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This is an updated question of one of my previous ones, I am specifically interested in finding invertible in $ℤ$ matrices $C$ and $D$ so that $CAD$ is diagonal. $A = $$ \begin{pmatrix} 2 & 1 & 2 \\\ 1 & 2 & 2 \\\ 2 & 3 & 0 \end{pmatrix} $ I dont know how to find specifically invertible ones.

Alexander Kraynov
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  • By invertible in $\mathbb Z$ do you mean $C,D\in M_n(\mathbb Z)$ are invertible and $C^{-1},D^{-1}\in M_n(\mathbb Z)$? – Aweygan Apr 09 '18 at 00:17
  • Yes, like that . – Alexander Kraynov Apr 09 '18 at 00:27
  • https://en.wikipedia.org/wiki/Smith_normal_form The point is that you cannot guess the final form of $C$ or of $D,$ you gradually improve $A$ by multiplying on the left by an elementary matrix $C_j$ and on the right by an elementary matrix $D_k$ The book I have is Integral Matrices by Newman. – Will Jagy Apr 09 '18 at 00:50

1 Answers1

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Here is a typeset Smith Normal Form

This question from 2012 has a very complete answer, including a 4 by 4 example Computing the Smith Normal Form

That worked out well. The eventual diagonal matrix, $ \; A_{43} \; ,$ is $$ L_4 L_3 L_2 L_1 A R_1 R_2 R_3 $$

parisize = 4000000, primelimit = 500000
?  a = [ 2,1,2; 1,2,2; 2,3,0]
%1 = 
[2 1 2]

[1 2 2]

[2 3 0]

?  a00 = [ 2,1,2; 1,2,2; 2,3,0]
%2 = 
[2 1 2]

[1 2 2]

[2 3 0]

? L1 =  [ 1,-1,0; 0,1,0; 0,0,1]
%3 = 
[1 -1 0]

[0  1 0]

[0  0 1]

? a10 = L1 * a00
%4 = 
[1 -1 0]

[1  2 2]

[2  3 0]

? L2 =  [ 1,0,0; -1,1,0; 0,0,1]
%5 = 
[ 1 0 0]

[-1 1 0]

[ 0 0 1]

? a20 = L2 * a10
%6 = 
[1 -1 0]

[0  3 2]

[2  3 0]

? L3 =  [ 1,0,0; 0,1,0; -2,0,1]
%7 = 
[ 1 0 0]

[ 0 1 0]

[-2 0 1]

? a30 = L3 * a20
%8 = 
[1 -1 0]

[0  3 2]

[0  5 0]


? R1 = [ 1,1,0; 0,1,0; 0,0,1]
%11 = 
[1 1 0]

[0 1 0]

[0 0 1]

? a31 =  a30 * R1
%12 = 
[1 0 0]

[0 3 2]

[0 5 0]

? R2 = [ 1,0,0; 0,1,0; 0,-1,1]
%13 = 
[1  0 0]

[0  1 0]

[0 -1 1]

? a32 =  a31 * R2
%14 = 
[1 0 0]

[0 1 2]

[0 5 0]

? R3 = [ 1,0,0; 0,1,-2; 0,0,1]
%15 = 
[1 0  0]

[0 1 -2]

[0 0  1]

? a33 =  a32 * R3
%16 = 
[1 0   0]

[0 1   0]

[0 5 -10]


? L4 = [ 1,0,0; 0,1,0; 0,-5,1]
%17 = 
[1  0 0]

[0  1 0]

[0 -5 1]

? a43 = L4 *  a33 
%18 = 
[1 0   0]

[0 1   0]

[0 0 -10]
Will Jagy
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