An endomorphism $T:V\to V$ of a finite dimensional $\mathbb{C}$-vector space endows $V$ with a $\mathbb{C}[X]$-module structure defined by $X\cdot v = T(v)$. From the Structure theorem for finitely generated modules over a principal ideal domain we get that $V$ is a sum of $\mathbb{C}[X]/(X-\lambda_i)^{m_i}$, and, as in each of these modules $X$ has a Jordan block matrix, a matrix $T$ can be recovered as a sum of Jordan blocks.

On the other hand, in order to prove the Structure theorem constructively, one can say that if $M$ is fg, then it is finitely presented and compute the Smith normal form associated to the presentation.

My question is if one can use the Smith normal form to find the Jordan normal form of $T$, at least in a couple of examples. I believe the problem amounts to find a presentation for V as a $\mathbb{C}$-module --this is what I am confused with.


For general field $\mathbb{F}$, we can assume $V=\mathbb{F}^n$.

Let $A$ be a $n\times n$ matrix over a field $\mathbb{F}$ corresponding to the linear transform $T$. Then invariant factors can be recovered from the Smith Normal Form of $xI-A$. More precisely, if $S(xI-A)S'=\textrm{Diag}(f_1,\cdots,f_n)$ for some invertible matrices $S, S'$ (over $\mathbb{F}[x]$) and $f_i\mid f_{i+1}$, then $f_i$ are the invariant factors of $A$.

This enables us to write $V$ as a sum of $\mathbb{F}[x]/(f_i)$. Then proceed with rewriting each of these $f_i$ as products prime ideals.

In case $\mathbb{F}$ is algebraically closed, e. g. $\mathbb{C}$, we have the Jordan Normal Form because only possible monic irreducible polynomials in $\mathbb{C}[x]$ are $x-\alpha$ for some $\alpha\in\mathbb{C}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.