I'm asked to prove that if $T$ is a linear operator on the vector space $V$, with $V=\bigoplus_{i=1}^k V_i$ ($V_i$ being $T$-invariant) and $f$ is a polynomial over $F$, and we define $f \alpha=f(T)\alpha$ for all $\alpha \in V$, then $fV=fV_1\bigoplus \cdots \bigoplus fV_k$.
I've already proved that $fV=\sum fV_i$ and now I'm trying to show that the sum is direct. For that, I assumed that there were some $\alpha_i \in V_i$ such that $\sum f(T)\alpha_i=0$ and I'm trying to verify that $f(T)\alpha_i =0$ for all $i$. Last equation can be rewritten as $f(T)\sum \alpha_i =0$. I have one question: is there a way to conclude that $f(T)\sum \alpha_i =0$ implies that $\sum \alpha_i=0$? This would mean that $\alpha_i=0$ and therefore $f(T)\alpha_i=0$ and the proof would be complete. Is this the correct approach?
Any kind of hint would be appreciated.
P.S. This is not homework, I'm just trying to learn by myself. This is an exercise from Hoffman & Kunze 2d. edition.
If $W_i$ is a subspace of $V_i$, then the sum $\sum_i W_i\subseteq\bigoplus V_i$ is direct, since if $\sum w_i=0$ for $w_i\in W_i$ then each $w_i=0$ as $w_i\in V_i$ and $\bigoplus V_i$ is direct. Yours is the case where $W_i=fV_i$.