Re: Linear algebra with eigenvalue AB.

On Jun 9, 7:06 pm, Bill Dubuque <w...@xxxxxxxxxxxxxxxxxxxx> wrote:
Tonico <Tonic...@xxxxxxxxx> wrote:
Bill Dubuque <w...@xxxxxxxxxxxxxxxxxxxx> wrote:
Tonico <Tonic...@xxxxxxxxx> wrote:
On Jun 8, 4:29?pm, "mina_world" <mina_wo...@xxxxxxxxxxx> wrote:
"mina_world" <mina_wo...@xxxxxxxxxxx> wrote:

AB and BA have same eigenvalues, for A,B in Mat_n(R).
Is this true ?   Why ?

Oh, Sorry. easy.
|xI - AB| = |xA.A^{-1} - AB| = |A.(xA^{-1} - B)|
|(xA^{-1} - B).A| = |xI - BA|.

Yup, though there's a step that'd need clarification....
and what if A is NOT invertible and you have no A^(-1)?

The proof is easy:  take det of  (I-AB)A = A(I-BA), and then
cancel  det(A),  valid since we're in the poly ring  Z[Aij,Bij]
which is an integral domain. It works even for non-square matrices,
see all of my prior posts here for much further discussion:
http://google.com/groups?threadm=y8zsl3e3nuy.fsf%40nestle.csail.mit.edu

The OP asked a question about real matrices. I can't see how can you
cancel out det A if you're not sure whether det A is not zero, no matter
whether we're in an integral domain or not (btw, in this case det A is in
the real field).  For the case det A is not zero the OP already had a
solution, and for the general case the OP has already been given directions.

I explained the idea at length in various examples in my prior posts linked
to by the above link. Perhaps you are making the common mistake of thinking
about polynomials (here det's) only as  functions, versus purely _formal_
expressions. In the above proof the det's are elements in the polynomial
ring  Z[Ai,Bj], which, being a poly ring over a domain, is itself a domain,
so enjoys cancellation. In particular, the entries of the matrices Ai, Bj
are _indeterminates_, not real numbers. One can view the proof as verifying
the identity for "generic" matrices A,B (i.e. having indeterminate entries).
Since the identity is true generically, it certainly remains true when one
specializes the indet's  Ai,Bj  to specific values in any commutative ring.
The proof works precisely because the determinant is of _polynomial_ form
in its entries. So  A D = A D' -> D = D' because A,D,D' are polynomials.
Once you understand the basic idea the proof becomes absolutely trivial.
It's a nice example of the power of polynomials (formal vs. functional)
and the _universality_ of polynomial identities.

--Bill Dubuque-

}***************************************

I may be missing something, but I still cannot see how can you cancel
anything anywhere if that something is zero. Perhaps you'reaking the
common mistake of not understanding that the OP SPECIFIED that his
matrices are real, son det A is a real number and thus cannot be
cancelled unless it is not zero.
I can't see what's the advantage to get into weird polynomial rings
and stuff...and EVEN there you can't cancel unles the cancelled thing
isn't zero...

Regards
Tonio
.

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