An algebra problem by Hobart Pao

There are two such vector spaces: the zero subspace and the empty set. It is easy to verify that each one is a subspace. The reason why the zero subspace has dimension zero is because the zero vector is linearly dependent to itself, so it is not a basis vector (and the set of basis vectors is the empty set for the zero subspace). Since the empty set subspace is empty, it can't possibly have any basis vectors. What about some other sets that only have a single vector? Could these have a dimension 0? Well, there are no such sets with these criteria. It can be shown that for any vector space $W$ over the field $F$ containing vectors $\vec{a}$ and $\vec{b}$ , $\vec{a} + k \vec{b}, k \in F$ is also contained in $W$ . So that means, if we have a set $W$ only containing a vector $\vec{a}$ , and want $W$ to be a subspace, then $W$ must contain all scalar multiples (any scalar in $F$ )of $\vec{a}$ which now is no longer a set with a single vector, so we have shown that only the zero subspace and empty set are vector spaces with dimension zero.