# Vector Subspace test for set with polynomial

$$U = P(R)$$ and $$M =$${ $$p(t) ∈ P_2 : p(0) > 0$$ }. Is $$M$$ a subspace of $$U$$? I'm a bit stuck here because I'm not quite sure if I'm applying the test correctly.

Check for Zero Vector:

$$0$$ in $$U$$ is defined by $$O = 0 + 0t+0t^2+ ... +0t^n$$. And $$0$$ in $$M$$ is given by $$p(0) = p_0+p_1t+p_2t^2$$, where $$t=0$$, so $$p(0) = p_0 > 0$$ as given by the definition of $$M$$. So since $$p_0 > 0$$, does that mean that zero vector of $$U$$ is not in $$M$$? So $$M$$ is not subspace? I'm doubting this reasoning a bit, since just because $$p(0)>0$$, does not directly mean there isn't some other $$t$$ where $$p(t) = 0$$. I'm not quite how to show that though.

Check for closure under addition and multiplication: Suppose $$p,qEM$$.

So, $$p(t) = p_0+p_1t+p_2t^2$$ and $$q(t) = q_0+p_1t+q_2t^2$$.

Consider $$p+q$$:

$$p(t) +q(t) = (p_0+p_1t+p_2t^2) + ( q_0+p_1t+q_2t^2)$$

$$= (p_0+q_0) + (p_1t+q_1t) + (p_2t^2+q_2t^2)$$

$$= (p_0+q_0) + t(p_1+q_1) + t^2(p_2+q_2)$$

And I get stuck here, I'm not sure how to finish this off.

Closure under multiplication:

Let $$aER$$.

So, $$ap(t) = ap_0+ap_1t+ap_2t^2$$ And I'm stuck here again on how to finish it off here.

I'm not sure if $$M$$ is subspace or if my proofs are even in the right direction.

• Does the identically zero polynomial $p\equiv 0$ belong to $M$? Commented Sep 25, 2021 at 21:18
• @ÁtilaCorreia I'm not sure, that's all the info I have. Commented Sep 25, 2021 at 21:21
• As @JoséCarlosSantos has pointed out, the null vector ($p\equiv 0$) does not belong to $M$. So $M$ cannot be a vector subspace of $U$. Commented Sep 25, 2021 at 21:22

As you wrote, the null polynomial in $$P_2(\Bbb R)$$ is the polynomial $$0+0\times t+0\times t^2$$. And this polynomial maps $$0$$ into $$0$$. So, it does not belong to $$M$$, and therefore $$M$$ is not a subspace.
It is also easy to see that if $$p(t)\in M$$, then $$-p(t)\notin M$$. This also proves that $$M$$ is not a subspace.
• I don't completely understand how $p(t)∈M$ , then $−p(t)∉M$. And how it proves $M$ is not a subspace. Is it because $p(t)>0$, so $-p(t)$ will be less than $0$? How would that proof work? Commented Sep 26, 2021 at 0:31
• If $p(t)\in M$, then $p(0)>0$. Therefore, $-p(0)<0$. So, $-p(t)\notin M$. Commented Sep 26, 2021 at 20:26