minimal polynomial and invariant subspace I'm trying to solve the following problem:
Let $T$ be a linear transformation on a finite dimensional vector space $W$. Suppose the minimal polynomial of $T$ is $p=g_1g_2$, where $g_1, g_2$ are relatively prime, show that
(a) $W=W_1 \oplus W_2$, where $W_i=\{\alpha \in V|g_i (T)\alpha=0\}$.
(b) $W_1, W_2$ are $T$-invariant.
(c) If $T_i$ is the restriction of $T$ on $W_i$,  then the minimal polynomial of $T_i$ is $g_i$.
I have solved (a) and (b), basically using the idea that polynomials in $T$ commute with $T$, and that there exist polynomial $r_1, r_2$ such that $g_1r_1+g_2r_2=1$. However, how should I proceed to show (c)? by (a) we have $g_i(T_i)=0$, so the minimal polynomial of $T_i$ divides $g_1$.

... continuing from your last sentence: now assume that the minimal polynomial of $T_1$ is $h_1$. Then for any $w\in W$ we have
$$
h_1(T)g_2(T)w=h_1(T)g_2(T)(w_1+w_2)=g_2(T)\underbrace{h_1(T)w_1}_{=h_1(T_1)w_1=0}+h_1(T)\underbrace{g_2(T)w_2}_{=g_2(T_2)w_2=0}=0,
$$
so $h_1g_2$ annihilates $T$ and then $g_1$ divides $h_1$ too (you have already mentioned that $h_1$ divides $g_1$). It gives that $h_1=g_1$. 
P.S. The argument for $g_2$ is the same.
