Find $a_{ij}$ in $v=\sum\limits_{1\le i<j\le n} a_{ij}e_i\wedge e_j$ If $v=\sum_{i=1}^n\lambda_{1i}e_i$ and $w=\sum_{i=1}^n\lambda_{2i}e_i$ are $2$ vectors in $\mathbb R^n$, then one can write $v\wedge w$ as a linear combination of $\{e_i\wedge e_j:1\le i&lt;j\le n\}$, what are then coefficients in that case, i.e. $v=\sum\limits_{1\le i&lt;j\le n} a_{ij}e_i\wedge e_j$, how can I express $a_{ij}$ with $\lambda_{1i}$ and $\lambda_{2i}$ ?


Well there are $\binom{n}{2}$ possibilities for $e_i\wedge e_j$ so at least the combination of  $\lambda_{1i}$ and $\lambda_{2i}$ cannot be just the product
Why don't we try a little example and see if we can figure this out?

Suppose $v = a e_1 + b e_2$ and $w = c e_1 + d e_2$ (so either $n = 2$ or we decide all the other coefficients are zero). Then
\begin{align*}
v \wedge w
&amp;= (a e_1 + b e_2) \wedge (c e_1 + d e_2)
\\
&amp;= ac e_1 \wedge e_1
+ ad e_1 \wedge e_2
+ bc e_2 \wedge e_1
+ db e_2 \wedge e_2
\\
&amp;= (ad - bc) e_1 \wedge e_2
\end{align*}
because $e_2 \wedge e_1 = - e_1 \wedge e_2$ (and $e_i \wedge e_i = 0$ for all $i$).

Can you now work out the general case? If not try another example with only three nonzero coefficients. It'll come.
You're right that there are $\binom{n}{2}$ possibilities. However, remember that $e_k$ are all basis vectors (This is an assumption I'm making based on notation. If it were not for this, then I doubt there is a neat expression). Therefore, all of the $$a_{ij} = \lambda_{1i}\lambda_{2j}e_i \wedge e_j$$
Will be $0$ when $i \neq j$ since $e_i \wedge e_j = 0$

However, when $i = j$, then $e_i \wedge e_j = 1$

Therefore,

$$
a_{ij} = \lambda_{1i} \lambda_{2j} \ \ \ \text{when} \ \ \ i = j \\
a_{ij} = 0 \ \ \ \text{when} \ \ \ i \neq j
$$ 

However, since your index runs in a way that $i &lt; j$, you can never have $i = j$. Therefore, the entire product collapses to $0$