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$S: (1, x^2, 2+x^2)$. I'm not sure how to test if this set (see left) is a subspace of $\mathbb{P}^2$.The book said that only vectors of the form $s(x^2) +t(1)$ are in spans(S). How did they know that? Also, could anyone recommend any other resources I could use online to help me with Linear Algebra? My textbook leaves a lot of open-ended questions I don't have the concepts down well enough to just figure out myself.

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  • $\begingroup$ Don't get thrown off by the fact that these are polynomials instead of more "traditional" vectors... Do you know how to solve the problem if it were instead talking about $\Bbb R^3$ and $\text{Span}\{(1,0,0),(0,0,1),(2,0,1)\}$? $\endgroup$
    – JMoravitz
    Jun 20 '18 at 4:14
  • $\begingroup$ You can write $2+x^2 = 2 \cdot 1 + x^2$, so the span of $1,x^2$ is the same as the span of $1,x^2,2+x^2$. $\endgroup$
    – copper.hat
    Jun 20 '18 at 4:16
  • $\begingroup$ The only thing I think I know how to do at this point is to write the vectors as a matrix and see if it has a determinant or not. $\endgroup$
    – kds
    Jun 20 '18 at 4:38
  • $\begingroup$ Are you asking whether $\{1,x^2,2+x^2\}$ is a subspace of $\mathbb P^2$, or whether the span of the bracketed set is a subspace of $\mathbb P^2$? $\endgroup$
    – Math1000
    Jun 20 '18 at 5:15
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I think you mean to prove $\operatorname{Span}(S)$ is a subspace of $P^2$, in fact, any element $p$ in $\operatorname{Span}(S)$ is $$p=a+bx^2+c(2+x^2)=(a+2c)+(b+c)x^2$$ that is why only vectors of the form $s(x^2)+t(1)$ are in $\operatorname{Span}(S)$.

If you are new to linear algebra, try to prove things by following definition. Like, a subset $W\subset U$ is a subspace of $U$ if and only if $W$ contains zero vector and $u+v\in W, \lambda u\in W$ for $u,v\in W, \lambda\in F$. Try to prove $\operatorname{Span}(S)$ is a subspace of $P^2$ by checking three conditions in the definition.

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  • $\begingroup$ where did the "s" and the 1 in s(x^2) +t(1) come from exactly? How do you interpret (a+2c) + (b+c)x^2 as s(x^2) +t(1)? Not to good a proofs, would prefer something more intuitive. $\endgroup$
    – kds
    Jun 20 '18 at 13:10
  • $\begingroup$ $s$ and $t$ are coefficients of basis vector $x^2$ and $1$, but any polynomial in $\operatorname{Span}(S)$ can be written as $p=(a+2c)\cdot 1+(b+c)\cdot x^2$, i.e., it is just combination of $1$ and $x^2$. If you don't really get it, try to look what vector space consisting of polynomials is, it is a very classical example for vector space. $\endgroup$
    – H-H
    Jun 20 '18 at 16:10

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