$$
\begin{aligned}
\operatorname{rank}(X+Y)
&=\operatorname{rank}(X)+\operatorname{rank}(Y)\\
&=\operatorname{rank}(AA^T)+\operatorname{rank}(A^TA)\\
&=2\operatorname{rank}(A).
\end{aligned}
$$
Now we are done.
