Second exterior power of a complex vector space Suppose we are given a $k$-dimensional complex vector space. Consider the second exterior power of $V$, that is $\Lambda^2V$. Denote $X=\{v_1\wedge v_2\colon v_1,v_2\in V\}$.
Now I have two questions:
1) Is it true that if $k=4$, then $x\in X$ if and only if $x\wedge x=0\in \Lambda^4 V$?
2) Let $k>4$. Does there exist $x\in \Lambda^2 V\setminus X$ such that $x\wedge x\neq 0$?
Thank you,
C.
 A: As a general remark, you are asking how to recognize perfect wedges in $\bigwedge^2 V$ via equations in higher exerior powers.    This is related to the Plucker relations for Grassmanians.  
E.g. a non-zero perfect wedge $v_1\wedge v_2$ in $\bigwedge^2 V$ corresponds to a $2$-plane in $V$ (the span of $v_1$ and $v_2$), are you are asking whether or not the single equation $x \wedge x = 0$ cuts out the locus of $2$-planes as a subvariety of $\bigwedge^2 V$.
The answer is yes when $\dim V = 4$, but no in general.
A: I think that the following argument answers your question #1 in the affirmative. We need a bit more detailed description of those elements of $\Lambda^2V$ that are not in the set $X$.
Step 1. If the four vectors $v_1,v_2,v_3,v_4$ are linearly dependent, then the element
$$
z=v_1\wedge v_2+v_3\wedge v_4\in X.
$$
Proof. Assume that, for example, $v_4=av_1+bv_2+cv_3$ for some constants $a,b,c.$ By plugging this in we can easily rewrite $z$ in the form
$$
z=(v_1+bv_3)\wedge v_2+cv_3\wedge v_1.
$$
If here $b=0$, then we get $z=v_1\wedge(v_2-cv_3)\in X$. If $b\neq0$, then we can rewrite $z$ as
$$
z=(v_1+bv_3)\wedge (v_2+\frac{c}{b}v_1)
$$
again verifying the claim.
Step 2. If $\dim V=4$, then any element of $\Lambda^2V$ of the form
$$
z=v_1\wedge v_2+v_3\wedge v_4+v_5\wedge v_6
$$
with $\{v_1,v_2,v_3,v_4\}$ linearly independent can be written as a sum of two elements of $X$.
Proof. The first four vectors form a basis of $V$, so we can write
$$
v_5=u_1+u_2,\ \text{where}\ u_1=av_1+b v_2,\quad u_2=cv_3+dv_4
$$
for some constants $a,b,c,d$. If $u_1=0$, then the vectors $\{v_3,v_4,v_5,v_6\}$
are linearly dependent, and the claim follows from the result of Step1. Similarly
we see that the case $u_2=0$ is easy, so we concentrate on the case $u_1\neq0\neq u_2$.
Next we rewrite the wedge products $v_1\wedge v_2$ and $v_3\wedge v_4$ using $u_1$ and $u_2$ as the other factor. This is, again, easy. Say, if $a\neq0$, then
$$
v_1\wedge v_2=(\frac1a u_1-\frac bav_2)\wedge v_2=u_1\wedge(\frac1a v_2),
$$
and if $a=0, b\neq0$ we proceed analogously. So we can write $v_1\wedge v_2=u_1\wedge v_1'$ and $v_3\wedge v_4=u_2\wedge v_2'$. Putting all this together we can write
$$
z=u_1\wedge(v_1'+v_6)+u_2\wedge(v_2'+v_6)
$$
as a sum of two elements of $X$.
The following is now obvious by induction on the number of summands from $X$.
Lemma. If $k=4$, then the elements $z=\Lambda^2V\setminus X$ can be written in the
form 
$$z=v_1\wedge v_2+ v_3\wedge v_4$$
with $v_1,v_2,v_3,v_4$ linearly independent.
We can then wrap up question #1.
If $z\in \Lambda^2V\setminus X$, then $z=v_1\wedge v_2+ v_3\wedge v_4$, where the vectors 
$v_i,i=1,2,3,4,$ form a basis of $V$. Then
$$
z\wedge z=2 v_1\wedge v_2\wedge v_3\wedge v_4\neq0,
$$
as an even number of swaps are needed to bring $v_3\wedge v_4\wedge v_1\wedge v_2$ into the form $v_1\wedge v_2\wedge v_3\wedge v_4$.
