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I'm confused on how these substitutions are occurring. On the first line, it looks like $p(t)=p(t)+q(t)$ and $q(t)=t^2[p(t)+q(t)]$. Then on the second line, it looks like they're subbing $q(t)$ for $t^2p(t)$.

Let $T : \mathbb{P}_2$ -> $\mathbb{P}_4$ be the transformation that maps a polynomial $p(t)$ into the polynomial $p(t) + t^2 p(t)$. Show that $T$ is a linear transformation.

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stumped
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  • They are not subbing $q(t)$ with $t^2[p(t)+q(t)]$, they are treating $p(t)+q(t)$ as $p(t)$ in the general case, and applying $T$ to it. – 高田航 Mar 30 '17 at 03:10
  • They aren't substitutions. The function $T$ maps a polynomial to a different polynomial, namely itself plus $t^2$ times itself. The first line is just applying this rule to the polynomial $p+q$. The second line is just rearranging things. – symplectomorphic Mar 30 '17 at 03:10
  • The notation is horrible & misleading. It would be better to write $(T(p))(t)$ instead of $T(p(t))$. – copper.hat Mar 30 '17 at 05:20

1 Answers1

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\begin{eqnarray} (T(p+q))(t) &=& (1+t^2)((p+q)(t)) \\ &=& (1+t^2)(p(t)+q(t)) \\ &=& (1+t^2) p(t) + (1+t^2) q(t)\\ &=& (T(p))(t)+(T(q))(t) \end{eqnarray} And so $T(p+q) = T(p)+T(q)$.

Similarly, \begin{eqnarray} (T(cp)(t)) &=& (1+t^2)((cp)(t)) \\ &=& (1+t^2)(c p(t)) \\ &=& c (1+t^2) p(t) \\ &=& c (T(p))(t) \end{eqnarray} And so $T(cp) = c T(p)$.

copper.hat
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