Question about the K-Module $\cdot _\varphi : K[X]\times V \to V$

Question

I managed to show several properties about the following mapping:

Let $K$ be a field, $V$ a finite dimensional $K$-Vectorspace and $\varphi \in \text{End}_K(V)$. $$ \cdot_\varphi : \begin{cases}K[X] \times V & \longrightarrow V \\ (P, v) & \longmapsto P(\varphi)(v)= \sum_{i=1}^n a_i\varphi ^i(v) \end{cases} $$

I managed to show three statements (which might be helpful for my missing one):

1. The triple $(K[X], V, \cdot_\varphi)$ defines a $K[X]$-Module, denote it as $V_\varphi$.
2. $U \subset V_{\varphi}$ is a $K[X]$-submodule $\iff$ $U \subset V$ is a $\varphi$-invariant $K$-vector subspace.
3. No element $v \in V$ is free over $K[X]$ (thanks to Cayley-Hamilton).

Problem: Let $\Omega \in \text{End}_K(V)$. Prove that: $$V_\varphi \cong V_\Omega \text{ as } K[X]-\text{Modules} \iff \exists \alpha \in \text{Aut}_K(V) : \varphi = \alpha \circ \Omega \circ \alpha^{-1} $$

For $"\implies"$ I tried to construct such an $\alpha$ with the definition of $\cdot_\varphi$ above but had no luck and for $"\Longleftarrow"$ I am utterly puzzled. I've been scanning my scripts and linear algebra books for useful lemmas but couldn't find one.

I don't need a full solution (if it is as complicated as I suppose it to be) but I would appreciate hints for both directions to get me going.

Answer 1

"$\Rightarrow$": Let $\Phi: V_\varphi \to V_\Omega$ be a module isomorphism. Then by definition of module isomorphisms $$ \forall p, q \in K[x], v, w \in V : \Phi(p \cdot_\varphi v + q \cdot_\varphi w) = p \cdot_\varphi \Phi(v) + q \cdot_\varphi \Phi(w). $$ In particular, for $w = 0$ and $p(x) = x$: $$ \Phi(\varphi(v)) = \theta(\Phi(v)), $$ and thus $\varphi(v) = \Phi^{-1}(\theta(\Phi(v)))$ by applying $\Phi^{-1}$ on both sides, and hence $\alpha = \Phi^{-1}$ is a possible choice, since from the definition of module isomorphisms we can infer the linearity of $\Phi = \alpha^{-1}$ and hence of $\alpha$, since the inverse of a bijective (remember $\Phi$ is an isomorphism) linear mapping is itself linear.

"$\Leftarrow$": Let $\alpha \in \text{Aut}_K(V)$ such that $\varphi = \alpha \circ \vartheta \circ \alpha^{-1}$. We define $\Phi: V_\varphi \to V_\vartheta$ by $$ \Phi(v) := \alpha^{-1}(v) $$ and observe that by the linearity of $\alpha^{-1}$, $$ \Phi \left( \sum_{j=1}^n a_j \varphi^j(v) + \sum_{j=1}^k b_j \varphi^j(v) \right) = \sum_{j=1}^n a_j \Phi (\varphi^j(v)) + \sum_{j=1}^k b_j \Phi (\varphi^j(v)) \\ = \sum_{j=1}^n a_j \alpha^{-1} (\varphi^j(v)) + \sum_{j=1}^k b_j \alpha^{-1} (\varphi^j(v)) \\ = \sum_{j=1}^n a_j (\alpha^{-1} \varphi \alpha)^j(\alpha^{-1}v) + \sum_{j=1}^k b_j (\alpha^{-1} \varphi \alpha)^j(\alpha^{-1}v) \\ = \sum_{j=1}^n a_j (\Omega)^j(\Phi(v)) + \sum_{j=1}^k b_j (\Omega)^j(\Phi(v)), $$ where we used that $$ \alpha^{-1} \varphi^j = \overbrace{(\alpha^{-1} \varphi^j \alpha) (\alpha^{-1} \varphi^j \alpha) \cdots (\alpha^{-1} \varphi^j \alpha)}^{j \text{ times}} \alpha^{-1}. $$