Let $k$ be an algebraically closed field. Any $k$-linear map $\phi:k^n\to k^n$ imposes an extra $k[x]$-module structure on $k^n$ by defining $p(x)\cdot v = p(\phi)(v).$ Clearly then $k^n$ is a finitely generated $k[x]$-module.

As $k[x]$ is a PID, applying the structure theorem gives $k[x]$-module isomorphism $$k^n\simeq \bigoplus_{j=1}^{m}\bigoplus_{i=1}^{s_j} k[x]/(x-a_j)^{d_{i,j}}.$$ Note that $k[x]$ cannot occur on right hand side as $k^n$ is finitely generated. For simplicity let right hand side be W.

Then I asked myself, "what does a $k[x]$-module homomorphism $f: V\to W$ mean?" Let $\phi: V\to V$ and $\varphi: W\to W$ be the linear map associated to the $k[x]$-module structure of $V$ and $W$ respectively. Then $f(\phi(v)) = f(x\cdot v) = x\cdot f(v) = \varphi(f(v))$ for every $v\in V$, in other words, the $k[x]$-module homomorphism is the commutative diagram relation $f\circ \phi = \varphi\circ f$(where vertical arrows stands for $f$ below:)

$\require{AMScd}$ \begin{CD} V @>{\phi}>> V\\ @VVV @VVV \\ W @>{\varphi}>> W \end{CD}

In particular when $f$ is an isomorphism, $\phi$ and $\varphi$ are conjugate, meaning their corresponding matrices are similar.

Back to the question, iff $\phi$ stands for the map $\phi: k^n\to k^n$ and $\varphi$ stands for the linear map $W\to W$ from the $k[x]$-module structure. Clearly $\varphi$, written in matrix form with a suitable basis is the Jordan Normal Form. And the statement is then excatly $\phi$, the original matrix, is conjugate(similar) to the Jordan Normal Form of it.

I haven't seen anyone mentioning the "isomorphism implies similar" statement to me. Is this avoidable in other ways? I only did this once by myself.



You must log in to answer this question.