Re: Linear algebra with eigenvalue AB.

On Jun 9, 11:03 pm, Bill Dubuque <w...@xxxxxxxxxxxxxxxxxxxx> wrote:
Tonico <Tonic...@xxxxxxxxx> wrote:
Arturo Magidin <magi...@xxxxxxxxxxxxxxxxx> wrote: (*edited*)
Tonico <Tonic...@xxxxxxxxx> wrote:
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.

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, so  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...

Nobody says you can. 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
S = {a,b,c,d,x,y,z,w} is a set of indeterminates. Then you can define
"det(A)" to be a polynomial in F[S] = F[a,b,c,d,x,y,z,w], the polynomial
ring in 8 indeterminates with  coefficients in F, namely "det(A)" = ad-bc,
a polynomial in F[S]. 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
F[S] and its "determinant" will therefore be itself a polynomial in F[S]

In other words, do not plug in actual values for  a,b,c,d,x,y,z,w,
but rather treat them as indeterminant variables. In F[S], the polynomial
det(A) = ad-bc is not the zero polynomial in F[S], so you can cancel it
from equations 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 real values.

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?

Not true. The only other proof mentioned here for the case det(A)=0 was
a reference by Robert Israel to a prior thread, where Robin Chapman says
that it can be proved by a topological argument (but gave no details).
Such a proof is far more complex than the trivial one-line algebraic
proof I presented above. It seems you have not followed my suggestion
to read my prior posts where I gave many more examples (some simpler)
illustrating such universal techniques. This is a very simple yet very
powerful technique in algebra that every mathematician should learn
to master. Often times one will see analysts struggle with obfuscated
topological density arguments when all that is required is a simple
algebraic argument exploiting the universality of polynomials. It
really is completely and utterly trivial once you can break through
the mental barrier imposed by analytic vs. algebraic thinking.

This should be learned in a first course in abstract algebra but,
alas, it seems many people don't. The fact that you are continuing to
struggle to comprehend something that is utterly trivial only serves
as further evidence reinforcing my oft-stated claim here about how
widely misunderstood is the simple algebraic notion of a polynomial.

I've probably written over 50 posts discussing many variations on
this theme, so please take the time to read a few of them if you
honestly want to learn something about it. Below is the link again.
FOLLOW THE LINKS, in my prior posts below, then follow links in these
posts, etc, etc, to traverse the entire tree of said prior posts.
There you'll find many similar examples, e.g. the Cayley-Hamilton
theorem;  adj(A) = det(A)^(n-1);  Leibniz's product rule for the
derivative of polynomials;  computing partial fraction expansions...
You should soon realize the technique has widespread application.http://google.com/groups?threadm=y8zsl3e3nuy.fsf%40nestle.csail.mit.edu

--Bill Dubuque-
*******************************************************

Thanx for the links. it's a little messy to follow those (I already
reached once 1999 and other time 2000...), so: do you have some site
where these techniques are written? For example, I tried to read about
the |Adj A| = |A|^(n-1) but I only found links to other posts with
links to other post with links to...*pant*.

If all this is more or less concentrated in some place, book or
articles, it'd be great.

Regards
Tonio
.

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