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Let $A$ be a $2 \times 2$ matrix over $\Bbb{Z}$ with characteristic polynomial $x^2 + 1$. Determine whether $A$ is similar to $$\pmatrix{0 & -1 \\ 1 & 0}$$ over $\Bbb{Z}$.

I'm so used to working with matrices over fields $\Bbb{Q}$ and $\Bbb{C}$ that I'm not sure how it differs when we switch to non-fields like $\Bbb{Z}$. Any advice would be much appreciated.

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up vote 5 down vote accepted

Notice that any such $A$ is an element of order $4$ in $SL_2({\bf Z})$. But $SL_2({\bf Z})\cong {\bf Z}_4*_{{\bf Z}_2}{\bf Z}_6$.

In an amalgamated product, the only elements of finite order are those which are conjugate to the elements of amalgamated groups, so every such $A$ is conjugate to the $1\in {\bf Z}_4$ or $3\in{\bf Z}_4$ in $SL_2({\bf Z})$.

$1$ corresponds to the matrix you provided (with the usual identification of $SL_2({\bf Z})$ with the amalgamated product), while $3$ is its inverse. They are not conjugate in $SL_2({\bf Z})$, but they are in $GL_2({\bf Z})$ by $\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Since any $A$ is conjugate to one of them, all the $A$ are similar.

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Dear tomasz: Are you sure that $A$ is conjugate in $\text{SL}(2,\mathbb Z)$ to $-A$? – Pierre-Yves Gaillard Aug 25 '12 at 20:25
@Pierre-YvesGaillard: Right, right, my mistake, I forgot that ${\bf Z}_4$ has two generators. ;) Fixed. – tomasz Aug 25 '12 at 21:39
Thank you Tomasz! – Conan Wong Aug 27 '12 at 7:40

Update: the proof below shows a necessary but not sufficient similarity over $\Bbb{Z}$. I will leave here for pedantic reasons. In the following paragraph, I will summarize; please consult references linked.

Similarity over $\Bbb{Z}$ implies similarity over $\Bbb{Q}$ but the converse is not true in general. The particular case for $2\times 2$ integral matrices having a characteristic polynomial $f(x) = x^2+1$ is covered by Example $1$ pp. $11$ in these (PDF) notes: Lectures on Integer Matrices Thomas J. Laffey. The proof follows from a theorem of Latimer and MacDuffee relating similarity classes to the number of ideal classes of $\Bbb{Z}[\theta]$ where $\theta$ is a complex root of the characteristic polynomial. And in the case of $x^2+1$, $\Bbb{Z}[i]$ is a PID and has class number 1 (see the PDF). Also relevant is this post in MathOverflow: Ideal classes and integral similarity.

Two matrices $A, B$ are similar over $\Bbb{Z}$ if $x I - A$ and $x I - B$ have the same Smith normal form over $\Bbb{Q}[x]$.

Recall that the $i$th invariant factor is the ratio between the gcd of all $i\times i$ minors, to the gcd of all $i-1 \times i-1$ minors. So, the SNF of $xI - B = \pmatrix{x & 1 \\ -1 & x}$ is $\pmatrix{1 & 0 \\ 0 & x^2+1}.$

It's not hard to show that if $A$ has a characteristic polynomial $\lambda^2+1$ then $xA - I$ has a characteristic polynomial $\lambda^2 + (-2x) \lambda + (x^2 + 1).$ Hence, the determinant of $xA - I$ is $x^2 + 1$, which is also the $2$nd determinantal divisor. But $x^2 + 1$ does not factor over $\Bbb{Q}$, and that pretty much forces the $2$nd invariant factor of $xA - I$ to be $x^2 + 1$, and the first invariant factor to be $1$ (since $1$ is the only element in $\Bbb{Q}[x]$ which divides $x^2 +1 $). So the SNF of $xA-I$ is also $\pmatrix{1 & 0 \\ 0 & x^2+1}.$ Hence $A$ and $B$ are similar over $\Bbb{Z}$.

More in the book Integral Matrices by M. Newman (chapter III section 14).

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The Smith normal form is usually defined for principal ideal domains – Cocopuffs Aug 25 '12 at 15:59
@Cocopuffs SNF is well-defined for matrices over $\Bbb{Z}[x]$ (see the book by Newman) and in general PIRs (in a paper by Kaplansky. Elementary divisors and modules). – user2468 Aug 25 '12 at 16:01
@JenniferDylan: ${\bf Z}[x]$ is PIR? What about $(2,x)$? – tomasz Aug 25 '12 at 16:03
@JenniferDylan In your reference it looks like you use the Smith normal form over $\mathbb{Q}[x]$ and claim that the equality is a necessary (but not necessarily sufficient) condition for similarity over $\mathbb{Z}$. But I could be wrong. – Cocopuffs Aug 25 '12 at 16:08
Oops. Sorry. Cocopuffs you're right. I will edit my question. The same argument carries since $x^2+1$ does not factor over $\Bbb{Q}$. (in my comments above I was thinking all the time about SNF being well-defined over $\mathsf{F}[x]$. Sorry for the mix up with Z[x].) – user2468 Aug 25 '12 at 16:11

The following argument is inspired by tomasz's answer. Set $$ B:=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix},\quad G:=\text{SL}(2,\mathbb Z),\quad H:=G/\{\pm\ I\}, $$ ($I$ being the identity matrix), and let $\pi:G\to H$ be the canonical projection.

Since $B$ and $-B$ are conjugate in ${GL}(2,\mathbb Z)$, it suffices to verify that $\pi(A)$ and $\pi(B)$ are conjugate in $H$.

The action of $G$ on the upper half-plane $U$ of $\mathbb C$ by Möbius transformations induces a faithful action of $H$ on $U$. It is not hard to check that an element of finite order of $H$ has a fixed point on $U$. Then, using Theorem VII.1 of Serre's A Course in Arithmetic, one sees that the order two elements of $H$ are conjugate, as required.

EDIT. The following is implicit above and in tomasz's answer:

Two elements of $\text{SL}(2,\mathbb Z)$ having the same finite order are conjugate in $\text{GL}(2,\mathbb Z)$.

Let me also point out that the proof of Theorem VII.1 in Serre's A Course in Arithmetic is short, self-contained, and (is it necessary to add?) wonderfully written.

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I thought my answer was out there, involving $SL_2$, but yours goes to $PSL_2$. :-) Still, it is nice to see different ways to look at the same problem. Even if neither yours nor mine looks like it would easily generalize to arbitrary matrices... – tomasz Aug 25 '12 at 21:46
Thank you Pierre-Yves. – Conan Wong Aug 27 '12 at 7:43

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