Is the zero vector in $\mathbb{R}^n$ by itself a subspace of $\mathbb{R}^n$? W is a subspace of $\mathbb{R}^n$ iff


*

*The zero vector ∈ W.

*X + Y ∈ W for any X, Y ∈ W.

*aX ∈ W for any X ∈ W and a ∈ R.


So, given W = { X : X = [x1...], x1 = 0, x2 = 0, ... xn = 0 } ∈ Rn


*

*The zero vector ∈ W because each X in W is the zero vector by definition,

*X + Y ∈ W because [0...] + [0...] = [0...]

*aX ∈ W because a[0...] = [0...]


So if I understand this correctly, the zero vector is itself a subspace of $\mathbb{R}^n$. Is this correct? 
In addition, can this be extended to say that for any W = { X : X = [$x_1$...] } ∈ $\mathbb{R}^n$, assuming W is a subspace of $\mathbb{R}^n$, $x_i$ (an arbitrary component of W) can only be a constant if it is 0? (I.e., $x_i$ can't be 1 or 2, but can be 0)
 A: Yes the set containing only the zero vector is a subspace of $\Bbb R^n$. It can arise in many ways by operations that always produce subspaces, like taking intersections of subspaces or the kernel of a linear map. It has dimension$~0$: one cannot find a linearly independent set containing any vectors at all, since $\{\vec0\}$ is already linearly dependent (taking $1$ times that vector is a nontrivial linear combination that gives the zero vector). The subspace is isomorphic to $\Bbb R^0$. Like any vector space of dimension$~k$, and hence like $\Bbb R^k$, it has a basis consisting of $k$ vectors; since $k=0$ such a basis is the empty set. Indeed, rather exceptionally, this is the unique basis for $\{\vec0\}$. That $\vec0$ is in the span of the empty set might seem strange, but the unique linear combination one can form of the empty set is the linear combination with no terms at all, and the value of such an empty sum (computed in a vector space) is by convention the zero vector.
In the comments to your question you chose the wrong one of the two available subsets $\emptyset$ and $\{0\}$ to be candidate for the basis; the latter has $1$ element which is too much for dimension$~0$, so naturally you find that it is linearly dependent.
A: Indeed, $\{0\}$ is a subspace. In fact, for any vector space $V$ and any $x\in V$, we have that $\{x\}$ is a subspace of $V$ if and only if $x$ is the zero vector of $V$, since any subspace of $V$ must contain the zero vector of $V$, and the subset of $V$ containing only the zero vector of $V$ is necessarily closed under addition and scalar multiplication (as you've shown).
A: Yes. Took me sometime to get this. 


*

*Zero is a member of ${0}$.

*${0}$ is closed under addition. Let u and v be members of ${0}$. $u=v=0$ hence $u+v=0$.

*${0}$ is closed under scalar multiplication. 

