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

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


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

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 $(-1,-3)\in S$?

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  • $\begingroup$ Thank you! Is there a way to formally prove this without recurring to examples? Or maybe if the Professor asks to "determine" and not give proof that's enough. $\endgroup$ – Cesare Nov 29 '19 at 18:06
  • $\begingroup$ Proving what exactly? Something about the first or something about the second set? $\endgroup$ – User Nov 29 '19 at 18:06
  • $\begingroup$ Proving that any negative $(x,y)$ doesn't belong to $S$. $\endgroup$ – Cesare Nov 29 '19 at 18:07
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    $\begingroup$ It suffices. A subspace must have additive inverses by definition. Maximilian has produced an element in S without an additive inverse, so S is not a subspace. $\endgroup$ – Nicholas Roberts Nov 29 '19 at 18:08
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    $\begingroup$ 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 $$x<0\text{ or } y<0\implies (x,y)\not\in S$$ $\endgroup$ – User Nov 29 '19 at 18:12

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