rank of $(A+A^{T})$ when $A$ is skew symmetric

Question

If $A$ is skew-symmetric and rank of $A$ is $2$. Then what can we say about $\operatorname{rank}\left(A+A^{T}\right)$? I don't find any theorem giving me a concise idea about this. (Maybe I am missing something).

Answer 1

Any vector in the image of $A+A^T$ is clearly the sum of a vector in the image of $A$ and a vector in the image of $A^T$ (up to here, this works with any $B$ instead of $A^T$). Consequently, $$\operatorname{rank}(A+A^T)\le \operatorname{rank}(A)+\operatorname{rank}(A^T) =2\operatorname{rank}(A)$$

Note that for any $k$ with $2k\le n$, we find an example $$ A=\begin{pmatrix}0_{k\times (n-k)}&I_k\\0_{(n-k)\times(n-k)}&0_{(n-k)\times k}\end{pmatrix}$$where the inequality is sharp. On the other hand, even for $k=n-1$, we find an example $$ A=\begin{pmatrix}0&1&0&\cdots\\-1&0&1&0&\cdots\\0&-1&0&1&\\\vdots&0&-1&0&\ddots\\ &\vdots&&\ddots&\ddots\end{pmatrix}$$ where the resulting rank is $0$.