Integer matrices with determinant equal to $1$

Question

Integer matrices with determinant equal to $1$ are quite useful in many situations. Take, for example, this question. For the $2 \times 2$ case it's easy to find many such matrices, e.g.,

$$\begin{bmatrix} 2 & 3 \\ 3 & 5 \\ \end{bmatrix}$$

$$\begin{bmatrix} 4 & 3 \\ 5 & 4 \\ \end{bmatrix}$$

• But how to construct the procedure for generation integer matrix with arbitrarily chosen dimension $n \times n$?
• Is it a method which is as general as it is possible?
• I'm also interested in the answer how many degrees of freedom has an integer matrix with determinant equal 1 (or other perhaps number) ? Without determinant constraint $n \times n$ matrix has of course $n^2$ degrees of freedom.. how many is lost when we constrain it with determinant?

Answer 1

You can just start with the identity matrix and apply transformations that don't change the determinant:

1. Adding to column (row) another column (row) multiplied by an integer.
2. Performing an even permutation of the columns (rows).

Hart to tell what is degrees of freedom for a discrete set. Its dimension is zero. But you can think of it a cutting all the $n^2$ dimensions that you had by one equation. So, $n^2-1$.

Answer 2

To motivate what might be said about "degrees of freedom" in the more general $n\times n$ case, let's look at the $2\times 2$ case in some detail.

One might informally say that the degrees of freedom in the special linear group $SL(2,\mathbb{Z})$ has an intuitive sense of "the number of coordinates needed to specify an instance".

This is a firmer notion when the coordinates involved are real numbers than when, as here, the coordinates are only discrete integers. The problem is related to the possibility that a pair of integers might be coded together as a single integer, so that the counting "how many coordinates are needed" becomes muddled. In the case of real numbers we are saved by imposing a requirement that any "coding" has to involve continuous functions (in a suitably restricted domain) that are continuously invertible (decoding). This prevents a pair of real numbers from being combined into a single real number.

Acknowledging that we are walking on slippery ground, let's consider a couple of "natural ways" to parameterize $SL(2,\mathbb{Z})$. The first involves making an arbitrary choice of the two diagonal entries $a,d$:

$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$

Since we desire that $ad - bc = 1$, we have (in order to get an element of $SL(2,\mathbb{Z})$) only to solve:

$$ bc = ad - 1 $$

Apart from the peculiar case that $ad = 1$ (which admits an infinite number of solutions $b,c$ provided at least one of them is zero), we find that there will be only finitely many $b,c$ which "factor" $ad - 1$.

This seems to say that there are (loosely speaking) two degree of freedom, since (with the two exceptions $a = d = \pm 1$) the values $a,d$ can be chosen arbitrarily and leave a finite number of additional "choices" (about the ways $ad - 1$ will factor).

On the other hand we might begin with choosing $a$ and $b$. Now the valid choices are those pairs $a,b$ which are coprime (no common divisor greater than one). Although not every pair is satisfactory in this sense, the relative fraction of coprime pairs $a,b$ in an expanding region $[-M,M]\times [-M,M]$ as $M\to \infty$ converges to $6/\pi^2$, which is roughly $61\%$.

So again it seems that choosing $a,b$ requires two degrees of freedom. Further, as $a,b$ are coprime, there exist coefficients $c,d$ such that $ad-bc=1$. Then we can introduce an additional integer coordinate $k$ because:

$$ \det \begin{bmatrix} a & b \\ c+ak & d+bk \end{bmatrix} = 1 $$

By this reckoning we would have three degrees of freedom at our disposal!

Answer 3

Say that we want to generate the integer matrix $$A=\begin{pmatrix}a_{11}&a_{12}&\dots&a_{1n}\\ a_{21}&a_{22}&\dots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\dots&a_{nn}\end{pmatrix}$$ so that $\det(A)=1$. We start by randomizing the rows $2,\dots, n$. Now a cofactor expansion in row 1 shows that the problem is equivalent to the one of finding solutions to the diophantine equation $$c_{11}a_{11}+c_{12}a_{12}+\dots+c_{1n}a_{1n} = 1$$ where $c_{11},\dots,c_{1n}$ are the cofactors of the first row of $A$. Since the rows $2,\dots,n$ have been set, the cofactors can easily be calculated. The diophantine equation is solvable if $\gcd\{c_{11},\dots,c_{1n}\}=1$ and in that case the solutions depend on $n-1$ parameters. An advantage of this method is that you can control the size of the entries of $A$.

I have developed a Python program that randomizes an integer matrix with determinant as a parameter. You can find it here: https://github.com/andis854/matrix_rdn_det.