$\mathbb{R}^n$ contains at least two subspaces. True or false? I was helping a friend with her linear algebra homework and I was wondering if my response for this question is right and a more elaborate answer of this question because I feel like I myself don't understand fully about the concept of a subspace; here it is though:
$\mathbb{R}^n$ contains at least two subspaces. True or false?
I put true (which it is) because any subspace requires that it must need itself and a zero subspace, but for some reason I'm doubting this answer because I can't physically picture this because I'm doubting my knowledge of a subspace, which I think is just an area of collection of vectors?
 A: Your answer is correct unless $n=0$.  In that case itself and the zero subspace are the same.
A: The statement is true for $n \geq 1$, otherwise the statement is false.
We are in a subspace when if we add two vectors, we still are in the subspace and if we multiply the vector by any scalar multiple (including $0$), we are still in the subspace. Because we can multiply always by $0$, the null vector always needs to be a member of our subspace.
Any other subspace is invalid. Imagine you have a subspace containing of all vectors on the real line except the vector $[2]$, then this is invalid because the addition of vector $[1]$ for instance with vector $[1]$ gives vector $[2]$ which needs to be in your subspace, and which is not know and thus the space of all vectors on the real line minus the vector $[2]$ is not a subspace.

For $n = 1$ we have only $2$ subspace. We have the space containing only the zero vector, and we have the space containing the full $1$ dimensional line.
To see that why the first one is true, note that every zero vector plus any other zero vector gives just back your zero vector (which is in your subspace) and every zero vector times any scalar number gives Always your zero vector (which is in your subspace). 
For the second one, every vector lying on your real line added by an other random vector lying on your $1$ dimensional line, just gives a vector back on that line. Multiplying any vector on that line by any scalar number, just gives a vector back on that $1$ dimensional line you had. 

For $n = 2$ you have an infinite amount of subspaces. Every set of vectors containing all vectors lying on a line through the origin form a subspace and the set of vectors containing the full 2D space. 
To see why this is true let us first consider the set of vectors that lie on a lines. If you add two vectors that lie on such a line, they still lie on that line. If you multiply a vector lying on that line by any scalar multiple, the resulting vector still lies on that line. When this line where you take all vectors from now goes through the origin, you have a subspace. Why through the origin? Well, one of the rules we had with subspaces was that you need to be able to multiply a vector by any scalar number. So you need to be able to multiply any 2D vector on that line by $0$. Because you need to be able to multiply any 2D vector by $0$, the result of that multiplication, namely the 2D zero vector, needs to be in your subspace.
For the second one, it is obvious that if you add two vectors in the full 2D space, you still get a vector in the 2D space and that if you multiply a vector in the 2D space by any scalar, you still are in the 2D space.

For $n = 3$ we have also infinite amount of subspaces. All 3D vectors lying on a single line through the origin form a subspace, all 3D vectors lying on a plane through the origin form a subspace, as well as the full $3$ dimensional space.

For $n = 4$ and higher it becomes harder to visualize, but you can still see that we have an infinite amount of subspaces. Even having all lines going through the origin in your $n$-dimensional space is enough to have an infinite amount of subspaces.
