Is the empty set linearly independent or linearly dependent? Is the empty set linearly independent or dependent?

By definition, it is linearly independent, because it is not linearly dependent.
A set $S$ is linearly dependent if there exists a finite set of vectors $v_1,\dots, v_n$ and corresponding scalars $\alpha_1,\dots,\alpha_n$ such that there exists at least one $\alpha_i\neq0$ so that $$\sum_{i=1}^n \alpha_i v_i=0$$

Remark: (equivalently, we could demand that all $\alpha_i$ are nonzero, but then we would also need to demand that there exists at least one $\alpha_i$ that is nonzero. This is because the empty set satisfies the demand that every element from it is nonzero...)

Clearly, these is no finite collections of vectors from $\{\}$ that satisfies the above condition, because there is no collection at all.

Furthermore, the empty set is also a basis of the vector subspace $\{0\}$, because $\{0\}$ is the 

Smallest vector space that includes $\{\}$ and is a vector space.


In a way, you could also round-about your reasoning that $\{\}$ is linearly independent like this:


*

*You know that $\{0\}$ is a vector space.

*You know that every vector space has a basis.

*You know that the basis of $\{0\}$ is a subset of $\{0\}$, so the basis of $\{0\}$ can either be $\{\}$ or $\{0\}$

*You know that $\{0\}$ is not a basis because it is not linearly independent (because $1\cdot 0=0$)

*Therefore, $\{\}$ is a basis.

*Because all bases are linearly independent, so is $\{\}$


Note, this isn't really a "good" proof because it makes a sort of begging the question fallacy. This wasn't meant to be a proof in a mathematical sense, just a proof "to yourself" that you already know the empty set is linearly independent, because that's the only way every vector space can have a basis, and you know that that is true (and, in fact, you or someone else must have used the fact that the empty set is linearly independent while proving that fact)
