Let $C(f)$ be the companion matrix of a monic polynomial $f(t) \in \mathbb{F}[t]$. I need to show that the Smith Normal Form of $tI - C(f)$ is equal to the diagonal matrix $\mbox{diag}(1,1,1,\dots,f(t))$.

A little bit baffled on how to begin. I've constructed the companion matrix and written out my monic polynomial. I think there are some relationships between equivalent/similar matrices of the form $tI-A$ that might help. But I'm definitely scratching my head on understanding how to link the Smith Normal Form to all of this.

  • $\begingroup$ Surely the usual "reduce to Smith Normal Form" algorithm gives this at once? $\endgroup$ Jan 25, 2020 at 13:54
  • $\begingroup$ Perhaps? What is that algorithm? $\endgroup$
    – notadoctor
    Mar 22, 2022 at 20:03
  • $\begingroup$ If the Smith NF of a matrix $M$ is $d_1,d_2,\dots, d_n$ then $d_1d_2\dots d_k$ is the hcf of all the $k\times k$ minors of $M$. Apply this to $M=tI-C(f)$, it's clear that there's always a $k\times k$ minor $1$ for $k<n$. $\endgroup$ Mar 23, 2022 at 7:40


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