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for my math class, we have a problem set and the following question appeared.

Let $P_3 = \{p(t) = a+bt+ct^2+dt^3\}$ be the vector space of polynomials of degree of at most 3. Determine which (if any) of the following subsets are subspaces.

  1. $p(t) = a+bt+ct^2+dt^3$ with $a = 0$.
  2. $p(t) = a+bt+ct^2+dt^3$ with $a + b + c + d= 0$.
  3. $p(t) = a+bt+ct^2+dt^3$ with $a, b, c, d$ integers.
  4. $p(t) = a+bt$ with $a + 2b = 0$.

For the first one, I said that it is not a valid subspace, because it is not closed under addition. Now, this is where I am stuck. If that is the case (I understand it may not be), wouldn't all of them not be subspaces?

Thanks for the help!

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  • $\begingroup$ What makes you say the first one isn't closed under addition? If you have a $p_1(t)=a_1+b_1t+c_1t^2+d_1t^3$ and you have a $p_2(t)=a_2+b_2t+c_2t^2+d_2t^3$, then what is $p_1(t)+p_2(t)$? What is the constant term? Since $a_1=0$ and $a_2=0$ what is $a_1+a_2$? $\endgroup$
    – JMoravitz
    Mar 4, 2016 at 5:15
  • $\begingroup$ Alright, I thought that because I thought we were defining addition as p(x) + p(y), which would not be equal to p(x+y), because, for example, cx^2 + cy^2 is not equal to c(x+y)^2. I understand that would be closed under addition now. $\endgroup$
    – vsav
    Mar 4, 2016 at 5:18
  • $\begingroup$ Also to check: the zero vector is an element of the subspace (the zero polynomial in this case) and that it is closed under scalar multiplication (presumably the scalar field in this question is $\mathbb{R}$ or $\mathbb{C}$). In general, one could check all properties simultaneously by seeing if for any $p_1(t)$ and $p_2(t)$ and scalar $\alpha$ that $\alpha p_1(t)+p_2(t)$ is also an element. Remember that addition is component-wise. The parent vector space here is isomorphic to $\mathbb{R}^4$. $\endgroup$
    – JMoravitz
    Mar 4, 2016 at 5:19
  • $\begingroup$ So, the first would be a valid subspace correct? $\endgroup$
    – vsav
    Mar 4, 2016 at 5:21
  • $\begingroup$ I'll let you figure that out, but you seem to be on the right track. From what I see exactly one is not a subspace and the other three are. Presumably your work needs to be shown for full credit so it remains to be proven using the properties given. $\endgroup$
    – JMoravitz
    Mar 4, 2016 at 5:25

1 Answer 1

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Note that 1,2 &4 can be written as $S_\alpha = \{ p | p(\alpha) = 0 \}$ for some fixed $\alpha$.

The map $L_\alpha:P_3 \to \mathbb{R}$ defined by $Lp = p(\alpha)$ is linear and $S_\alpha = \ker L_\alpha$.

Since the kernel of a linear operator is a subspace, it follows that the $S_\alpha$ are subspaces.

For the third, consider a non zero polynomial $p$ with integer coefficients. Is $\sqrt{2} p$ a member of this set?

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