So I am studying for a midterm examination, and I was stuck on this practice problem.

$A \in \mathbb{R}^{n\times n}$ and $f_i$ are the diagonal entries in the Smith Normal Form of $xI - A$.

a) For which matrices $A$ is $f_1 \neq 1$?

b) For which matrices $A$ is $f_{n-1} = 1$?

So for part b I think its the matrices A such that A is not triangular because I feel this has to do with the fact that the characteristic and minimal polynomial do not factor into linear polynomials. So $f_n = \det(xI - A)$. Is that true?

And I am completely unsure how to proceed for part a.

Any help, and links on some clarification regarding the smith normal would be greatly appreciated.

  • $\begingroup$ The diagonal entries of SNF have very special properties. What are they? $\endgroup$ – vadim123 Nov 27 '16 at 3:24
  • $\begingroup$ the diagonal elements are the invariant factors, and the nontrivial ones are the elementary divisors. But I am still confused, I dont think I grasp the concept of elementary divisors and invariant factors very well ... So a bit more clarification would be very helpful. $\endgroup$ – ManikSin Nov 27 '16 at 3:37
  • $\begingroup$ That's just jargon; what are the properties that those elements have? See here. $\endgroup$ – vadim123 Nov 27 '16 at 3:55
  • $\begingroup$ Oh wow completely forgot about the property regarding the minors. Yeah so if f_1 is not trivial then that means that all elements of xI - A are divisible by a polynomial, which results that A is strictly diagonal, right ? and then for part b it shows that the gcd of all n-1 minors is 1, so then det(xI - A) = f_n which tells us that A is not triangular, right ? $\endgroup$ – ManikSin Nov 27 '16 at 4:15
  • 1
    $\begingroup$ @ManikSin The characteristic polynomial of $A$ is the product of invariant factors, that is, $P_A(X)=\prod_{i=1}^nf_i(X)$. (a) If $f_1(X)\ne1$, then $\deg f_1\ge 1$, and since $f_1\mid\cdots\mid f_n$ we get $\deg f_i\ge1$ for all $i$. Since $n=\deg P_A=\sum_{i=1}^n\deg f_i\ge n$ we get $\deg f_i=1$ for all $i$, hence $A$ is diagonalizable. (b) If $f_{n-1}=1$ then $f_1=\cdots=f_{n-1}=1$, and $f_n=P_A$. In this case $A$ is similar with a companion matrix. $\endgroup$ – user26857 Nov 27 '16 at 9:00

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