Jordan Canonical Theorem stated that:
Let $K$ be an algebra closed field. Let $V$ be a nonzero, finite dimensional vector space over $K$, and let $\psi \in \operatorname{End}_K(V)$. Then there exises a basis $B$ for V relative to which the matrix of $\psi$ is in Jordan canonical form. Moreover, the Jordan canonical form matrix which represents $\psi$ in this wa is uniquely determined by $\psi$ apart from the order in which the elementary Jordan matrices appear on the "diagonal".
I tried to prove this problem as the following:
- Prove that $V = V_1 \oplus V_2 \oplus \dots \oplus V_n$, where $V_i = \left \{ x \in V \text { s.t } \left (\psi - \lambda _i\right)^{k_i} = 0 \text { for some } k_i\right\}$
- Each $V_i$ is a finitely generated module over $K[X]$
- $V \cong K[X] \Big/ K[X](X-\lambda _1)^{k_1} \oplus K[X] \Big/ K[X](X-\lambda _2)^{k_2} \oplus \dots \oplus K[X] \Big/ K[X](X-\lambda _n)^{k_n}$.
- Each $K[X] \Big/ K[X](X-\lambda _i)^{k_i}$ is cyclic, so it has a basis for $V$ such that the matrix is Jordan.
In my proof, I confuse a little bit.
Is this true that each $X-\lambda _i$ is an irreducible element of $V$?
Is this true that each $K[X] \Big/ K[X](X-\lambda _i)^{k_i}$ is cyclic?
Is my proof true in general?
Thank you for your time.