I'm trying to prove that, for every field $F$, vector space $V$ over $F$, and linear mapping $T: V \to V$, if $T = T_1 \oplus T_2$, then $$ f(T_1 \oplus T_2) = f(T_1) \oplus f(T_2) $$
Going from RHS, each of the components must be independent, but I can't connect it to LHS.
I'm assuming that $T_i:V_i\rightarrow V_i$ are linear transformations for $i=1,2$ and $T_i^k$ means the composition of $T_i$ with itself $k$ times. Moreover, $V=V_1\oplus V_2$. I understand $T_1\oplus T_2$ to be $T_1\oplus T_2:V\rightarrow V$. In other words, $(v_1,v_2)\mapsto (T_1(v_1),T_2(v_2))$. Let $f=\sum a_ix^i$. Then, \begin{align*} f(T_1\oplus T_2)(v_1,v_2)&=\left(\sum a_i (T_1\oplus T_2)^i\right)(v_1,v_2)\\ &=\sum a_i(T_1\oplus T_2)^i(v_1,v_2)\\ &=\sum a_i\left(T_1^i(v_1),T_2^i(v_2))\right)\\ &=\sum \left(a_iT_1^i(v_1),a_iT_2^i(v_2))\right)\\ &=\left(\sum a_iT_1^i(v_1),\sum a_iT_2^i(v_2))\right)\\ &=\left(f(T_1)(v_1),f(T_2)(v_2)\right)\\ &=(f(T_1)\oplus f(T_2))(v_1,v_2). \end{align*}