According to C.H. Edwards' Advanced Calculus of Several Variables: The dimension of the subspace $V$ is defined to be the minimal number of vectors required to generate $V$ (pp. 4).
Then why does $\{\mathbf{0}\}$ have dimension zero instead of one? Shouldn't it be true that only the empty set has dimension zero?
Best Answer
A vector by itself doesn't have a dimension. A subspace has a dimension. Why $\{\mathbf{0}\}$ is considered as having dimension $0$? Because of consistency with all other situations. For instance $\mathbb{R}^3$ has dimension $3$ because we can find in it a linearly independent set with three elements, but no larger linearly independent set. This applies to vector spaces having a finite spanning set and so of subspaces thereof.
What's the largest linearly independent set in $\{\mathbf{0}\}$? The only subsets in it are the empty set and the whole set. But any set containing the zero vector is linearly dependent; conversely, the empty set is certainly linearly independent (because you can't find a zero linear combination with non zero coefficients out of its elements). So the only linearly independent set in $\{\mathbf{0}\}$ is the empty set that has zero elements.