If $a\mathbf{x}+b\mathbf{y}$ is an element of the non-empty subset $W$, then $W$ is a subspace of $V$ Okay, so my text required me to actually prove both sides; The non-empty subset $W$ is a subspace of a $V$ if and only if $a\mathbf{x}+b\mathbf{y} \in W$ for all scalars $a,b$ and all vectors $\mathbf{x},\mathbf{y} \in W$. I figured out one direction already (that if $W$ is a subspace, then $a\mathbf{x}+b\mathbf{y}$ is an element of $W$ since $a\mathbf{x}$ and $b\mathbf{y}$ are in $W$ and thus so is their sum), but I'm stuck on the other direction.
I got that if $a\mathbf{x}+b\mathbf{y} \in W$, then $c(a\mathbf{x}+b\mathbf{y}) \in W$ as well since we can let $a' = ca$ and $b' = cb$ and we're good, so $W$ is closed under scalar multiplication. But for closure under addition, my text states that I can "cleverly choose specific values for $a$ and $b$" such that $W$ is closed under addition as well but I cannot find any values that would work. What I'm mostly confused about is how choosing specific values for $a$ and $b$ would prove anything, since $a, b$ can be any scalars and $\mathbf{x},\mathbf{y}$ can be any vectors, so setting conditions like $a = b$, $a = -b$, $a = 0$ or $b = 0$ don't seem to prove anything.
Also something I'm not sure about is if they're saying that $a\mathbf{x}+b\mathbf{y} \in W$, am I to assume that that is the only form? So if I'm testing for closure under addition, I have to do something like $(a\mathbf{x}+b\mathbf{y})+(c\mathbf{z}+d\mathbf{w})$?

Consider $a=b=1$ and $a=-b=1$.
