Re: efficient foursquare() and/or threesquare1m4() functions

[hermann@stamm-wilbrandt.de]

On 2023-10-08 10:19, Bill Allombert wrote:
> On Sun, Oct 08, 2023 at 01:25:07AM +0200, hermann@stamm-wilbrandt.de 
> wrote:
> > Because I already implemented generation of ternary quadratic form Q 
> > for n
> > - that represents n
> > - and has determinant 1.
>
> > Now I need to figure out how to determine matrix M, such that
> > M~*Q*M is diagonal matrix. The diagonal entries of M~*Q*M
> > are three square representation of n.
>
> See qfgaussred
>
> Cheers,
> Bill.
Thanks, but that matrix is not the matrix M I search for.

For ternary quadratic form representing n=62 with determinant 1 ...

? Q=[41,50,1;50,61,0;1,0,62];
? qfeval(Q,[0,0,1])
62
? matdet(Q)
1
?

... the result for M~*Q*M is not a diagonal matrix:

? M=qfgaussred(Q)

[41 50/41 1/41]

[ 0  1/41  -50]

[ 0     0    1]

? M~*Q*M

[  68921          2100       -102418]

[   2100   107561/1681 -5245900/1681]

[-102418 -5245900/1681    6250045/41]

?


Since 62 = 6^2 + 5^2 + 1^2, the matrix M I need would
result in M~*Q*M being a diagonal matrix with entries {1,5,6}.

So what is this matrix named in linear algebra?
How can it be computed?


Regards,

Hermann.


P.S:
Characteristic polynomial of Q is not product of distinct linear terms.

? factor(charpoly(Q))

[x^3 - 164*x^2 + 6324*x - 1 1]

?

So diagonalizability by H^(-1)*Q*H should not be possible [H^(-1) 
instead M~ above].
But all non-diagonal entries are very close to 0?!?!?

? [L,H]=mateigen(Q,1);
? H^(-1)*Q*H

[0.00015812841567833090336360740761774181802 8.557414429477184768 E-41 
5.977755398451150209 E-38]

[3.761581922631320025 E-37 61.999596775635851601749685190428479366 
-7.523163845262640051 E-37]

[1.6502838659591020717 E-37 -9.183549615799121156 E-41 
102.00024509594847006734695120216390289]

?


But that does not help to get  matdiagonal([1,5,6]) as result for Q.