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"?

  • $\begingroup$ 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. $\endgroup$
    – Sant97
    Commented May 31, 2020 at 16:38
  • $\begingroup$ @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 $\lambda \in K$ $\endgroup$
    – Anyway142
    Commented May 31, 2020 at 16:44

1 Answer 1


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.


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