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Suppose $A$ is an $n \times n$ matrix over the field $F$ with only one invariant factor. Show any $n \times n$ matrix $B$ over $F$ that commutes with $A$ is a polynomial in $A$.

Andreas Caranti
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Alexandra
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  • Any $n \times n$ matrix commutes with the identity matrix, and yet the only polynomials in the identity matrix are the scalar multiples of the identity matrix... – Branimir Ćaćić Mar 24 '13 at 08:50
  • Does this hold? If $A$ is the identity matrix, then it commutes with all the matrices, but polynomials in $A$ are all scalar. Also $1$ is the sole invariant factor of $A$. Not that invariant factors make much sense over a field. Undoubtedly you mean something else (and should not mention a key assumption in the title alone anyway). – Jyrki Lahtonen Mar 24 '13 at 08:51
  • It might be a matter of wording, but I take the assumption to mean that the underlying vector space is a cyclic $F[A]$-module. – Andreas Caranti Mar 24 '13 at 08:53
  • In other words, all invariant factor but one are $1$. – Andreas Caranti Mar 24 '13 at 09:09
  • @Andreas: So you are talking about the invariant factors of $xI-A$ as opposed to $A$? That does make the question meaningful. Well decpihered :-) – Jyrki Lahtonen Mar 24 '13 at 09:27
  • @JyrkiLahtonen, right, thanks, let me spell this out in my answer, with credit where it's due ;-) – Andreas Caranti Mar 24 '13 at 09:29
  • @MarcvanLeeuwen, just noticed you let that oldish question come back to the surface. – Andreas Caranti Mar 24 '13 at 09:33
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    @AndreasCaranti: Yes, I changed its title to be more informative – Marc van Leeuwen Mar 24 '13 at 09:34
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    @JyrkiLahtonen I think the invariant factor in the question is to be interpreted is the one of $F^n$ as $F[X]$ module, with $X$ acting by $A$; see modules over a PID. In fact I don't see many other modules over a PID here. – Marc van Leeuwen Mar 24 '13 at 09:42
  • @Marc: I quite agree! Feeling a bit silly for not realizing it right away. – Jyrki Lahtonen Mar 24 '13 at 09:57
  • I think this is an edge case where a Question narrowly avoids being a duplicate by being confused. Does it make sense to leave it open essentially for the value of sorting out what the Question really means? – hardmath Mar 24 '13 at 13:50

1 Answers1

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I take the assumption to mean that the underlying vector space $V$ is a cyclic $F[A]$-module. (Only one invariant factor of $x I - A$ is different from $1$. Thanks to Jyrki Lahtonen for a clarifying remark in a comment to OP.)

Suppose $v_{0}$ is a generator of $V$ as such a $F[A]$-module. Then $v_{0} B = v_{0} f(A)$ for some polynomial $f \in F[x]$ (apologies, my operators act on the right).

Now given any $v \in V$, there is $g \in F[x]$ such that $v = v_{0} g(A)$. Thus $$ v B = v_{0} g(A) B = v_{0} B g(A) = v_{0} f(A) g(A) = v_{0} g(A) f(A) = v f(A). $$ It follows $B = f(A)$.

Andreas Caranti
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