I'm learning about proving whether a subset of a vector space is a subspace.
It is my understanding that to be a subspace this subset must:
- Have the $0$ vector.
- Be closed under addition (add two elements and you get another element in the subset).
- Be closed under multiplication by a scalar (one element by any scalar yields another element in the subset).
Have a subset in $\mathbb{R}^3$:
$$S = \{(t-2s,-s,t) \ / \ t,s \in \mathbb{R}\}$$
Prove that it is a subspace of $\mathbb{R}^3$.
Does it have the $0$ vector?
Well yeah, take $t = s = 0$ and you get $(0,0,0).$
Is it closed under addition?
We want to show that
$$x + y \in S$$
For some $x,y \in S$.
So we got
$$x = (t_1-2s_1,-s_1,t_1)\\ y = (t_2-2s_2,-s_2,t_2)$$
Then
$$x + y = (t_1-2s_1,-s_1,t_1) + (t_2-2s_2,-s_2,t_2)\\ (t_1+t_2-2s_1-2s_2 , -s_1-s_2, t_1+t_2)\\$$
Now let
$$a = t_1+t_2\\ b = s_1+s_2\\$$
So the above expression would be
$$(a-2b, -b, a)$$
Since $a,b \in \mathbb{R}$, we prove that it is closed under addition.
Is it closed under multiplication?
We want to show that
$$x\cdot r \in S$$
For some $x\in S, r \in \mathbb{R}$.
Have
$$x = (t-2s, -s, t)$$
Then
$$x \cdot r = (r\cdot(t-2s) , r\cdot -s, r\cdot t)\\ (tr-2sr , -sr , tr)$$
Let
$$a = tr\\ b = sr$$
We would have
$$(a-2b, -b, a)$$
Since $a,b\in\mathbb{R}$, this is also closed under multiplication.
I have a doubt:
Somebody also solved this, but instead of proving the addition and the multiplication closures separately, they did it in a single proof rather than two. They proved something like
Have $x,y \in S$ and some $r \in \mathbb{R}$, we want to prove that $$x + r\cdot y \in S$$
Then they proceeded to do something similar as any of the two proofs I did above (evaluate and then assign constants $a$ and $b$). Does that mean that, for future proofs, does it suffice to prove $x + r\cdot y \in S$ to prove that it is a subspace? Is there no need to do it separately?
Yes. If r=1 then you are proving that it is closed under addition and if x=0 you are proving that it is closed under product by scalars. So it is a way to unify both questions.