I was working on a question which defined a new operation between vectors.
Let $\vec{u}=(u_i)_{i\in\lbrack n\rbrack},\vec{v}=(v_i)_{i\in\lbrack n\rbrack}\in\Bbb R^n$, define $\star:\Bbb R^n\times\Bbb R^n\to\mathcal{M}_n(\Bbb R)$ where $\vec{u}\star\vec{v}=A_{ij}=(u_iv_j)$ as a matrix.
The question is as follows:
If $\vec{u},\vec{v}$ are linearly independent, find the rank of the matrix $\vec{u}\star\vec{v}-\vec{v}\star\vec{u}$.
What I did was try to conjecture the rank with a couple of common l.i. vectors. I chose $e_1, e_2$. This gave me the matrix with entries $1$ in $(1,2)$, $-1$ in $(2,1)$, and $0$ elsewhere. Thus if the question is true, the matrix's rank should be $2$.
To prove this I tried checking what happened when $\vec{u},\vec{v}$ were linearly dependent. The only possibility is that $\vec{v}$ is a scalar multiple of $\vec{u}$, then $\vec{v}=\lambda\vec{u}$ for a real $\lambda$.
Should $\lambda$ be $0$, the matrix's rank would be 1. Else the matrix would be the zero matrix.
Now to prove the result, I think that if I prove that the rank is less than or equal to $2$ then I've got the proof. This is because the other cases are covered in what I just wrote.
Still, I feel that even if it were true, my proof would be too flimsy and not elegant. I'm looking for any hints and help to approach the solution.
We have $u \star v= uv^T$.
$$uv^T-vu^T = \begin{bmatrix}u &-v \end{bmatrix} \begin{bmatrix} v^T \\ u^T\end{bmatrix}= \begin{bmatrix}u &v \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix} \begin{bmatrix} v^T \\ u^T\end{bmatrix}$$
Hence the rank is at most $2$.
Let $B$ be the right inverse of $\begin{bmatrix} v^T \\ u^T\end{bmatrix}$ which exists since the rows are linearly independent.
$$B^T(uv^T-vu^T)B=\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$$
Hence the rank is $2$.