A.B= m1, B.A=m2, m1 and m2 diagonal matrices

[arivero]

can anybody refresh me on this? I would like to parametrize the pairs of n x n matrices such that the products (A . B) and (B . A) give diagonal matrices of real coefficients. I get lost already for n=3.

 

[arivero]

Hmm, ok, if m1 and m2 are diagonal and having no zero eigenvalues, it seems easy to proof that they have the same eigenvalues.

Now, what happens if there are not simultaneusly diagonal, ie when the commutator [m1,m2] is different of zero? Is it possible to have an A,B pair? Under what conditions?

 

[matt grime]

What does "an A,B pair" mean?

 

[arivero]

What does "an A,B pair" mean?

Srill as in the subject and the first post: a pair of square nxn matrices such that AB=m1, BA=m2.

The case with the commutator [m1,m2]=0 seems rather trivial, but I am still wondering how to attack the general case when [m1,m2] <>0.

Surely we can assume m1 diagonal, without losinge generality.

 

[matt grime]

Erm, all diagonal matrices commute with each other, so it can't be like "still as in the subject" and of course you can't assume that m1 is diagoneal, or m2. Why do you think AB must be diagonalizable? As I said, you've changed to some other unspecfied task: classify pairs of matrices A and B such that ... only yo'ure not saying what the ... is.

 

[arivero]

You are right, matt :-) overnight I noticed that the case in the subject was rather trivial, amounting to a permutation of the eigenvalues. So I tweaked the requeriment to have only one diagonal, say m1, and asking [m1,m2]<>0 (thus the other is not diagonal). Still, A and B are defined as in the subject, AB=m1, BA=m2. I have removed the other condition, simultaneus diagonal of m1, m2.

Which I am asking is to find and parametrize pairs of matrices A, B. Or, equivalently, to classify pairs of matrices A and B such that "there exist m1, m2 with AB=m1, BA=m2, [m1,m2]<>0 and m1 diagonal".

 

[arivero]

Ok I figured it, at least for the non degenerate case with |A| and |B| different of zero. Then AB and BA have the same spectrum of eigenvalues. (I didn't know it, I thought they shared just Det and Tr)

BTW, has this theorem got a name?

 

[matt grime]

I don't know if it has a special name, but you don't require both A and B to be invertible. It suffices for one of them to be invertible. Wlog A is invertible, then

AB-xI = A(BA-xI)A^{-1}

taking dets on both sides gives me x is an eigenvalue of AB iff it is an eigenvalue of BA.

Thus AB=m(1) and BA=m(2) implies A^{-1}m(1) =m(2)A^{-1} or equivalentky the m's are conjugate. This completely classfies the case when one of A or B is invertible.

 

[arivero]

This completely classfies the case when one of A or B is invertible.

Yep, that is. So we have proved that any family of massive Weyl bi-spinors is composed of pairs of equal mass spinors (AB and BA being the mass square of the right and left spinors). I'll think about the degenerate case (which could happen if still a neutrino has zero mass).

 

[arivero]

taking dets on both sides gives me x is an eigenvalue of AB iff it is an eigenvalue of BA

Furthermore, as AB and BA have the same characteristic polynomial, the same can be said of any cyclic pair A.B.C...E.F, B.C...E.F.A. Thus the ciclic property of the trace is generalised to any of the coefficients of the characteristical polinomial: writing
$$|Q- \lambda I_n|= \lambda^n + \tau_{n-1}[Q] \; \lambda^{n-1}+...+ \tau_1[Q] \; \lambda+ \tau_0[Q]$$
we get n "ciclic invariants". Guess $$\tau_{n-1}[Q]$$ is (a multiple of) the trace and $$\tau_0[Q]$$ the determinant, isn't it? Or do they appear as more complicated combinations?