I am trying to characterize matrices with a certain property : Define $U$ as an $n \times n$ matrix (over C or R; you can also assume that it is unitary or orthogonal if it helps). Now take $n$ unknowns and transform by $U$ : $y=Ux$; I want the product $y_1 y_2 \cdots y_n = \sum c_i x_i^n$; so basically I don't want any "cross terms". In the case of $n=2$ this would put a restriction on $U=((a,b),(c,d))$ : $ad+bc=0$; and $y_1y_2$ would be $c_1x_1^2+c_2x_2^2$; My question is : are such matrices known in the literature by some name? and can they be characterized in any nice way? (minimal number of parameters,...)
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
I'm guessing that these don't have a name, as they are fairly degenerate matrices.
If more than two of the $c_i$ are nonzero, then no matrices satisfy these criteria. This follows since the Fermat curve $$X^n + Y^n = Z^n$$ is irreducible and a decomposition as you describe would make it a union of lines.
If two of the $c_i$ are nonzero, then all such matrices of this form have two columns $j, j'$ with nonzero entries and have $M_{ij}/M_{ij'}$ going through all $n^{th}$ roots of some nonzero complex number.
If one $c_i$ is nonzero, then only one column has nonzero entries, and all its entries are nonzero.
If all the $c_i$ are zero, then some row of $M$ is zero.
-
$\begingroup$ Thanks for the response. The motivation for the question came from a totally different angle (optics); I didn't think of the problem in terms of factoring. If I follow your answer correctly this would imply that adding the requirement that the matrix is invertable would mean that only the case $n=2$ has solutions. This negative result would have value on its own. Do you have a good reference for the irreducibality of the fermat curve? also the fermat curve is in only 3 variables, can I assume that the general $\sum_1^n x_i^n$ is also irreducible? $\endgroup$– unknownCommented Sep 10, 2015 at 19:40