Tensor algebra wedge Let $k$ be a field and $V$ a vector space of dimension $n$. Let $P$ denote the image of the universal
alternating map $V \times V \to \bigwedge^2(V)$. (Thus $P$ consists of pure products of the form $v_{1} \bigwedge v_{2}$.)
How to show that  $P$ is a union of some one-dimensional sub-spaces of $\bigwedge^2(V)$? Also if $n>2$, is $P = \bigwedge^2(V)$?
Try: I have shown that the elements of $P$ look like $(a_1v_1+a_2v_2+\cdots+a_nv_n) \bigwedge (b_1v_1+\cdots+b_nv_n)=(a_1b_2-a_2b_1)(v_1 \bigwedge v_2)+$ similar terms, where $v_1,\dots, v_n$ are basis of $V$. 
 A: To show $P$ is a union of $1$-dimensional subspaces, we need to show that for every $v_1\wedge v_2\in P$, it is contained in a $1$-dimensional subspace $W\subset\wedge^2 V$ with $W \subseteq P$. If $v_1\wedge v_2\neq 0$, you could let $W=\{rv_1\wedge v_2\}$ where $r$ ranges over your field.
In regards to the second question, when $n=3$ they are the same. The identity $$(bv_1+c v_2)\wedge(\frac{a}{b}v_2+v_3)= av_1\wedge v_2+bv_1\wedge v_3+cv_2\wedge v_3$$
allows you to realize any linear combination of $v_i\wedge v_j$ as long as $b\neq 0$. If $b=0$, use a slightly different identity. Thus $P$ is a $3$-dimensional vector space, and in particular, is not a finite union of its $1$-dimensional subspaces, as conjectured in the comments.
To get an example where $P\neq \wedge^2V$, we need to go to $n=4$. I claim that $v_1\wedge v_2+v_3\wedge v_4\not\in P$. Using your equation, we would have $a_1b_4=a_4b_1$ and $a_1b_3=a_3b_1$. Assuming either of these two equations has nonzero terms, we can divide one into the other and get $b_3a_4=a_3b_4$, but this would contradict that the coefficient of $v_3\wedge v_4$ is $1$. Thus we can assume $$a_1b_4=a_4b_1=a_1b_3=a_3b_1=0.$$ Since $a_1b_2-a_2b_1=1$, if $a_1=0$, $b_1\neq 0$. So if $a_1=0$, we have $a_4=a_3=0$, by the above equation. On the other hand, if $a_1\neq 0$, we have $b_4=0$, and by similar reasoning, $b_1=b_3=0$. So in both cases we have $v_2\wedge w$ for some $w$, which it is not hard to show cannot equal $v_1\wedge v_2+v_3\wedge v_4$.
