Re: Sums and products of eigenvalues

On 16-11-2008 16:17, Mariano Suárez-Alvarez wrote:

Let B and C be two square matrices (not necessarily with the same size)
whose entries are in a sub-ring A of an algebraically closed field _k_.
Let _x_ (in _k_) be an eigenvalue of B and let _y_ (also in _k_) be an
eigenvalue of C. Is there some way of getting a square matrix S (with
coefficients in A) from B and C such that x + y is an eigenvalue of S?
And is there some way of getting a square matrix P (again with
coefficients in A) from B and C such that x.y is an eigenvalue of P?
If I have understood this question correctly, then I think the
matrices are something of a distraction.
The assumptions on B and C are equivalent to x and y being integral
over the ring A. That is, they are roots of monic polynomials with
coefficients in A. It is a standard result in commutative algebra that
the elements of _k_ integral over A form a subring of _k_, so x+y and
xy are also integral over A, which means that the matrices that you
are seeking certainly exist, as the companion matrices of the minimal
polynomials of x+y and xy.
I know all this. And my intention was precisely to *prove* the "standard
result" which says that "the elements of _k_ integral over A form a
sub-ring of _k_", and to prove it using linear algebra and eigenvalues.
Meanwhile, I was able to see a similar approach which works, based upon
endomorphisms of k^n rather then matrices. Indeed, let _f_ and _g_ be
endomorphisms of k^m and k^n respectively (with _m_ and _n_ naturals).
Suppose that there are vectors _v_ and _w_ in k^m and k^n which are
eigenvectors of _f_ and _g_ respectively with eigenvalues _x_ and _y_
respectively. I will use _x_ to denote the tensor product. Then
v _x_ y is an eigenvector of f _x_ id + id _x_ g with eigenvalue
x + y and it is an eigenvector of f _x_ g with eigenvector x.y.
I don't know how you calculate these minimal polynomials in general,
but I expect somebody does! I am sure there are algorithms for special
cases, like rings of integers of number fields.
It is not hard to get a monic polynomial with coefficients in A with
x + y as a root. Let P(z) and Q(z) be monic polynomials with
coefficients in A such that P(x) = 0 and that Q(y) = 0. Write, in _k_,
P(z) = (z - x_1) ... (z - x_n)
and
Q(z) = (z - y_1) ... (z - y_m).
Now, take the polynomial
R(z) = prod_{1 <= i <= n, 1 <= j <= m} (z - x_i - y_j).
Then x + y is clearly a root of R(z). The problem with this approach is
that it is not easy to prove that the coefficients of R(z) all belong
to A. It can be built in a similar way a polynomial S(z) with
coefficients in A of which x.y is a root.
Consider one of the coefficients of R as a function of
x_1, ..., x_n, y_1, ..., y_n. Since it is obviously
simmetric in the x's, it really is a polynomial
on the elementary symmetric functions of the x's
and the y's. But it is also symmetric on the y's, so it
it really a polynomial on the symmetric functions of the
x's and, separately, the y's. Therefore it is in A.
I know this, thanks. I did not express myself correctly. When I wrote
that "it is not easy to prove that the coefficients of R(z) all belong
to A" what I had in mind was that I would like to prove, with as little
background as possible, that the algebraic integers form a ring and that
the algebraic numbers form a field. But the students to whom I would
like to teach that do not know the theorem that you mentioned about
symmetric polynomials. On the other hand, they have some knowledge about
Linear Algebra and I was trying to see if I could use that knowledge in
order to prove what I want to prove.

I would say that, when presenting the result
to students, it's best to use the argument
involving symmetric functions, even though
it may take some time to prove the fundamental
theorem on symmetric functions, for that argument
very much puts to the front the real reason
x y and x + y are also integral (and because
the same trick is useful in many, many other
situations) while the one using tensor products
is cool but rather unenlightening: it is the
kind of slick, quick proof which ends up
being classified as `unattainable dark magic'
by students. If they are sufficiently
familiar with tensor products, I would add
an exercise to their problem list providing
some hints for them to reach the proof using
tensor products by themselves.

My students are not familiar with tensor products. What I would like to
do is to re-write this approach using bilinear maps instead of tensor
products. Using, for instance, the fact that the dual of V _x_ W is
"the same thing as" the space of all bilinear maps from V x W into the
base field.

Best regards,

Jose Carlos Santos
.

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