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

A standard solution

[prof-yura.livejournal.com]

The answer: N is an every positive integer that is divisible by 3. The proof is based on the irreducibility of the cubic polynomial x^3+x+1 over the rationals that could be easily checked (the only divisors of the constant term are 1 and -1 and they are not the roots). Namely, let us consider the set K of all matrices of the form a+b A+c A^2 with a,b,c rational numbers. Since A^3=-(1+A), the set K is closed under multiplication. Now the irreducibility implies that actually one may divide in K (of course, by non-zero elements). In other words, K is a field (like the fields of rational, reals or complex numbers.) What's interesting is that K carries a natural "structure of 3-dim'l vector space over the field of rational numbers Q, "i.e.", one may add up elements of K and multiply them by rational numbers. Now the action of the matrix A on the coordinate vector space Q^n provides Q^n with a natural structure of K-vector space, whose K-dimension is N/3 and therefore 3 divides N.

For N=3 one may take as A the matrix B

0 0 -1
1 0 -1
0 1 0

If N=3M then one should take the block-diagonal M x M matrix, whose
"diagonal" entries are B.

Re: A standard solution

[ny-quant.livejournal.com]

С грустью убедился, что толком Вашего решения не понимаю. Видимо, действительно надо было правильно специализироваться.