The definition of a subspace in linear algebra

Question

I'm trying to learn linear algebra on my own but I am stuck on the definition of a linear subspace.

Let's assume I want to find out if $S$ is a subspace of $\mathbb{R}^2$, where $ S = [X_1 , X_2] $ and $ X_1 > 0 $.

$S$ would not be a subspace since it violates rule 3 of the definition. A negative one times the entire matrix would produce a vector that is not in the subspace.

But I would say it is quite obvious that the area where X is positive is a SUBSPACE of $\mathbb{R}^2$? It seems intuitive.

Why did mathematicians choose the three rules they chose? What's the motivation?

Answer 1

I think the fact that you find intuitive is that the set $\{(x_1,x_2):x_1>0\}$ is a subset of $\mathbb{R}^2$. This is certainly correct: every element of the former set is also an element of the latter.

But being a subspace means more than just being a subset. In addition, it requires the subset to possess the same kind of structure that the larger space does: in particular, to be closed under scalar multiplication. The half-plane fails this requirement.

The motivation for insisting on this is that when we want to do linear algebra, we need things to be linear spaces. An arbitrary subset of a linear space, like, say, a Cantor set, has nothing to do with linear algebra methods, so the definition is made to exclude such things.

Answer 2

The definition of a subspace is a subset that itself is a vector space. The "rules" you know to be a subspace I'm guessing are

1) non-empty (or equivalently, containing the zero vector)

2) closure under addition

3) closure under scalar multiplication

These were not chosen arbitrarily. If you look through the definition of a vector space you'll notice that every condition will automatically be satisfied by vectors in any subset except these three.

What I mean is, for instance, if $U$ is a subset of $V$ and $V$ is a vector space, then I already know that $u_1+u_2=u_2+u_1$ for any $u_1,u_2\in U$ because they are also elements of $V$, and this property holds for elements of $V$.

Therefore what you are thinking of as some random "rules" to be a subspace are really just the minimal requirements for a subset of $V$ to itself be a vector space.

Answer 3

That can't be right because the set {(x1,x2):x1>0}{(x1,x2):x1>0} does not contain the 0 vector due to the condition.