# Similarity of matrices over $\Bbb{Z}$

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.

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.

• Dear tomasz: Are you sure that $A$ is conjugate in $\text{SL}(2,\mathbb Z)$ to $-A$? Aug 25, 2012 at 20:25
• @Pierre-YvesGaillard: Right, right, my mistake, I forgot that ${\bf Z}_4$ has two generators. ;) Fixed. Aug 25, 2012 at 21:39

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).

• The Smith normal form is usually defined for principal ideal domains Aug 25, 2012 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, 2012 at 16:01
• @JenniferDylan: ${\bf Z}[x]$ is PIR? What about $(2,x)$? Aug 25, 2012 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. Aug 25, 2012 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, 2012 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.

• 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... Aug 25, 2012 at 21:46