I was introduced to Exterior Algebra in a very abstract way, and I am wondering if I got it right, concrete.
Consider $ R^3 $ and let $ V = \{e_1,e_2,e_3 \} $ be basis for it.
The Exterior space, $ \bigwedge^3V $, is of dimension $ 2^n = 8 $ and a basis consists of the vectors
- $ e_1,e_2,e_3, $
- $ e_{12} = e_1 \wedge e_2, \ e_{13} = e_1 \wedge e_3, \ e_{23} = e_2 \wedge e_3 $
- $ e_{123} = e_1 \wedge e_2 \wedge e_3 $
Q: I should get one more basis vector in $ \bigwedge^3 V $, and the only one I can think of is $ e_\varnothing $? If it is $ e_\varnothing $, is $ e_\varnothing = 0?$ Why do we have to include the zero vector in the basis?
Q: Furthermore, since $ A \wedge A = 0 $, the only operations with the given basis we can do thats not zero is $$ e_i \wedge e_j = e_{ij}, \ i \neq j $$ and $$ e_i \wedge e_{jk} = e_{ijk}, \ i \neq j \neq k$$
One starting comment: instead of $\wedge^3 V$ you want $\wedge^3 \mathbb{R}^3$, since exterior products apply to vector spaces, not sets of basis vectors for them.
The basis for $\wedge^3 \mathbb{R}^3$ is indexed by subsets of $\{1,2,3\}$ of cardinality 3, so there's one basis vector: what you have called $e_{123}$. Thus, it's a one-dimensional vector space. More generally, the dimension of $\wedge^d \mathbb{R}^n$ is the binomial coefficient $\binom{n}{d}$.
The exterior algebra does indeed have the 8 basis vectors you listed above, although $e_\varnothing = 1 \in \mathbb{R}$; this is the "empty" exterior power, so you can think of this as analogous to $c^0 = 1$ for $c \in \mathbb{R}$ (take $c \ne 0$ if you want $0^0$ to be something besides 1 of course!). The exterior algebra contains all of the exterior powers $\wedge^d \mathbb{R}^3$ for $d \ge 0$. For $d > 3$ this is zero, by the binomial dimension calculation earlier.