# Gaining Linear Algebra Intuition — Subspaces

So I aced linear algebra over the fall semester, though I'm deeply troubled in that I struggle to really describe what I did. I cannot say with confidence what it all meant, nor do I have any sort of intuition with the subject. I don't think i ever really "got it" or internalized it. Therefore, while i still have a week before summer classes, I started re-reading the text to gain some insights.

I'm starting with subspaces and am struggling with why a subspace must contain the zero vector. The closure under linear combinations is something i understand, being that my go to example is a plane in $\mathbb{R}^3$. But why must $\mathbf{0} \in V$?

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The subspace must be a vector space on its own, and every vector space contains the 0 vector. – user99680 May 14 '14 at 2:05
A bit more intuition to you: a set of linearly independent vectors in $\mathbb{R}^3$ not always spans $V^3$ so maybe this is one of the reasons to define a subspace. – Lucas Zanella May 14 '14 at 2:06
careful: there is no such thing is the zero vector. – Ittay Weiss May 14 '14 at 2:11
Another question about subspaces: How does a subspace $V$ relate with the ambient space? For example, if $V \subset \mathbb{R}^n$ is a subspace, does this mean that $V$ is a set of particular solutions to the matrix equation $A \textbf{x} = \textbf{b}$ ? What about $V^{\perp}$? – David D. May 14 '14 at 2:38
This set of lectures helped me a lot: ocw.mit.edu/courses/mathematics/…. – Tpofofn May 14 '14 at 2:49

A subspace must be closed under scalar products. And, a subspace must be a non-empty subset. So, if you have a subspace, then you have at least one vector $v$ in it. Then, you also have the scalar product $0\cdot v$ in the subspace. But, it follows from the distributivity axioms in a vector space, $0\cdot v=0$ always. Thus, $0$ must be in the subspace.
Alternatively, $0=v-v$. – lhf May 14 '14 at 2:33
@lhf that does not prove that it's the same $0$ as in the ambient space though. – Ittay Weiss May 14 '14 at 8:56
not quite, since you would need to prove first that the $-v$ in the subspace and in the ambient space agree. For that, of course, you'd first need to show that the $0's$ agree. – Ittay Weiss May 14 '14 at 9:28
That's just a consequence of the definition, it's not part of the definition itself. It's closed under linear combinations. So if $x$ is in there, $-x$ will be too, and from there $x + (-x)$.