I'm seeking some guidance on the exercise above. I know that to prove the set has closure under addition and scalar multiplication, addition of any two elements must be found in the set and a scalar multiple of any element must also be in the set, but I'm confused with some particulars with this. I have no clue what a $4$-tuple is. I count $4$ elements in the set, $-2\alpha -2\beta$, $\alpha$, $\beta$, $\alpha$, although I find it bizarre that $\alpha$ just seems to appear twice. If what I described are indeed elements of the set, I can take a stab at trying to evaluate closure:
Closure under addition
For closure under addition, I'll take two arbitrary elements, let's say $-2\alpha -2\beta$ and $\alpha$.
Then, for there to be closure under addition, $(-2\alpha -2\beta) + (\alpha)) \in S$
$(-2\alpha -2\beta) + (\alpha)) = -\alpha -2\beta \notin S$.
So.. I guess there isn't closure? Because this can't be written in terms of any of the elements they gave me.
Closure under scalar multiplication
For scalar multiplication, I need to see if scaling an element by some constant creates an element still in that set.
Let's pick $\alpha$ and $\beta$ first.
$$\alpha \in S \subset \mathbb R$$ $$c\alpha = \alpha_2 \in S \subset \mathbb R$$
And the same applies to $\beta$.
Now, for the element $(-2\alpha -2\beta)$.
$$c(-2\alpha -2\beta) = -2c\alpha -2c\beta$$ $$c\alpha = \alpha_2$$ $$c\beta = \beta_2$$ $$\therefore c(-2\alpha -2\beta) = -2\alpha_2 - 2\beta_2 \in S$$
Basically, I feel like my way of evaluating closure here is absolutely wrong and I'm looking for some guidance to approach this.
A $4-$tuple is an ordered list of four numbers. It is like an ordered pair, but has four elements instead of two. Here we think of them as coordinates in $\Bbb R^4$. The subset is defined as all those $4-$tuples that can be expressed as $(-2\beta-2\alpha, \alpha, \beta, \alpha)$, so they are all the points where the second and fourth coordinates are the same and the first component is $-2$ times the sum of the second and third. To demonstrate closure under addition, you take two arbitrary elements of the set, add them componentwise, and show that the sum is also in the set. Our two elements might be $(-2b-2a,a,b,a)$ and $(-2d-2c,c,d,c)$. If you add these, what do you get? Can you find $\alpha$ and $\beta$ that show the sum is in the set? For scalar multiplication, you do the same thing. Take an arbitrary element of the set and a arbitrary real number and multiply them, then show that the product is in the set.