[proofplan]
We first check that each scalar multiplication map is an $F$-linear endomorphism of $V$. We then prove centrality by evaluating both composites $T_aS$ and $ST_a$ on an arbitrary vector and using $F$-linearity of $S$. Finally, we verify directly that $a \mapsto T_a$ preserves addition, scalar multiplication, multiplication, and the unit, and we prove injectivity in the nonzero case by evaluating at one nonzero vector.
[/proofplan]
[step:Check that scalar multiplication maps are $F$-linear endomorphisms]
Fix $a \in F$. The map $T_a: V \to V$ is defined by $T_a(v)=av$ for every $v \in V$. Let $v,w \in V$ and let $c \in F$. By the [vector space](/page/Vector%20Space) distributive and associativity axioms,
\begin{align*}T_a(v+w)=a(v+w)=av+aw=T_a(v)+T_a(w).\end{align*}
Also,
\begin{align*}
T_a(cv)=a(cv)=(ac)v=(ca)v=c(av)=cT_a(v),
\end{align*}
where commutativity of multiplication in the field $F$ is used in the equality $ac=ca$. Hence $T_a$ is $F$-linear, so $T_a \in \operatorname{End}_F(V)$.
[/step]
[step:Show that each scalar endomorphism commutes with every endomorphism]
Fix $a \in F$ and $S \in \operatorname{End}_F(V)$. For every $v \in V$,
\begin{align*}
(T_aS)(v)=T_a(S(v))=aS(v).
\end{align*}
Since $S$ is $F$-linear,
\begin{align*}
(ST_a)(v)=S(T_a(v))=S(av)=aS(v).
\end{align*}
Therefore $(T_aS)(v)=(ST_a)(v)$ for every $v \in V$, and hence $T_aS=ST_a$ as maps $V \to V$. Thus every scalar endomorphism lies in the center of $\operatorname{End}_F(V)$.
[guided]
Fix $a \in F$ and let $S \in \operatorname{End}_F(V)$. To prove that $T_a$ is central, we must prove equality of two endomorphisms $T_aS: V \to V$ and $ST_a: V \to V$. Equality of maps is checked pointwise, so let $v \in V$.
First compute the composite in which $S$ is applied before $T_a$:
\begin{align*}
(T_aS)(v)=T_a(S(v))=aS(v).
\end{align*}
Now compute the composite in the other order:
\begin{align*}(ST_a)(v)=S(T_a(v))=S(av).\end{align*}
The defining property of $S \in \operatorname{End}_F(V)$ is $F$-linearity. Applying that property to the scalar $a \in F$ and the vector $v \in V$ gives
\begin{align*}S(av)=aS(v).\end{align*}
Thus
\begin{align*}(ST_a)(v)=aS(v)=(T_aS)(v).\end{align*}
Since this equality holds for every $v \in V$, the two composites are equal as maps. Hence $T_aS=ST_a$, so $T_a$ commutes with every element of $\operatorname{End}_F(V)$.
[/guided]
[/step]
[step:Verify that the scalar assignment is a unital $F$-algebra homomorphism]
Define the map $\Phi: F \to \operatorname{End}_F(V)$ by $\Phi(a)=T_a$ for every $a \in F$. The codomain is an $F$-algebra under pointwise addition, pointwise scalar multiplication, and composition as multiplication, as stated in the theorem.
Let $a,b,c \in F$. For every $v \in V$,
\begin{align*}T_{a+b}(v)=(a+b)v=av+bv=(T_a+T_b)(v),\end{align*}
so $\Phi(a+b)=\Phi(a)+\Phi(b)$. Also,
\begin{align*}T_{ca}(v)=(ca)v=c(av)=(cT_a)(v),\end{align*}
so $\Phi(ca)=c\Phi(a)$. For multiplication, using composition in $\operatorname{End}_F(V)$,
\begin{align*}(T_aT_b)(v)=T_a(T_b(v))=T_a(bv)=a(bv)=(ab)v=T_{ab}(v),\end{align*}
so $\Phi(ab)=\Phi(a)\Phi(b)$. Finally,
\begin{align*}T_1(v)=1v=v=\operatorname{id}_V(v),\end{align*}
so $\Phi(1)=\operatorname{id}_V$. Therefore $\Phi$ is a unital homomorphism of $F$-algebras.
[/step]
[step:Prove injectivity when $V$ is nonzero]
Assume $V \neq \{0\}$. Choose $v_0 \in V$ with $v_0 \neq 0$. Let $a,b \in F$ and suppose $\Phi(a)=\Phi(b)$. Then $T_a=T_b$, so evaluating at $v_0$ gives
\begin{align*}av_0=bv_0.\end{align*}
Subtracting $bv_0$ from both sides gives
\begin{align*}(a-b)v_0=0.\end{align*}
If $a-b \neq 0$, then $a-b$ has an inverse in the field $F$, and multiplying by $(a-b)^{-1}$ gives
\begin{align*}v_0=(a-b)^{-1}((a-b)v_0)=(a-b)^{-1}0=0,\end{align*}
contradicting the choice of $v_0$. Hence $a-b=0$, so $a=b$. Thus $\Phi$ is injective.
[/step]
[step:Handle the zero vector space and identify the central copy]
If $V=\{0\}$, then there is exactly one function $V \to V$, namely the identity map on the singleton set $\{0\}$. Hence $T_a=T_b$ for all $a,b \in F$.
If $V \neq \{0\}$, the preceding steps show that $\Phi: F \to \operatorname{End}_F(V)$ is an injective unital $F$-algebra homomorphism, and its image is
\begin{align*}\Phi(F)=\{T_a : a \in F\}.\end{align*}
The centrality step shows that every element of $\Phi(F)$ commutes with every element of $\operatorname{End}_F(V)$. Therefore $\Phi(F)$ is an $F$-subalgebra of the center of $\operatorname{End}_F(V)$, and since $\Phi$ is injective, this subalgebra is isomorphic to $F$.
[/step]
