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Suppose $T: P_5(\Bbb R) \to P_5(\Bbb R)$ is a linear transformation, where $P_5(\Bbb R)$ denotes the vector space of all polynomials over the reals such that their highest degree cannot exceed $5$. Also suppose that $\operatorname{null}(T')= \operatorname{span}(R)$, where $T'$ denotes the dual map of $T$ and $R$ is the linear functional on $P_5(\Bbb R)$ defined by $R(p)= p(8)$. Prove that $\operatorname{range}T = \{p\in P_5(\Bbb R) | p(8)=0\}$.

Here is my reasoning:

We know that $\operatorname{null}R = \{p\in P_5(\mathbb R) | R(p)=0\}$. By the definition of $R$, we have $\operatorname{null}R = \{p\in P_5(\Bbb R) | p(8)=0\}$. Now we just need to show that $\operatorname{range}T$ is equal to $\operatorname{null}R$. To do that, I have also managed to show that these two spaces have the same dimension. But it remains to show that $\operatorname{range}T$ is a subspace of $\operatorname{null}R$. How do I show this? Thanks so much.

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You need to show that for every polynomial $p\in P_5(\Bbb R)$, $T(p)\in \operatorname{null}R$, or equivalently, $R(T(p))=0$. By definition $R(T(p))=(T'(R)) (p)$, and since $R\in \operatorname{null}T'$, $T'(R)=0$; hence $R(T(p))=0$ for all $p$.

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