Need to determine if a subset is a vector space in $\mathbb R^2$

Question

Its been about a year since I have done anything with vector spaces and subsets of vector spaces, and now I've found that I have forgotten a lot of material.

I am given this: Suppose $$\mathbf x \equiv \binom{x_1}{x_2}, \mathbf {x+y} \equiv \binom{x_1+y_1}{x_2+y_1}, \alpha \mathbf x \equiv \binom{\alpha x_1}{\alpha x_2} $$ with all variables real numbers. Now I am given various subsets of $\mathbb R^2$ and am being asked whether or not they are vector spaces. Some of them I can easily do such as: $$\{\mathbf x \in \mathbb R^2 : x_1=1\}$$

I know that this is not a vector space because, among other things, it isn't closed under vector addition. There are several similar to this which I can do, but here is one I don't know how to approach: $$\{\mathbf x \in \mathbb R^2 : x_1+4x_2=0\}$$ Here is my attempt:

Addition of two vectors becomes: $$\mathbf {x+y} \equiv \binom{x_1+y_1}{4(x_2+y_1)}, $$but does that necessarily mean that $(x_1+y_1)+4(x_2+y_2)=0$? I want to say yes because this can be rewritten as $(x_1+4x_2)+(y_1+4y_2)$ and since both terms equal $0$ this must also equal zero. I understand the multiplication with a scalar. The unique zero vector is just $\binom{\mathbf 0}{\mathbf 0}$. I am having trouble with the inverse, is it just $\mathbf x = \binom{-x_1}{4(-x_2)}$? I think I can handle the axioms about scalar multiplication being distributive and there being an identity.

Thanks for any help in advance and sorry if the post/attempt is kind of rambling!

Answer 1

The usual proposition is

A subset $S$ of $\Bbb R^2$ (or $\Bbb R^n$) is a subspace if, and only if: it contains the null vector, whenever it contains two vectors $x,y$, it also contains their sum, $x+y$, and finally, whenever it contains a vector $x$, and $\lambda$ is a scalar, it also contains $\lambda x$.

We can see $(0,0)$ is a solution of $ x_1+4x_2=0$. Now, suppose that $ x_1+4x_2=0$ and $ y_1+4y_2=0$. What happens when you sum them? And if $ x_1+4x_2=0$; what happens when you multiply this by a scalar? The first sentence tells us that $\vec 0\in S$, the second that if $x,y\in S$, then $x+y\in S$, the third, that if $x\in S$, then $\lambda x\in S$. Remember, you have to assume that $x,y\in S$, see what this means, and use that to prove their sum, say, is also in $S$.

In general, any set that is a solution to something of the form $$a_1x_1+a_2x_2+\cdots+a_nx_n=0$$ will be a subspace of $\Bbb R^n$, exactly by the same reasoning from above.

If you know that the interesection of subspaces is a subspace, this tells you that any system of equations $$\begin{cases}a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n=0\\ a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n=0\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\vdots\ \\a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n=0 \end{cases}$$ will also be a subspace of $\Bbb R^n$.