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?
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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