I am reading Rational Canonical form from The Abstract Algebra book by Dummit and Foote. I have some doubt in Smith normal form. Smiths normal for says for any $n\times n$ square matrix $A$ over an arbitrary field $F,$ $xI-A$ is equivalent to diagonal matrix in $F[x]$ whose diagonal elements are either $1$ or the invariant factors of the pair $(F^n,A)$. But after looking at other references it seems to me that $xI-A$ is not only equivalent but Similar to such diagonal matrix in $F[x].$ I can't understand how they are similar. I need some help to understand the similarity.

And also I want to know if there are references for the Canonical form in the modern approach by what I mean using the results of modules over PID.

Thank you.

  • 3
    $\begingroup$ In Dummit and Foote the argument for why they are similar is at the end of exercise 19 of section 12.1. Details are left to the reader, but essentially everything is there. $\endgroup$ – user565560 Jun 2 '18 at 11:45
  • 1
    $\begingroup$ You might want to know there is a name for the inverse of the negative this matrix: The resolvent: en.wikipedia.org/wiki/Resolvent_formalism $\endgroup$ – Benedict W. J. Irwin Jun 18 '18 at 9:27
  • $\begingroup$ Could you add which "other references" seem to suggest that $xI - A$ is already similar to its Smith normal form? $\endgroup$ – Jendrik Stelzner Jun 18 '18 at 17:50

I don't want to set up all the machinery related to this problem but only give some basic ideas.

Let $A\in M_n(F)$, where $F$ is a field, $m$ be its minimal polynomial and $\chi_A$ be its characteristic polynomial. If $p\in F[x]$, then $C_p$ denotes the companion matrix of $p$.

Note that $F^n$ can be viewed as a finitely generated module over the PID $F[A]$: $p(A).v=p(A)v$. Then there is the so called "structure theorem" cf.


Step 1. The Frobenius normal form of $A$ over the field $F$. For every $v\in F^n$, there is $q$, a divisor of $m$ of degree $r$ s.t. $q(A)v=0$ and s.t. $v,Av,\cdots,A^{r-1}v$ is linearly independent. By this way, we can construct a basis of $F^n$, using polynomials $p_1,\cdots,p_k=m$ s.t. $p_i$ divides $p_{i+1}$ and $p_1\cdots p_k=\chi_A$; thus $A$ is similar to its Frobenius form $diag(C_{p_1},\cdots,C_{p_k})\in M_n(F)$. Moreover,

(*) the polynomials $(p_i)_i$ uniquely define the similarity class of $A$ over $F$.

Step 2. The smith normal form of $xI_n-A$ over the PID $F[x]$. There are $S,T\in M_n(F[x])$, where $\det(S),\det(T)\in F^*$ s.t. $S(xI-A)T=diag(I_r,p_1,\cdots,p_k)$ with $r=n-k$, the Betti number associated to $xI-A$. Thus $xI-A$ is equivalent to its Smith normal form. The key of the proof is to show that if $A=C_p$, then the Smith normal form of $xI-A$ is $diag(I_{n-1},p)$. cf. for example


Roughly speaking, Frobenius and Smith say pretty much the same thing.

Conclusion. According to (*),

$xI-A,xI-B\in M_n(F[x])$ have same Smith normal form IFF $A,B$ are similar over $F$.

Note that $xI-A$ is absolutely not similar to its Smith normal form (in general $xI-A$ is not diagonalizable).

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