Why is the empty family linearly independent? Why is the empty family linearly independent? 
How can you prove this? Is it right to say that it can not be written as a linear combination of the others vectors?
 A: Any subset of a linearly independent set is linearly independent. This is almost obvious for non empty subsets; why shouldn't it be for the empty subset?
In order to prove that a set is linearly dependent you have to exhibit vectors in the set and coefficients (not all $0$) so the linear combination gives $0$. Can you for the empty set?
If you want to see it differently, can a vector in the empty set be written as linear combination of the other vectors in the empty set? Of course not. For the same reason, a single non zero vector forms a linearly independent set, because it can't be written as a linear combination of the other vectors (there's none remaining).
But it's easier if one takes a step further and defines that the only linear combination of the empty set is the zero vector. So a singleton $\{v\}$ with $v\ne0$ is linearly independent, because the only linear combination of the vectors in the set different from $v$ is $0\ne v$; conversely $\{0\}$ is linearly dependent, because $0$ is a linear combination of the “other” vectors, namely of the empty set.
If you don't trust this (but you should), take it as a definition by convention, so also the zero vector space has a basis (the empty set) and every set containing the zero vector is linearly dependent.
A: It vacuously satisfies the condition for linear independence and it serves as a basis for the zero vector space.
A: If it were linearly dependent, then there would be a nontrivial linear combination of the vectors in the family that added up to the zero vector.  Since the empty family is empty, no such linear combination exists!
The definition of linear dependence is simpler than that of linear independence (since "linearly independent" is defined as "not linearly dependent"), so in a tricky conceptual question like this, it may pay to turn it around and turn it into a question of linear dependence.
