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.

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    $\begingroup$ Hint: Can you find an expression for the column vectors of $\vec u\ast\vec v-\vec v\ast\vec u$ and show that they are linearly independent? This would conclude that the rank is $n$. $\endgroup$ – Prasun Biswas Mar 13 '18 at 23:03
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    $\begingroup$ Note that $u \star v = u v^T$. Hence ${\cal R} (u \star v) = \operatorname{sp} \{ u \}$ if $v \neq 0$. $\endgroup$ – copper.hat Mar 13 '18 at 23:07

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 $2$.

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  • $\begingroup$ $\begin{bmatrix} u & v\end{bmatrix} \in \mathbb{R}^{n \times 2}$ right? so the matrix in the middle should in $\mathbb{R}^{2 \times 2}$? $\endgroup$ – Siong Thye Goh Mar 14 '18 at 22:48
  • $\begingroup$ Sorry, brain fart. ;-) $\endgroup$ – egreg Mar 14 '18 at 22:50
  • $\begingroup$ no worries, i make lots of such mistakes too. $\endgroup$ – Siong Thye Goh Mar 14 '18 at 22:57

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