С днём рождения, kdv2005

[ny_quant]

После некоторых сомнений, решил поздравить именинника нестандартной (как мне кажется) задачей.

Для каких N существуют матрицы размера NxN с рациональными элементами, удовлетворяющие уравнению A^3+A+I=0

Комментарии пока что буду скринить, чтобы всем было интереснее решать.

Comments

[misha-b.livejournal.com]

N delitsya na tri?

[ny-quant.livejournal.com]

Правильно. Чуть позже расскриню решение. Если не трудно, пока что напишите своё.

[misha-b.livejournal.com]

First the size needs to be divisible by 3.
All eigenvalues of A are roots of the polynomial x^3+x+1=0
det A is the product of eigenvalues. It is not hard to see that the only way for it to be rational is for eigenvalues to go in triples.

On the other hand it is not hard to construct a matrix of size 3 (hence any size divisible by three):
0 -1 -1
1 0 0
0 1 0
(from the recursive equation).

Spasibo, zabavnaya zadacha.

[misha-b.livejournal.com]

Chut' pogoryachilsya. It is intuitively clear that eigenvalues go in triples, but i would have to see how to formulate the exact argument. Chto-to is teorii Galois, kazhetsya.

[ny-quant.livejournal.com]

Я как раз хотел было написать, что не въехал. Моя интуиция молчит как воды в рот набрала.

[misha-b.livejournal.com]

I think the following is true:

if P is any polynomial of three variables, P of the roots is rational if and only if it is contained in the algebra generated by the three symmetric polynomials x_1+x_2+x_3, x_1x_2+x_2x_3 + x_1 x_3 and x_1 x_2 x_3 (corresponding to the coefficients of the cubic). Since det is a product of the roots it needs to be a power of x_1 x_2 x_3

It's been a long time since i studied these things :)

[ny-quant.livejournal.com]

Я, к сожалению, вообще этого не проходил, поэтому торможу по полной :(

Когда Вы говорите "P of the roots", Вы имеете подстановку трёх собственных чисел искомой матрицы или что-то другое? Если да, то Вы уже предполагаете, что их три?

Чтоб я хоть примерно понял откуда ветер дует, как бы выглядел базис алгебры полиномов, если бы исходное уравнение включало и квадратичный член? Я это спрашиваю потому, что мне непонятно как это тройка соответствует коэффициентам уравнения.

Since det is a product of the roots it needs to be a power of x_1 x_2 x_3

Этого тоже не понял. Это обобщение на случай, когда размерность ужене ровно три а кратна трём?

[misha-b.livejournal.com]

Sorry, chto-to ya nevnyatno izlagaju.

1. All eigenvalues of the matrix are roots of x^3+x+1=0.
Let's call them a,b,c

2. det A is a product of these eigenvalues and therefore has the form
a^k b^l c^n for some integers k,l,n. Clearly det A is a rational number (all coefficients of the matrix are rational).

3. Let us now consider a polynomial of three variables P(x,y,z).
Claim:
P(a,b,c) is rational if and only if P can be written as a sum
(with rational coefficients) of powers of the elementary symmetric polynomials xyz, xy+xz+zy, x+y+z.

4. Consider now P(x,y,z)= x^k y^l c^n . We know that P(a,b,c) is rational.
Therefore P is a sum of powers of the elementary symmetric polynomials.
By some additional simple argument, which I omit, the only polynomial
which can appear is xyz. Therefore, it is a power of xyz and roots go in triples.

Mne kazhetsya, primerno tak.

[prof-yura.livejournal.com]

The polynomial x^3+x+1 is irreducible over the rationals; in particular, all its roots are Galois-conjugate over the field Q of rational numbers. Since the degree N characteristic polynomial of the matrix has rational coefficients, all the roots of x^3+x+1 have the samr multiplicities as roots of the characteristic polynomial. This implies that N is divisible by 3.

[misha-b.livejournal.com]

> Since the degree N characteristic polynomial of the matrix has rational coefficients, all the roots of x^3+x+1 have the samr multiplicities as roots of the characteristic polynomial.

How do you see this?

[prof-yura.livejournal.com]

Well, it's similar to the situation with complex-conjugate roots of a polynomial with real coefficients: they have the same multiplicity. Here the roots of x^3+x+1 are Galois-conjugate over the rationals and therefore if one of them has say, multiplicity M as a root of a polynomial f(x) with rational multiplicities then all other roots of x^3+x+1 have the same multiplicity as roots of f(x).

[misha-b.livejournal.com]

Thanks! Of course, it is just needs to be invariant under conjugation.
I was making it far too complicated.

You are welcome

[prof-yura.livejournal.com]

a polynomial f(x) with rational multiplicities

multiplicities => coefficients