Prove the modules isomorphism theorem

Question

$(\Rightarrow)$

Given $V_1\cong V_2$, there exist an isomorphism $\gamma:V_1\to V_2$.

Since $\alpha_1\in \operatorname{End}_k(V_1)$ and $\alpha_2\in \operatorname{End}_k(V_2)$, $\alpha_1:V_1\to V_1$ and $\alpha_2:V_2\to V_2$ are homomorphism.

I am confused how to prove this theorem. How to get $\alpha_1=\gamma^{-1}\alpha_2\gamma$? Is it because $\gamma$ is isomorphism so there exist $\gamma^{-1}$? How to associate $\gamma$ with $\alpha_1$, $\alpha_2$?

Answer 1

We have an isomorphism $\gamma$ which also respects the $k[x]$-module structure.

In $V_i$, the module structure is given by $p\cdot v:=p(\alpha_i)v$ where for a polynomial $p=c_0+c_1x+c_2x^2+\dots$ and a linear transformation $\alpha:V\to V$, $$p(\alpha)=c_0\cdot\mathrm{id}+c_1\cdot \alpha +c_2\cdot\alpha^2+\dots$$ So, in particular, $\gamma$ will satisfy $\gamma(x\cdot v)=x\cdot\gamma(v)$ for every $v\in V$, which appeals to $$\gamma(\alpha_1(v))=\alpha_2(\gamma(v))$$ which leads to the desired connection between $\gamma$ and $\alpha_i$.

See if you can prove the other direction.