Trivial Vector Space iff Zero Dimension

Theorem

Let $V$ be a vector space.

Then $V = \set \bszero$ if and only if $\map \dim V = 0$, where $\dim$ signifies dimension of vector space.

Proof

Necessary Condition

Suppose $V = \set \bszero$.

We have that $V$ has no $\mathbf v \in V, \mathbf v \ne \bszero$.

Thus there exists no $\set {\mathbf v} \subseteq V$ such that $\mathbf v \ne \bszero$.

Such a $\set {\mathbf v}$ would be a linearly independent set.

Hence $\O$ is the only possible basis for $V$.

Hence $\map \dim V = \card \O = 0$.

$\Box$

Sufficient Condition

Suppose $\map \dim V = 0$.

Then by definition of dimension, $0 = \map \dim V = \card B$, where $B$ is a basis for $V$.

Thus $B = \O$.

Suppose there are sets $\set {\mathbf v}$ such that $\mathbf v \in V, \mathbf v \ne \bszero$.

By Singleton is Linearly Independent, any such $\set {\mathbf v}$ would be a linearly independent set.

So $V$ has no such $\mathbf v \ne \bszero$.

Thus $V = \set \bszero$.

$\blacksquare$