# Verify if the following sets are linear subspaces

I'm asked to determine if the following sets are linear subspaces:

$$\{{(x,y)\in\mathbb{R}:x=t^5, y=3t^5,t\in\mathbb{R}}\}$$

and

$$\{{(x,y)\in\mathbb{R}:x=t^2, y=3t^2,t\in\mathbb{R}}\}$$

To verify this, I first check if $$(0,0)$$ belongs to both sets. By setting $$t=0$$, it does.

However, I don't really know how to check for the sum or multiplication by a scalar. For example, for $$t=1$$, in both sets we get (1, 3). However, I don't think $$(1,3)+(1,3)=(2,6)$$ belongs to either set.

My textbook says the first set is a subspace, while the second one isn't. In both cases I guess I could factor the parameter like this: $$t^5(1,3)^{tr}$$ (which happens to be a basis for the first set). Any hints on how to check if they are subspaces?

For the first set: Since the function $$\mathbb R\to\mathbb R: t\mapsto t^5$$ is bijective, the first set is the same as $$\{{(x,y)\in\mathbb{R}:x=t, y=3t,t\in\mathbb{R}}\}.$$
For the second set: Let $$S:=\{{(x,y)\in\mathbb{R}:x=t^2, y=3t^2,t\in\mathbb{R}}\}$$ We have $$(1,3)\in S$$, but do we have $$(-1,-3)\in S$$?
• Proving that any negative $(x,y)$ doesn't belong to $S$. – Cesare Nov 29 '19 at 18:07
• As at @NicholasRoberts pointed out, my counter-example is, even in the most formal context, a valid proof. You can also note that $t^2\geq 0$ for all $t\in\mathbb R$ which means that $$x<0\text{ or } y<0\implies (x,y)\not\in S$$ – User Nov 29 '19 at 18:12