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$U = P(R)$ 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.

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

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – eddie
    Commented Sep 26, 2021 at 0:31
  • $\begingroup$ If $p(t)\in M$, then $p(0)>0$. Therefore, $-p(0)<0$. So, $-p(t)\notin M$. $\endgroup$ Commented Sep 26, 2021 at 20:26

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