Direct sum of a vector space V and a field R

Question

What does it mean, that the exterior algebra of a vector space V over R is a superset of direct sum of real numbers and real vector space V? I thought that given vector spaces must have trivial intersection (zero vector), but R and V has not generally intersection in terms of linear algebra. So, how can https://i.stack.imgur.com/ZLc6l.png be a direct sum?

Answer 1

Looking at the link you sent on Wikipedia, you have the following: $$ \bigwedge(V)= \bigwedge^0(V) \oplus \bigwedge^1(V) \oplus \cdots \oplus \bigwedge^n(V) $$ where $n=\dim(V)$ (so the sum is infinite if $V$ has infinite dimension vector over $\mathbb{R}$); and $\bigwedge^k(V)$ is the $k-$th exterior power of $V$ with itself.

However, by convention you have that $\bigwedge^0V=\mathbb{R}$, and $\bigwedge^1V=V$, so on the equation above you get: $$ \mathbb{R}\oplus V \subset \mathbb{R}\oplus V \oplus \bigwedge^2(V)\oplus \cdots =\bigwedge(V)$$

Addendum For your last question, note that if $V$ is a real vector space, then you can do the exterior direct sum $\mathbb{R}\oplus V$, which is a new real vector space with elements the sums $\lambda + v$, with $\lambda\in \mathbb{R}$ and $v\in V$. In this case, we don't require them to have empty intersection. If you look at $W=\mathbb{R}\oplus V$, then inside of $W$ you have two subspaces with empty intersection, one of dimension $1$ (which is $\mathbb{R}$) and the other one, $V$.

Sorry if there are any mistakes! Hope it helps.