[proofplan] We first verify that every power $T^k$ is an $F$-linear endomorphism of $V$, using the identity map for $k=0$ and induction on $k$ for the positive powers. Once this is known, each scalar multiple $a_kT^k$ is again an $F$-[linear map](/page/Linear%20Map) from $V$ to $V$. Finally, we prove linearity of the finite sum defining $p(T)$ directly by evaluating it on $\alpha v+\beta w$ and distributing through the sum. [/proofplan]

[step:Show that every power $T^k$ is an $F$-linear endomorphism]For $k=0$, the map $T^0=\operatorname{id}_V:V\to V$ is $F$-linear, since for all $\alpha,\beta\in F$ and all $v,w\in V$, \begin{align*} \operatorname{id}_V(\alpha v+\beta w)=\alpha v+\beta w=\alpha\operatorname{id}_V(v)+\beta\operatorname{id}_V(w). \end{align*} We prove by induction that $T^k\in \operatorname{End}_F(V)$ for every integer $k\geq 0$. The case $k=0$ was just proved. Assume that $T^k:V\to V$ is $F$-linear for some integer $k\geq 0$. By definition, $T^{k+1}=T\circ T^k:V\to V$. For all $\alpha,\beta\in F$ and all $v,w\in V$, the linearity of $T^k$ and then the linearity of $T$ give \begin{align*} T^{k+1}(\alpha v+\beta w)=T(T^k(\alpha v+\beta w))=T(\alpha T^k(v)+\beta T^k(w)). \end{align*} Applying linearity of $T$ to the last expression gives \begin{align*} T^{k+1}(\alpha v+\beta w)=\alpha T(T^k(v))+\beta T(T^k(w))=\alpha T^{k+1}(v)+\beta T^{k+1}(w). \end{align*} Thus $T^{k+1}$ is $F$-linear. By induction, $T^k\in \operatorname{End}_F(V)$ for every integer $k\geq 0$.[/step]

[guided]The first point is that the expression $p(T)$ only makes sense as an endomorphism-valued expression if every power $T^k$ is itself a map from $V$ to $V$ and is $F$-linear. We start with $k=0$. By definition, $T^0=\operatorname{id}_V:V\to V$. For arbitrary scalars $\alpha,\beta\in F$ and arbitrary vectors $v,w\in V$, the identity map satisfies \begin{align*} \operatorname{id}_V(\alpha v+\beta w)=\alpha v+\beta w. \end{align*} Since $\operatorname{id}_V(v)=v$ and $\operatorname{id}_V(w)=w$, this becomes \begin{align*} \operatorname{id}_V(\alpha v+\beta w)=\alpha\operatorname{id}_V(v)+\beta\operatorname{id}_V(w). \end{align*} Therefore $T^0=\operatorname{id}_V$ is $F$-linear. Now fix an integer $k\geq 0$ and assume as the induction hypothesis that $T^k:V\to V$ is $F$-linear. The next power is defined by composition: \begin{align*} T^{k+1}=T\circ T^k. \end{align*} This is a map $V\to V$ because $T^k$ maps $V$ into $V$ and $T$ also maps $V$ into $V$. To prove that $T^{k+1}$ is $F$-linear, take arbitrary $\alpha,\beta\in F$ and arbitrary $v,w\in V$. First use the definition of composition: \begin{align*} T^{k+1}(\alpha v+\beta w)=T(T^k(\alpha v+\beta w)). \end{align*} The induction hypothesis says that $T^k$ is $F$-linear, so \begin{align*} T^k(\alpha v+\beta w)=\alpha T^k(v)+\beta T^k(w). \end{align*} Substituting this into the previous expression gives \begin{align*} T^{k+1}(\alpha v+\beta w)=T(\alpha T^k(v)+\beta T^k(w)). \end{align*} Since $T$ is $F$-linear by the hypothesis $T\in\operatorname{End}_F(V)$, we may distribute $T$ across this linear combination: \begin{align*} T(\alpha T^k(v)+\beta T^k(w))=\alpha T(T^k(v))+\beta T(T^k(w)). \end{align*} Using again the definition $T^{k+1}=T\circ T^k$, this becomes \begin{align*} T^{k+1}(\alpha v+\beta w)=\alpha T^{k+1}(v)+\beta T^{k+1}(w). \end{align*} Thus $T^{k+1}$ is $F$-linear. The induction proves that $T^k$ is an $F$-linear endomorphism of $V$ for every integer $k\geq 0$.[/guided]

[step:Check that each scalar multiple $a_kT^k$ is $F$-linear] For each integer $k$ with $0\leq k\leq m$, define \begin{align*} S_k:V&\to V, & v&\mapsto a_kT^k(v). \end{align*} Since $T^k$ is $F$-linear by the previous step, for all $\alpha,\beta\in F$ and all $v,w\in V$, \begin{align*} S_k(\alpha v+\beta w)=a_kT^k(\alpha v+\beta w). \end{align*} Using linearity of $T^k$ and distributivity in the [vector space](/page/Vector%20Space) $V$, we obtain \begin{align*} S_k(\alpha v+\beta w)=a_k(\alpha T^k(v)+\beta T^k(w))=\alpha a_kT^k(v)+\beta a_kT^k(w)=\alpha S_k(v)+\beta S_k(w). \end{align*} Thus $S_k$ is $F$-linear for each $k$ with $0\leq k\leq m$. [/step]

[step:Compute the finite sum to prove that $p(T)$ is linear] Let $\alpha,\beta\in F$ and let $v,w\in V$. By definition of $p(T)$, \begin{align*} p(T)(\alpha v+\beta w)=\sum_{k=0}^{m}a_kT^k(\alpha v+\beta w). \end{align*} Since each $T^k$ is $F$-linear, this becomes \begin{align*} p(T)(\alpha v+\beta w)=\sum_{k=0}^{m}a_k(\alpha T^k(v)+\beta T^k(w)). \end{align*} Using distributivity in $V$ and the fact that finite sums in a vector space may be separated term by term, we get \begin{align*} p(T)(\alpha v+\beta w)=\alpha\sum_{k=0}^{m}a_kT^k(v)+\beta\sum_{k=0}^{m}a_kT^k(w). \end{align*} By the definition of $p(T)$ on $v$ and on $w$, the right-hand side is \begin{align*} \alpha p(T)(v)+\beta p(T)(w). \end{align*} Therefore \begin{align*} p(T)(\alpha v+\beta w)=\alpha p(T)(v)+\beta p(T)(w). \end{align*} Because $\alpha,\beta\in F$ and $v,w\in V$ were arbitrary, $p(T):V\to V$ is $F$-linear. Hence $p(T)\in \operatorname{End}_F(V)$. [/step]