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Let V be a finite dimensional vector space over a subfield of $\Bbb C$ and $f$ a skew-symmetric bilinear form on V. I was trying to show that $f$ has rank 2 if and only if there exist linearly independent linear functionals $L_{1} ,L_{2}$ on V such that $$f(u,v) = L_{1}(u) L_{2}(v) – L_{1}(v) L_{2}(u) , \forall u,v \in V$$ I have been able to prove that $f$ has rank 2 implies there exist linearly independent linear functionals $L_{1} ,L_{2}$ on V . But unable to prove the other side, please help me.

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  • $\begingroup$ It would be nice if you may share your approach to prove the first part. $\endgroup$ – Majid May 26 '18 at 0:04
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Assume that $$f(u,v) = L_{1}(u) L_{2}(v) – L_{1}(v) L_{2}(u) ,\ \forall u,v \in V$$ and consider $L_f$ as the linear operator from $V$ to $V^*$ which assigns to each $u\in V$ the function $L_fu(v) = f(u,v)\ , \ \forall v\in V$.

We know that $$rank(f) = rank(L_f).$$

Now, fix $u\in V $ and let $a =L_1(u)$ and $b=L_2(u)$, then we have $$L_fu(v) = f(u,v) = a L_{2}(v) – bL_{1}(v)$$ which shows that $$Img(L_f) = <L_1, L_2>$$ and so as $L_1$ and $L_2$ are linearly independent, $rank(L_f)=2$ which completes the proof.

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