Does $\text{dim}W = 0$ imply $W = \{\textbf{0}\}$?

Question

If we have $W$ and $X$ as subspaces of $\mathbb{R}^8$, with dim$W$ = 3, dim$X$ = 5 and $W + X = \mathbb{R}^8$, then this implies that dim($W \cap X) = 0$. Does this then imply that $W \cap X = \{\textbf{0}\}$, where $\textbf{0} \in V$? Why / why not?

Answer 1

Yes. One of (probably many) ways to say this is that if $W \cap X$ contained a non-zero vector $v,$ then the set $\{v\}$ would be linearly independent, and so the dimension of $W \cap X$ would be at least $1.$

Answer 2

Yes, it does imply $W = \lbrace 0 \rbrace$. Basically by definition, the only vector space with dimension $0$ is the trivial subspace.

The definition of dimension is the length of all of the bases. Many texts don't work with empty bases, but define dimension zero for the trivial subspace as a special case. If you don't define a special case, and are happy to work with empty bases (and hence, empty sums), it still works out fine.

Here's another way to think about it: if there was a non-zero vector $v \in W$, then $\operatorname{span}(v)$ is one-dimensional and contained in $W$, so $\operatorname{dim} W \ge 1 > 0$.