Let $W$ be a $n$-dimensional vector space with basis $w_1,...,w_n$ and $f:W\times W\rightarrow\bigwedge^2W$ with $f(x,y)=x\wedge y$.
1 For $n=3$, how do I express $f(aw_ 1+bw_2+cw_3,dw_ 1+ew_2+fw_3)$ in the basis vectors $\{w_1\wedge w_2,w_1\wedge w_3,w_2\wedge w_3\}$ of $\bigwedge^2W$?
2 How do I prove that $f$ is surjective for $n=3$?
3 How do I prove that $f$ is not surjective for $n=4$?
What I have done:
1 I know that $f(aw_ 1+bw_2+cw_3,dw_ 1+ew_2+fw_3)=aw_ 1+bw_2+cw_3\wedge dw_ 1+ew_2+fw_3$, but how do I express this in the basis vectors?
Is it $$(ad)w_1\wedge w_1+(ae)w_1\wedge w_2+(af)w_1\wedge w_3+(bd)w_2\wedge w_1+(be)w_2\wedge w_2+(bf)w_2\wedge w_3+(cd)w_3\wedge w_1+(ce)w_3\wedge w_2+(cf)w_3\wedge w_3=$$ $$(ae-bd)w_1\wedge w_2+(af-cd)w_1\wedge w_3+(bf-ce)w_2\wedge w_3?$$
2 I have no idea where to look for this question, unfortunately.
3 I think that I should find an element in $\bigwedge^2W$ that does not have an original in $W\times W$, which one could that be?
For # 1: You are correct here. This is just from bilinearity and skew-symmetry of the wedge product.
For # 2: Notice that what's going on here is the standard cross product in $\mathbb{R}^3$. Write the basis vectors of $\mathbb{R}^3$ as $w_1 = \vec{i} = \left<1,0,0\right>$, $w_2 = \vec{j} = \left<0,1,0\right>$ and $w_3 = \vec{k} = \left<0,0,1\right>$, and the standard basis vectors of $\bigwedge^2 \mathbb{R}^3$ (of which there are also three!) as $v_1 = w_2 \wedge w_3$, $v_2 = w_3 \wedge w_1 = -(w_1 \wedge w_3)$, and $v_3 = w_1 \wedge w_2$. If we abuse notation a little and also refer to $v_1,v_2,v_3$ as $\vec{i},\vec{j},\vec{k}$, then this is the standard fact about the cross product in $\mathbb{R}^3$ that $\vec{i} \times \vec{j} = \vec{k}$, $\vec{j} \times \vec{k} = \vec{i}$, $\vec{k} \times \vec{i} = \vec{j}$, etc...
The moral of the story is that your map $f$ is surjective because for any vector $v \in \mathbb{R}^3$, you can find two other vectors $u,w$ so that $u \times w = v$.
For #3: You are right. You should find an element of $\bigwedge^2 \mathbb{R}^4$ (which is $\binom{4}{2}=6$-dimensional) which has no pre-image under $f$. Here's an idea (though not a complete solution): Take generic elements $v_1 = \sum_{i=1}^4 a_i w_i$ and $v_2 = \sum_{j=1}^4 b_j w_j$ (where $w_i$ is the standard basis as above), and look at what the wedge looks like:
$$ \begin {eqnarray} v_1 \wedge v_2 &=& \sum_{i,j} a_ib_j \, w_i \wedge w_j \\ &=& \sum_{i<j} (a_ib_j - a_jb_i) \, w_i \wedge w_j \end {eqnarray} $$ Here is one way to think about what this means: the components/coordinates of the product are all the two-by-two minors/sub-determinants of the 2-by-4 matrix whose rows are $v_1 = (a_1,\dots,a_4)$ and $v_2 = (b_1,\dots,b_4)$. So try to come up with six numbers which cannot be obtained in this way; that is, find 6 numbers which ARE NOT the minors of any 2-by-4 matrix.