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 {0} have dimension zero instead of one? Shouldn’t it be true that only the empty set has dimension zero?

**Answer**

A vector by itself doesn’t have a dimension. A *subspace* has a dimension. Why {0} is considered as having dimension 0? Because of consistency with all other situations. For instance R3 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 {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 {0} is the empty set that has zero elements.

**Attribution***Source : Link , Question Author : Qingtian , Answer Author : egreg*