Re: Linear algebra with eigenvalue AB.

On Jun 9, 9:09 pm, magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
In article <3c5b06d9-d914-469d-8b24-89c5f007b...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,





Tonico  <Tonic...@xxxxxxxxx> wrote:
On Jun 9, 7:06=A0pm, 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 ? =A0 Why ?

Oh, Sorry. easy.
|xI - AB| =3D |xA.A^{-1} - AB| =3D |A.(xA^{-1} - B)|
|(xA^{-1} - B).A| =3D |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: =A0take det of =A0(I-AB)A =3D A(I-BA), and then
cancel =A0det(A), =A0valid since we're in the poly ring =A0Z[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=3Dy8zsl3e3nuy.fsf%40nestle.csail.mit...
u

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 i=
n
the real field). =A0For the case det A is not zero the OP already had a
solution, and for the general case the OP has already been given directi=
ons.

I explained the idea at length in various examples in my prior posts linke=
d
to by the above link. Perhaps you are making the common mistake of thinkin=
g
about polynomials (here det's) only as =A0functions, versus purely _formal=
_
expressions. In the above proof the det's are elements in the polynomial
ring =A0Z[Ai,Bj], which, being a poly ring over a domain, is itself a doma=
in,
so enjoys cancellation. In particular, the entries of the matrices Ai, Bj
are _indeterminates_, not real numbers. One can view the proof as verifyin=
g
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 =A0Ai,Bj =A0to specific values in any commutative =
ring.
The proof works precisely because the determinant is of _polynomial_ form
in its entries. So =A0A D =3D A D' -> D =3D D' because A,D,D' are polynomi=
als.
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.

Nobody says you can, but Bill is not cancelling the VALUE of the
determinant.

Here is an example. Suppose that you are working with 2x2
matrices. Consider FORMAL matrices A=(a,b;c,d) and B=(x,y;z,w). Here
a, b, c, d, x, y, z, and w are indeterminates. Then you can define
"det(A)" to be a polynomial in F[a,b,c,d,x,y,z,w], the polynomial ring
in 8 indeterminates with coefficients in F, namely

"det(A)" = ad - bc, the polynomial.

Likewise, det(B) = xw-yz.

Any matrix obtained by adding, multiplying, or scalar multiplying the
matrices A and B will yield a matrix whose entries are polynomials in
a, b, c, d, x, y, z, w, and its "determinant" will therefore be itself
a polynomial in a, b, c, d, x, y, z, and w.

In other words, do not plug in actual values for a, b, c, d, x, y, z,
and w, but rather treat them as indeterminant variables. the
polynomial ad-bc is not the zero polynomial in F[a,b,c,d,x,y,z,w], so
you can cancel it from expressions in that polynomial ring.

Only after you are done simplifying the expression in the polynomial
ring will you proceed to use an evaluation map to obtain actual real
values.

--
======================================================================
"It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org-

*******************************************************
Fair enough: something like is what I suposed he was trying to do, and
again the question is: what for? Mina's question was very specific and
simple, and he (together with with the help of others) already covered
all the possible cases in her problem: det A different from zero, and
equal to zero. Why then to make more messy simple stuff?
Perhaps there's something I'm missing, as I wrote before, but I can't
see the advantage in this specific case.

Regards
Tonio
.

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