Number of matrices n x n with n^2 different elements which have that same characteristic polynomial

[Artur]

Dear Max,

My doubtness are that first coefiicent of Viette (trace) and last  
coefiicent of Viette (determinant) aren't unique Viette coefficents for  
n>=3 and we can imagine situation that for some n will be existed case  
where
Let zeroes of characteristic polynomial of matrix A will be respectively  
(a1,a2,...an) and matrix B will be (b1,b2,...,bn)
now trace A is a1+a2+...an  trace B is b1+b2+...bn
det A =a1*a2*..*an     det B=b1*b2...*bn

and now your proof with number 2n! will be good if doesn't existed case
if (a1*a2*..*an=b1*b2...*bn) and (a1+a2+...an)=(b1+b2+...bn)
don't occured such case that  
(a1*a2+a1*a3+...+a1*an+a2*a3+...)<>(b1*b2+b1*b3+...+b1*bn+b2*b3+...)
These was verified by computer examination of all cases by me that in 3 x  
3 that counter sample doesn't exist but we don't know that doesn't exist  
also in 4 x 4 and 5 x 5 and ... matrices.

BEST WISHES
ARTUR



Dnia 14-01-2007 o 01:13:18 Max A. <maxale at gmail.com> napisał(a):

> The following statements (suggested by lofar in russian forum
> lib.mexmat.ru/forum/) are true:
>
> Let n x n matrices A and B contain the same n^2 variables as elements,  
> then
> 1) If |det(A)|=|det(B)| then B=PAQ or B=(PAQ)^T where P and Q are some
> permutation matrices.
> 2) If |det(A)|=|det(B)| and trace(A)=trace(B) then B=PAP^{-1} or
> B=(PAP^{-1})^T where P is some permutation matrix.
>
> Now it is easy to give an answer to your original question:
>
>> How many different matrices n x n with n^2 different elements occured  
>> which have that same characteristic polynomial?
>
> Note that if charpoly(A)=charpoly(B) then det(A)=det(B) and
> trace(A)=trace(B), implying that B=PAP^{-1} or B=(PAP^{-1})^T for some
> permutation matrix P. Therefore, the answer to your question is 2*n!
> (where n! stands for the number of different permutation matrices and
> 2 accounts for a possible transposition of the matrix).
>
> Max
>
> On 1/8/07, Artur <grafix at csl.pl> wrote:
>> Dear Seqfans,
>> I'm asking: How many different matrices n x n with n^2 different  
>> elements
>> occured which have that same characteristic polynomial
>> We have to count all permutations n^2 elements in n x n matrix and count
>> only these permutations
>> which don't changed starting polynomial
>> for 2 x 2 case
>> we have 4 matrices X^2-(a+d)X+ad-bc
>> a b   a c   d c   d b
>> c d   b d   b a   c a
>>
>> BEST WISHES
>> ARTUR
>>
>>