# Relation between Jordan normal form and cyclic module decomposition

Let $$V$$ be a vector space over $$\mathbb{C}$$, we consider it as a $$\mathbb{C}[x]$$-module by choosing a linear map $$\varphi:V\rightarrow V$$ and for $$f(x)\in\mathbb{C}[x]$$ and $$v\in V$$, define: $$f\cdot v=f(\varphi(v))$$ Suppose we know that $$V$$ is written as direct sum of cyclic modules as follows: $$V=\mathbb{C}[x]/\langle g_1(x)\rangle\oplus \cdots\oplus \mathbb{C}[x]/\langle g_r(x)\rangle \tag{*}$$ where $$g_i(x)$$'s are the annihilators of the cyclic modules.

My question is: Do $$g_1(x)\cdots g_r(x)$$ tell us anything about the Jordan blocks of the map $$\varphi$$? What about the minimal and characteristic polynomials? Furthermore, if we know the Jordan form of $$\varphi$$, can we get the decomposition as ($$*$$) directly?

• I guess you really mean $f\cdot v=f(\varphi)(v)$, that is, first compute the polynomial $f$ on $\varphi$, and then apply the result to the vector $v$. – Andreas Caranti Mar 26 '19 at 15:24

The characteristic polynomial will be the product of the $$g_{i}$$.
The minimal polynomial is slightly more complicated. First note that the decomposition (*) is not unique. But it becomes unique if you insist on the $$g_{i}$$ being monic, and require the divisibility conditions $$g_{1} \mid g_{2} \mid \dots \mid g_{r}.$$ Then $$g_{r}$$ is your minimal polynomial.
Coming to Jordan, you can also write (*) (recall, it is not unique) so that each $$g_{i}$$ is of the form $$g_{i}(x) = (x -\lambda_{i})^{e_{i}}$$. Then each such $$g_{i}$$ corresponds to a Jordan block of size $$e_{i}$$, with eigenvalue $$\lambda_{i}$$. This should also answer your last question.