Prove or disprove:
If $u,v \in \Bbb R^n$ are linearly independent column vectors, then $\operatorname{rank}(uv^T-vu^T)=2$.
I can see that $\operatorname{rank}(uv^T-vu^T) \leq 2$ since $\operatorname{rank}(A+B) \leq \operatorname{rank}(A)+\operatorname{rank}(B)$, but I could not make it always $=2$ or sometimes $<2$.
How can I confirm the number of linearly independent columns in $uv^T-vu^T$ is exactly 2?
Thanks in advance.
Hint:
The matrix $uv^t-vu^t$ is skew-symmetric and has trace zero. If its rank were $\le 1$, it would have to be equal to zero.
Suppose that $\operatorname{rank}(uv^t-vu^t) \le 1$. Notice that $$\operatorname{Tr}(uv^t-vu^t) = \operatorname{Tr}(uv^t)-\operatorname{Tr}(vu^t) = \operatorname{Tr}(v^tu) - \operatorname{Tr}(u^tv) = v^tu-u^tv = 0$$ since $v^tu$ and $u^tv$ are real numbers both equal to $\sum_{j=1}^n u_jv_j$. Moreover, we have $$(uv^t-vu^t)^t = vu^t-uv^t = - (uv^t-vu^t)$$ so $uv^t-vu^t$ is a skew-symmetric matrix and hence diagonalizable over $\Bbb{C}$. Because of rank it has at most one non-zero eigenvalue. However, its trace is equal to zero so it has to be $uv^t-vu^t = 0$.
In particular we have $$0 = (uv^t-vu^t)u = u(v^tu) - v(u^tu)$$ so by linear independence of $u,v$ it follows $u^tu = 0$ and hence $u=0$. But this is a contradiction with linear independence of $u,v$.
Define $A:=uv^T-vu^T$.
Let $w_1\in\mathbb{R}$ such that $v^Tw_1=0$ and $u^Tw_1=1$, then $Aw_1=-v$. Similarly, let $w_2\in\mathbb{R}$ such that $u^Tw_2=0$ and $v^Tw_2=1$, then $Aw_2=u$. Since $u$ and $v$ are linearly independent, then we have that $\mathrm{rank}(A)\ge2$.
Now, let $M=\begin{bmatrix}u^T\\v^T\end{bmatrix}$ and since $u$ and $v$ are linearly independent the kernel has dimension $n-2$ and for all vector $w$ in that kernel, we have that $Aw=0$, which implies that $\dim(\mathrm{ker}(A))\ge n-2$.
Since we have that $n=\mathrm{rank}(A)+\dim(\mathrm{ker}(A))$, this implies that $\mathrm{rank}(A)\le 2$, which proves that $\mathrm{rank}(A)= 2$.
