minimal polynomial and invariant subspace

Question

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 $g_1$.

Answer 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.

Answer 2

You already know the minimal polynomial of$~T_i$, call it $m_i$, divides $g_i$. Since $T_i$ is the restriction of$~T$ to$~W_i$ you know that $m_i(T)$ vanishes on$~W_i$. Now letting $g_j=\frac p{g_i}$ be the other factor than $g_i$ (so in fact $j=3-i$), one has $g_j(T_j)=0$, so $g_j(T)$ vanishes on$~W_j$. But then the $m_ig_j(T)= m_i(T)\circ g_j(T)$ vanishes on both $W_i$ and $W_j$, and therefore on the whole space. If $m_i$ were a strict divisor of $g_i$, then $m_ig_j$ would be a strict divisor of $g_ig_j=p$, contradicting that $p$ is minimal polynomial. So $m_i=g_i$.