# Relation between Jordan Normal Form and cyclic modules

I've just started reading about the relation between cyclic modules and Jordan Normal Form and, being honest, I've quite a doubt. The text I am using says that "clearly", the following assumption is true:

Let $$A=K[T]$$, where $$K$$ is a field and $$M=K^n$$ is a cyclic $$A$$-module and $$\phi: M \rightarrow M$$ such that $$\phi(\sum a_iv_i) = U.(a_i)$$, where $$(a_i)$$ is a row vector and $$U$$ is a $$r\times r$$ matrix with coefficients in K, then $$U$$ has a jordan normal form with only one Jordan block.

Could, anyone, help me understanding why this is to "trivial"?

• Sorry, I don't understand your question. If $\phi=id_M$ the matrix would be $I_{n\times n}$ and there would be $n$ different Jordan blocks. Commented May 31, 2020 at 16:38
• @NotPhiQuadro By "only one Jordan block" I mean that $U$ is conjugated to a matrix with $\lambda$ in the diagonal and $1$ in the line imediately below, for $\lambda \in K$ Commented May 31, 2020 at 16:44

We know that, by the proof of Theorem (Jordan Normal Form) that, for an endomorphism $$\varphi: M \rightarrow M$$, where $$\varphi$$ acts as $$T$$, $$\varphi$$ leaves the submodules $$M_i = K[T]/\langle (T-\lambda_i)^{e_i})\rangle$$ of $$M$$ invariant. Thus, to find the Jordan Normal Form we can concentrate on finding appropriate matrix representations $$U_i$$ for each restriction $$\varphi_i: M_i \rightarrow M_i$$ of $$\varphi$$ up to the submodules $$M_i$$ of $$M$$. Then, putting the bases $$\mathcal{B}_i = \{ [1],[T -\lambda_i], ... , [(T-\lambda_i)^{e_i-1}]\}$$ of $$M_i$$ together to a basis $$\mathcal{B} = \bigcup \mathcal{B}_i$$ of $$M$$, yields to a matrix representation for $$\varphi$$ as desired.
Since $$U$$ has only one Jordan block, there is only one $$\varphi := \varphi_1: M \rightarrow M$$ and only one eigenvalue $$\lambda$$, such that that $$\mathcal{B} = \{ [1],[T-\lambda], ... , [(T-\lambda)^{e})]\}$$ is a basis of $$M$$. By the other hand, note that $$\mathcal{B} = \langle T-\lambda \rangle$$ and, thus, $$M$$ is cyclic, as desired.