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?
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$\begingroup$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.
$\endgroup$ 7 $\begingroup$The sum over the empty set is the additive identity... in this case zero.
$\endgroup$ 2 $\begingroup$If $~0~$ is a part of your basis, and if it is the only vector, there are zero linearly independent vectors in it, since any non-negative multiple of $~0~$ is $~0~$.
Thus, the dimension of the space is $~0~$.
$\endgroup$ $\begingroup$Short answer: Because its basis is the empty set $\emptyset$.
If $V$ is a set with exactly one element and $F$ is a field, there is exactly one way to define addition and scalar multiplication such that $V$ is a vector space over $F$. In this case, $\emptyset$ is the only linearly independent subset. Since the span of $\emptyset$ - the intersection of all vector subspaces containing $\emptyset$ - is equal to $V$, $\emptyset$ is even a basis - the basis of $V$.
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