# Trivial Vector Space iff Zero Dimension

## Theorem

Let $V$ be a vector space.

Then $V = \left\{{\mathbf 0}\right\}$ iff $\dim \left({V}\right) = 0$, where $\dim$ signifies dimension.

## Proof

### Necessary Condition

Suppose $V = \left\{{\mathbf 0}\right\}$.

We have that $V$ has no $\mathbf v \in V, \mathbf v \ne \mathbf 0$.

Thus there exists no $\left\{{\mathbf v}\right\} \subseteq V$ such that $\mathbf v \ne \mathbf 0$.

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

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

Hence $\dim \left({V}\right) = \left|{\varnothing}\right| = 0$.

$\Box$

### Sufficient Condition

Suppose $\dim \left({V}\right) = 0$.

Then by definition of dimension, $0 = \dim \left({V}\right) = \left\vert|{B}\right\vert|$, where $B$ is a basis for $V$.

Thus $B = \varnothing$.

Suppose there are sets $\left\{{\mathbf v}\right\}$ such that $\mathbf v \in V, \mathbf v \ne \mathbf 0$.

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

So $V$ has no such $\mathbf v \ne \mathbf 0$.

Thus $V = \left\{{\mathbf 0}\right\}$.

$\blacksquare$