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February 4

Cubes of the form 3 x² ± x y + 5 y² with x, y coprime

Are there any cubes of the form $3x^{2}\pm xy+5y^{2}$ with x and y coprime ? I've found no solutions for $|x|,|y|\leq 10^{3}.$ — 79.113.247.229 (talk) 13:26, 4 February 2015 (UTC)[reply]

Work through the method here- I think you'll find that x must be 0 for any 3x^2 +- x*y + 5*y = k^3- so no. DTLHS (talk) 20:14, 4 February 2015 (UTC)[reply]

Characteristic polynomial without determinant

The usual definition of the characteristic polynomial of an endomorphism A of a finite dimensional vector space uses the determinant. I faintly remember that, way back in time at university, there was an approach that didn't need neither determinants nor eigenvalues. I can't find anything on the net, and I can't figure it out by myself.

As far as I remember, the trick was to decompose the vector space into subspaces that were stable under the endomorphism (corresponding to the Jordan blocks, but those were not yet defined at this stage of deduction).

It worked roughly by taking a basis b1, ..., bn, then computing b1 A^0, b1 A^1, ... until the set of vectors was no longer linearly independent. Then construct a polynomial from the coefficients of b1 A^k= sum ai b1 A^i, then find a new basis containing b1 A^0 to b1 A^(k-1) for the first subspace. Continue iteratively for the subspace consisting of the remaining vectors of the basis. Finally, multiply all the polynomials together.

Anyone out there who has seen such a thing? 93.132.26.165 (talk) 13:51, 4 February 2015 (UTC)[reply]

You could get the matrix into say, upper triangular form, read off the eigenvalues, and construct the characteristic polynomial from the eigenvalues. Down with Determinants! is a fun paper going into the math behind a determinant-free approach. --Mark viking (talk) 14:12, 4 February 2015 (UTC)[reply]

I had seen that before. That guy uses the concept of eigenvalues and, even worse, restricts himself to vector spaces over the complex numbers. The above approach would work for any field. Eigenvalues, a volume function (p(0)(-1)^n = det A) and Cayley-Hamilton would be for free, based solely on the concept of vector spaces and linear independence. 93.132.26.165 (talk) 14:52, 4 February 2015 (UTC)[reply]

What you're doing is constructing the characteristic polynomial from the cyclic subspaces of the matrix. Off the top of my head, I would check Hoffman and Kunze "Linear algebra". They emphasize the cyclic subspace perspective fairly heavily. Sławomir Biały (talk) 14:17, 4 February 2015 (UTC)[reply]

That's in the line of the thing ... but I'm fairly decadent ... . I was hoping for something on the internet. Well, probably I will step out into the cold and get the book, in physical form, from the library ;-) . 93.132.26.165 (talk) 14:52, 4 February 2015 (UTC)[reply]

You might want to check out Advanced Linear Algebra by Steven Roman (GTM 135) as well. The characteristic polynomial is defined as the product of the elementary divisors, and cyclic subspaces are clearly involved. (Disclaimer: I haven't read the chapter, but at least determinants are nowhere in sight.) YohanN7 (talk) 16:15, 4 February 2015 (UTC)[reply]

Thanks, this is even available online. 488 pages. This will take some time. 93.132.26.165 (talk) 16:43, 4 February 2015 (UTC)[reply]

Ouch, this starts with theorem 0.1 stating det A^t=det A. .... So determinants are a prerequisite, not something to be derived. Standard approach, not what I am looking for. 93.132.26.165 (talk) 17:28, 4 February 2015 (UTC)[reply]

February 4

Cubes of the form 3 x2 ± x y + 5 y2 with x, y coprime

Characteristic polynomial without determinant

Cubes of the form 3 x² ± x y + 5 y² with x, y coprime