Let $V$ be a subset of the vector space $\R^n$ consisting only of the zero vector of $\R^n$. Namely $V=\{\mathbf{0}\}$.
Then prove that $V$ is a subspace of $\R^n$.

Proof.

To prove that $V=\{\mathbf{0}\}$ is a subspace of $\R^n$, we check the following subspace criteria.

Condition (a) is clear since $V$ consists of the zero vector $\mathbf{0}$.

To check condition (b), note that the only element in $V=\{\mathbf{0}\}$ is $\mathbf{0}$. Thus if $\mathbf{x}, \mathbf{y} \in V$, then both $\mathbf{x}, \mathbf{y}$ are $\mathbf{0}$. Hence
\[\mathbf{x}+\mathbf{y} =\mathbf{0}+\mathbf{0}=\mathbf{0}\in V\] and condition (b) is met.

To confirm condition (c), let $\mathbf{x}\in V$ and $c\in \R$. Then $\mathbf{x}=\mathbf{0}$.
We have
\[c\mathbf{x}=c\mathbf{0}=\mathbf{0}\in V\] and condition (c) is satisfied.

Hence we have checked all the subspace criteria, and hence the subset $V=\{\mathbf{0}\}$ consisting only of the zero vector is a subspace of $\R^n$.

What’s the dimension of the zero vector space?

What’s the dimension of the subspace $V=\{\mathbf{0}\}$?

The dimension of a subspace is the number of vectors in a basis. So let us first find a basis of $V$.

Note that a basis of $V$ consists of vectors in $V$ that are linearly independent spanning set. Since $0$ is the only vector in $V$, the set $S=\{\mathbf{0}\}$ is the only possible set for a basis.

However, $S$ is not a linearly independent set since, for example, we have a nontrivial linear combination $1\cdot \mathbf{0}=\mathbf{0}$.

Therefore, the subspace $V=\{\mathbf{0}\}$ does not have a basis.
Hence the dimension of $V$ is zero.

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