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]

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