4
$\begingroup$

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 $f_i^{n_i}(t)$?

$\endgroup$
5
$\begingroup$

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 $f_i^{n_i}$.

$\endgroup$
  • $\begingroup$ Great answer. Many thanks. +1. $\endgroup$ – Spenser Apr 22 '13 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.