3

Lets consider 2x2 integer matrix with determinant equal 1: $$\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$$

I am working on the following:

How to extend this to 3x3 matrix in order to get another matrix with determinant 1: $$\left( \begin{array}{ccc} a & b & e \\ c & d & f \\ i & h & g \\ \end{array} \right)$$

And also is there any $a,b,c,d$ for which this extension is unique. I even have no idea how to start solving this. I have discovered the following so far on the web, but not sure how to use this:

Integer matrices with determinant equal to $1$

https://mathoverflow.net/questions/24131/is-the-semigroup-of-nonnegative-integer-matrices-with-determinant-1-finitely-gen

EDITED:

Actually I am looking for general algorithm, how to construct all 3x3 matricies from 2x2 matrix with determinant 1.

EDITED 2:

Some samples of such matricies:

$$\left( \begin{array}{ccc} 1 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & 0 & 2 \\ \end{array} \right) $$

$$\left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 5 & 9 \\ \end{array} \right)$$

$$\left( \begin{array}{ccc} 1 & 1 & 1 \\ -6 & -5 & -4 \\ 9 & 5 & 2 \\ \end{array} \right) $$

Gevorg Hmayakyan
  • 2,036

3 Answers3

3

You can just set $g = 1$ and $e,f,i,h = 0$:

$$1 = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = \begin{vmatrix} a & b & 0\\ c & d & 0\\ 0 & 0 & 1 \end{vmatrix}$$

Furthermore, the extension is never unique since:

$$1 = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = \begin{vmatrix} a & b & 0\\ c & d & 0\\ 0 & 0 & 1 \end{vmatrix} = \begin{vmatrix} a & b & 0\\ c & d & 0\\ a & b & 1 \end{vmatrix}$$

Note that $a$ and $b$ cannot both be $0$, so the two extensions are indeed different.

mechanodroid
  • 47,570
  • Thank you for answer. I am just wondering if there is any known algorithm of constructing such 3x3 matricies from the 2x2? And not sure if understand the part of uniqueness, could you please detalize abit? – Gevorg Hmayakyan Dec 24 '17 at 23:03
  • 1
    @GevorgHmayakyan Concerning uniqueness, the determinant is invariant under elementary row operations. So $$\begin{vmatrix} a & b & 0\ c & d & 0\ 0 & 0 & 1 \end{vmatrix} = \begin{vmatrix} a & b & 0\ c & d & 0\ a & b & 1 \end{vmatrix}$$ because the second matrix is obtained from the first by adding the first row to the third. Furthermore, these two matrices are not the same since $a$ and $b$ cannot both be $0$. If they were, then we would have: $$\begin{vmatrix} a & b \ c & d \ \end{vmatrix} = \begin{vmatrix} 0 & 0 \ c & d \ \end{vmatrix} = 0 \ne 1$$ – mechanodroid Dec 24 '17 at 23:07
  • 1
    You can in general construct such $3\times 3$ extensions by applying elementary row and column operations on the example above. – mechanodroid Dec 24 '17 at 23:08
  • Thank you a lot, but could you please describe the idea on the following matrix $\left( \begin{array}{ccc} 1 & 1 & 1 \ -1 & 0 & 1 \ -1 & 0 & 2 \ \end{array} \right)$ – Gevorg Hmayakyan Dec 24 '17 at 23:13
  • 1
    @GevorgHmayakyan By adding the first row multiplied by $n \in \mathbb{Z}$ to the third, and adding the second row multiplied by $m \in \mathbb{Z}$ to the third we obtain:

    $$\begin{vmatrix} 1 & 1 & 1 \ -1 & 0 & 1 \ -1 & 0 & 2 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \ -1 & 0 & 1 \ -1+n-m & n & 2+n+m \end{vmatrix}$$

     – mechanodroid Dec 24 '17 at 23:24
  • 1
    By further adding the first column multiplied by $k \in \mathbb{Z}$ to the third column, and adding the second column multiplied by $l \in \mathbb{Z}$ to the third, we obtain:

    $$\begin{vmatrix} 1 & 1 & 1 \ -1 & 0 & 1 \ -1+n-m & n & 2+n+m \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1+k+l \ -1 & 0 & 1-k \ -1+n-m & n & 2+n+m-k(-1+n+m)+ln \end{vmatrix}$$

     – mechanodroid Dec 24 '17 at 23:25
  • 1
    Also, you could first do the column operations, and then the row ones:

    $$\begin{vmatrix} 1 & 1 & 1 \ -1 & 0 & 1 \ -1 & 0 & 2 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1+k+l \ -1 & 0 & 1-k \ -1 & 0 & 2-k \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1+k+l \ -1 & 0 & 1-k \ -1+n-m & n & m(1+k+l)+n(1-k) + 2-k \end{vmatrix}$$ I hope you get the general idea.

     – mechanodroid Dec 24 '17 at 23:25
  • Yes the idea behind the uniqueness is clear thanks. – Gevorg Hmayakyan Dec 24 '17 at 23:31
2

$SL_2 \mathbb Z$ $$ \left( \begin{array}{cc} \alpha & \beta \\ \gamma & \delta \end{array} \right) $$

homomorphic image

$SL_3 \mathbb Z$ $$ \left( \begin{array}{ccc} \alpha^2 & 2 \alpha \beta & \beta ^2\\ \alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta\\ \gamma^2 & 2 \gamma \delta & \delta^2 \end{array} \right) $$

If you take the transpose of the 3 by 3, you get an anti-homomorphism.

Will Jagy
  • 146,052
  • This is exactly what was looking for. Could you please provide some reference regarding to this result? – Gevorg Hmayakyan Dec 25 '17 at 00:15
  • And also is it possible extend this on nxn matrix? – Gevorg Hmayakyan Dec 25 '17 at 00:17
  • 1
    @GevorgHmayakyan the earliest version I have is a book by Fricke and Klein (1897) https://books.google.com/books?id=H5kLAAAAYAAJ&printsec=frontcover&dq=Vorlesungen+%C3%BCber+die+Theorie+der+automorphen++Functionen&hl=en&sa=X&ved=0CCAQ6AEwAGoVChMIscb4rNvYxgIVSCiICh3qMwQX#v=onepage&q=muesli&f=false – Will Jagy Dec 25 '17 at 00:18
  • But this is still not exactly what I wanted, sorry and the problem is that I am looking for extension, while we have $\alpha^2$ and $2\alpha \beta$ while there should be $\alpha$ and $\beta$ – Gevorg Hmayakyan Dec 25 '17 at 00:20
  • Thank you for the book. – Gevorg Hmayakyan Dec 25 '17 at 00:20
1

Hint:

Consider a block-diagonal matrix, for which the first diagonal block is $\;\begin{matrix}a&b\\c&d\end{matrix}$.

Bernard
  • 179,256