Computing bases for direct, wedge, tensor products, etc., of given vector spaces I am filled with all kinds of vector space and I want to make sure I understand the basis for each kind of vector space.
Suppose $\{v_i\}_{i=1}^n$ is the basis for vector space $V$, $\{w_j\}_{j=1}^m$ is the basis for vector space $W$. 


*

*$V+W$: all linear combo of $\{v_i\}$, $\{w_j\}$, basis is $\{v_i,w_j\}$, dimension is $\dim(V)+\dim(W)-\text{number of overlap}$

*$V\oplus W$: Same as $V\times W$?

*$V\times W$: all $(v,w)$ with $v\in V$ and $w\in W$, basis $(v_i,w_j)$, dimension is $\dim(V)\dim(W)$

*$V\otimes W$: all linear combo of $v_i\otimes w_j$, basis $v_i\otimes w_j$, dimension is $\dim(V)\dim(W)$. By universal property of tensor, $V\times W$ is isomorphism to $V\otimes W$?

*$V\bigwedge V$ or $\bigwedge^rV$: all linear combo of $v_i\wedge v_j$, basis $v_i\wedge v_j$ and $i\neq j$, dimension is ${\dim V}\choose r$

*$V^{`}\cdot V$ or $S_rV$:(symmetric power) all linear combo of $v_i^{`}\cdot v_j$, basis $v_i^{`}\cdot v_j$ and $i\neq j$, dimension is ${\dim V +r-1} \choose r$.


So eventually, I want to understand what are the basis for the two vector spaces $\bigwedge^2(V\oplus W)$ and $ \bigwedge^2V\oplus(V\otimes W)\oplus\bigwedge^2W$(and finally I need to find an isomorphism between these two spaces, which need to prove it maps basis to basis, and that is why I need to make sure I know the basis). 
Let $\{(v_i,w_j)\}=\{e_k\}$, where $i=1,\cdots,n$, $j=1,\cdots,m$, $k=1,\cdots,nm$.
Is $\{e_i\wedge e_j\}_{i,j=1}^{nm}$ with $i\neq j$ the basis for $\bigwedge^2(V\oplus W)$? OMG, I am so confused.
BTW, it seems that these two spaces are not isomorphism, is the problem wrong?
 A: As in the question, we set $n := \dim V$ and $m := \dim W$ and assume both are finite (though many conclusions hold just as well when either or both are infinite), and pick arbitrary bases $A = (v_a)$ of $V$ and $B = (w_b)$ of $W$. Also, we usually regard bases as ordered sets of vectors; the below doesn't always specify orders, but of course these can be chosen arbitrarily to produce a bona fide basis.

*

*Sums $V + W$ Pick a basis $(x_c)$ of $V \cap W$ (denote $p := \dim (V \cap W) = p$), and extend it to a basis $(x_1, \ldots, x_p, y_1, \ldots, y_{m - p})$ of $W$. Then, $$(v_1, \ldots, v_n, y_1, \ldots, y_{m - p})$$ is a basis of $V + W$, which thus has dimension $n + m - p$; usually this is written $$\dim(V + W) = \dim V + \dim W - \dim (V \cap W).$$


*Direct sums $V \oplus W$ Via some identifications we can view this as a special case of (1), but we may as well be explicit: $$(A \times \{0\}) \cup  (\{0\} \times B) = \{(v_a, 0) : v_a \in A\} \cup \{(0, w_b) : w_b \in B\}$$ is a basis of $V \oplus W$, and so $$\dim (V \oplus W) = \dim V + \dim W.$$ For convenience, we often mildly abusively write $(v_a, 0)$ as $v_a$ and $(0, w_a)$ as $w_a$, in which case we can write the basis as
$$A \cup B = (v_1, \ldots, v_n, w_1, \ldots, w_m).$$


*Cartesian products $V \times W$ Since the product has a finite number of factors, this is indeed the same as $V \oplus W$ in (2); see The direct sum $\oplus$ versus the cartesian product $\times$ for more.


*Tensor products $V \otimes W$ $$\{v_a \otimes w_b : v_a \in A, w_b \in B\} \cong A \times B$$ is a basis for $V \otimes W$ and hence $$\dim (V \otimes W) = \dim V \cdot \dim W.$$


*Multivector spaces $\bigwedge^k V$
$$\{v_{a_1} \wedge \cdots \wedge v_{a_k} : 1 \leq a_1 < \cdots < a_k \leq n\},$$ and so $$\dim \bigwedge^k V = {{\dim V}\choose{k}}.$$ By convention, $\bigwedge^0 V$ is just the field $\mathbb{F}$ underlying $V$, and we take the empty wedge product to be $1 \in \mathbb{F}$, so $(1)$ is a basis of $\bigwedge^0 V$; $\bigwedge^k V$ is trivial for $k > m$, in which case the given basis is empty.


*Symmetric powers $\bigodot^k V$
$$\{v_{a_1} \odot \cdots \odot v_{a_k} : 1 \leq a_1 \leq \cdots \leq a_k \leq n\}$$ is a basis for $\bigodot^k V$, where $\odot$ defines the symmetric product, and one can then show $$\dim \bigodot^k V = {{\dim V + k - 1}\choose{k}}.$$
For completeness we should also mention one more:
Tensor powers $\bigotimes^k V$
$$\{v_{a_1} \otimes \cdots \otimes v_{a_k} : v_{a_1}, \ldots, v_{a_k} \in A\} \cong A \times \cdots \times A$$ is a basis for $\bigotimes^k V$, so that $$\dim \bigotimes^k V = (\dim V)^k.$$
Bivector space over a direct sum Finally, we can treat $\bigwedge^2 (V \oplus W)$. By (2), a basis for $V \oplus W$ is $$A \cup B = \{v_1, \ldots, v_n, w_1, \ldots, w_m\},$$
and then by (5), a basis of $\bigwedge^2 (V \oplus W)$ is
\begin{multline}
(v_1 \wedge v_2, \ldots, v_1 \wedge v_n, v_1 \wedge w_1, \ldots, v_1 \wedge w_m, v_2 \wedge v_3, \ldots, w_{m - 1} \wedge w_m) \\
= \{v_a \wedge v_{a'} : a < a'\} \cup \{v_a \wedge w_b\} \cup \{w_b \wedge w_{b'} : b < b'\}.
\end{multline}

The elements $v_a \wedge v_{a'}$ comprise a basis for $\bigwedge^2 V$, and likewise those of the form $w_b \wedge w_{b'}$ comprise a basis for $\bigwedge^2 W$. The forms $v_a \wedge w_b$ span a space that can be identified with $V \otimes W$ via the map characterized by the isomorphism $v_a \otimes w_b \mapsto v_a \wedge w_b$. This gives a natural decomposition $$\bigwedge^2 (V \oplus W) \cong \bigwedge^2 V \oplus (V \otimes W) \oplus \bigwedge^2 W.$$ Of course, the dimension formulas produced above lead to the same result for the dimension of $\bigwedge^2 (V \oplus W)$: $${{n + m}\choose 2} = \frac{1}{2}(n + m)(n + m - 1) = \frac{1}{2}n(n - 1) + nm + \frac{1}{2}m(m - 1) = {n \choose 2} + nm + {m \choose 2}.$$

