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?
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Study its skew-symmetry. – User Apr 10 '17 at 16:15
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6There should only be $i_1$ upto $i_k$. – hmakholm left over Monica Apr 10 '17 at 16:16
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oops, you're right – Nażysław Zbyłutowicz Apr 10 '17 at 16:21
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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 $$ \begin{array}{lll} \{\; x_1\wedge x_2, & x_1 \wedge x_3, & x_1 \wedge x_4, \\ & x_2 \wedge x_3, & x_2 \wedge x_4, \\ & & x_3 \wedge x_4 \;\} \end{array} $$
hmakholm left over Monica
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Thank you, that make sense. How would linear independence be established in this context? I feel like the standard method of setting up a matrix and row reducing it will not suffice here. – Nażysław Zbyłutowicz Apr 10 '17 at 16:29
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@SocraticaFan: Hmm, that depends on exactly how you have defined $\Lambda^k(V)$. If it's by a universal property, I think most straightforward would be to prove that the space of formal linear combinations of these postulated basis elements satisfies the universal property, and therefore this space is $\Lambda^k(V)$. – hmakholm left over Monica Apr 10 '17 at 16:49
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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.
Alekos Robotis
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