Let $K$ be a field and define two matrices:
\begin{align*}
A:= \begin{pmatrix} 1&1\\ 0&3\end{pmatrix} \text{ and } B:= \begin{pmatrix} 1&0\\ 0&3\end{pmatrix}
\end{align*}
We then define a $K[X]$-module on $V=K^2$ such that
\begin{align*}
K[X]\times V \rightarrow V, \quad (P,v)\mapsto P(A)\cdot v
\end{align*}
where $P(A)\in K^{2\times 2}$ is achieved by plugging in matrix $A$ into polynomial $P$ and $P(A)\cdot v$ is therefore the matrix-vector-multiplication. We call this module $V_A$. $V_B$ can be constructed similarly.
I want to show that for $K=\mathbb{Q}$ the modules $V_A,V_B$ are isomorphic.
My attempt:
This is a lot like the similarity problems for matrices. So I calculated
$$S=\begin{pmatrix} 1&1\\ 0&2\end{pmatrix}$$
so that
$$B = S^{-1}AS$$
Can I then define following? $$\phi: V_B\rightarrow V_A, \quad P(B)v \mapsto S^{-1}P(A)Sv$$
I guess the rest is just showing that this is an module-isomorphism.
\begin{align*}
&1.\quad \phi(B(v_1+v_2)) = S^{-1}AS (v_1+v_2) = S^{-1}ASv_1+ S^{-1}ASv_2 = \phi(Bv_1)+\phi(Bv_2)\\
&2.\quad \phi(r\cdot Bv)=rS^{-1}ASv = r\cdot \phi(Bv)
\end{align*}
for $v,v_1,v_2\in V$
Surjective: Let $P(A)v\in V_A$ for any $P\in K[X], v\in V$. Then it's easy to find the corresponding $\phi(SP(B)S^{-1}v)=P(A)v$
Injective: Assume $P_1(A)v=P_2(A)v$. Then $SP_1(B)S^{-1}v = SP_2(B)S^{-1}v \Rightarrow P_1(B) = P_2(B) $
Is that correct? My prof asked me to be especially careful with the notation since Linear Algebra and Modules are kinda fused here. Any remarks?
Thanks in advance for any help.
Best Answer
Actually one can define $\phi:V_B\to V_A$ by $\phi(v)=Sv$. This map is bijective and clearly satisfies $\phi(v_1+v_2)=\phi(v_1)+\phi(v_2)$.
We have to show now that $\phi(P(X)v)=P(X)\phi(v)$. But $$\phi(P(X)v)=\phi(P(B)v)=S(P(B)v)=(SP(B))v=(P(A)S)v=P(A)(Sv)=P(X)\phi(v).$$