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Suppose that $T(\alpha)=c\alpha$. If $f$ is any polynomial, then $f(T)\alpha= f(c)\alpha$.

My attempt: Let $f=\sum_{i=0}^na_i\cdot x^i$ be polynomial over $F$. Then $f(T)=\sum_{i=0}^na_i\cdot T^i$. So $[f(T)](\alpha)=(\sum_{i=0}^na_i\cdot T^i)(\alpha)=\sum_{i=0}^n a_i\cdot T^i(\alpha)$. We claim $\forall 1\leq j\leq n$, $P(j): T^j(\alpha)=c^j\alpha$. Base case: $j=1$. By hypothesis, $T(\alpha)=c\alpha$. Inductive step: Suppose $T^k(\alpha)=c^k\alpha$, for some $1\lt k\lt n$. Since $T$ is linear map, we have $T^{k+1}(\alpha)=T(T^k(\alpha))=T(c^k\alpha)=c^kT(\alpha)=c^k(c\alpha)=c^{k+1}\alpha$. Thus $T^{k+1}(\alpha)=c^{k+1}\alpha$. By mathematical induction, $P(j)$ holds $\forall 1\leq j\leq n$. By distributive property, $\sum_{i=0}^n a_i\cdot T^i(\alpha)= \sum_{i=0}^n a_i\cdot c^i\alpha= (\sum_{i=0}^n a_i\cdot c^i)\cdot \alpha =f(c)\cdot \alpha$. Hence $[f(T)](\alpha)=f(c)\cdot \alpha$. Is my proof correct?

user264745
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    That looks good. Just a comment; you need not have restricted your $k$ to be between $1$ and $n$. $n$, over here, is the degree of the polynomial. But you are just trying to prove that $T^m(\alpha) = c^m \alpha$ for every $m \in \mathbb{N}$ and this is true irrespective of the degree of the polynomial. – Abhijeet Vats Oct 30 '22 at 17:12
  • @AbhijeetVats yeah, that’s a good point. I had tunnel vision. – user264745 Oct 30 '22 at 17:16

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