$V = V_1\oplus V_2\oplus....\oplus V_k$, show that $f V =f V_1\oplus f V_2\oplus...\oplus f V_k$. Let $T$ be a linear operator on the vector space $V$ over the field $F$. If $f$ is a polynomial over $F$ and $a$ is in $V$, let $fa = f(T)a$. If $V_1, . . . , V_k$ are $T$-invariant sub-spaces and $V = V_1\oplus V_2\oplus....\oplus V_k$, show that $f V =f V_1\oplus f V_2\oplus...\oplus f V_k$.
Let $a = v_1 + v_2 +...+ v_k$ since $V_i's$ are T-invariant we have $Tv_i \in V_i$ and hence $fv_i = f(T)V_i \subset fV_i$. Can we conclujde from here...Please tell if my lolgic is wrong.

You almost completed one part of the question. Since the $V_i$ are $T$-invariant, they are also $f(T)$ invariant and so $fv_i = f(T)v_i \in f(T)(V_i) = fV_i$ which shows that any vector $fa \in fV$ can be written as a sum of vectors from $fV_i$. What is left is to show that the sum is in fact a direct sum, or, in other words, each $fa$ can be represented uniquely as a sum of vectors from $fV_i$. 
To see this, write $0 = w_1 + \dots + w_k$ with $w_i \in fV_i$. Since $V_i$ is $f(T)$-invariant, we have $fV_i \subseteq V_i$ and we also have a representation of $0$ as a sum of elements from $V_i$. Since the $V_i$ form a direct sum, we have $w_i = 0$ for all $i$ which shows that the $fV_i$ also form a direct sum.
