what is the dimension of $\wedge^k(V)$?

Question

Let's say $B = \{x_1, x_2, ... x_n\}$ is a basis for $V$, then $\wedge^k(V)$ has basis $\{x_{i_1} \wedge x_{i_2} \wedge ... \wedge x_{i_k} : i_1 < i_2 < ... < i_k\}$. I'm not exactly clear on what this basis is. How many elements does it have and how can these be determined?

Answer 1

The basis you describe has one element for each $k$-element subset of $\{1,2,\ldots,n\}$.

There are $\binom nk$ such subsets, so the dimension of $\Lambda^k(V)$ is $\binom{\dim(V)}{k}$.

For example, for $n=4, k=2$, the basis elements would be $$ \begin{array}{lll} \{\; x_1\wedge x_2, & x_1 \wedge x_3, & x_1 \wedge x_4, \\ & x_2 \wedge x_3, & x_2 \wedge x_4, \\ & & x_3 \wedge x_4 \;\} \end{array} $$

Answer 2

The basis for $\Lambda^k(V)$ consists of all choices of ascending $k-$tuples of indices contained in $\{1,\ldots, n\}$ where $n=\dim(V)$. This is equivalent to the combinatorial problem of choosing a subset of size $k$ from $\{1,\ldots, n\}$. So, there are $$ \dim(\Lambda^k(V))={n\choose k}={\dim(V)\choose k}$$ ways to do this.