TITLE: Primary Decomposition Theorem: What are the characteristic polynomials of the maps of the decomposition?
QUESTION [5 upvotes]: Let $T$ be an operator on a finite dimensional vector space $V$. Suppose that the characteristic polynomial of $T$ is
$$\chi(t)=f_1^{n_1}(t)\cdots f_k^{n_k}(t)$$
where $f_1,\ldots,f_k$ are distinct irreducible polynomials, and suppose the minimal polynomial of $T$ is 
$$m(t)=f_1^{m_1}(t)\cdots f_k^{m_k}(t)$$
Then, by the Primary Decomposition Theorem, we get that
$$V=W_1\oplus\cdots\oplus W_k$$
where $W_i:=\ker(f_i^{m_i}(T))$. Moreover, we have 
$$T=T_1\oplus\cdots\oplus T_k$$
where $T_i:=T|_{W_i}$, and the minimal polynomial of $T_i$ is $f_i^{m_i}(t)$.
Now, here is my Question: How can we show that the characteristic polynomial of $T_i$ is $f_i^{n_i}(t)$?

REPLY [6 votes]: The characteristic polynomial $\chi$ must be the product of the characteristic polynomials of the $T_i$. Moreover each irreducible factor $f_i$ of $\chi$ is relatively prime to the minimal polynomial of each $T_j$ with $j\neq i$, and therefore to its characteristic polynomial; it must therefore by present only in the characteristic polynomial of $T_i$. Since all factors $f_i$ end up there, and none of the other $f_j$, the characteristic polynomial of $T_i$ must be $f_i^{n_i}$.