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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.

user26857
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David W
  • 123

2 Answers2

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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 $\operatorname{rank}(A+A^\text{T})=2\operatorname{rank}(A)$.

Rócherz
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charmd
  • 6,050
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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.

user1551
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