Proof of tr(AB)=tr(BA)
Sikari
Would someone give me an elementary proof of following (without using
eigenvalues).Trace(A) is denoted tr(A).
Theorem
tr(AB)=tr(BA)
Regards Sikari
Robin Chapman
> Hi
> Would someone give me an elementary proof of following (without using
> eigenvalues).Trace(A) is denoted tr(A).
Erm, why would one want to use eigenvalues?
> Theorem
> tr(AB)=tr(BA)
Write the entries of A as a_{i,j} and those of B as b_{i,j}
and calculate explictly the entries of AB and BA and
hence their traces. Should end up the same.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen
Zdislav V. Kovarik
Matrices? Let C = A*B, D = B*A,
write in full the sum of C(i,i) and the sum of D(i,i),
and compare.
Linear transformations in a finite dimensional space?
Prove it first for A of rank 1, and then extend by
linearity.
Banach spaces and suitable operators? Prove for A of finite
rank first (as above), then extend by limits.
Cheers, ZVK(Slavek).
Charles Matthews
"Robin Chapman" wrote
> Erm, why would one want to use eigenvalues?
I wouldn't, in this case. There is a proof technique that uses some
Zariski-dense subset, on which you can give a more 'obvious' reason why a
purely algebraic identity is true.
For example, Cayley-Hamilton is obvious for diagonal matrices, thence
obvious for diagonalisable matrices. These are dense, for topologies a lot
stronger than Zariski's. So it is true generally.
I was once told this was the world's worst proof of Cayley-Hamilton. But it
tells _me_ why it's true.
Charles
Denis Feldmann
> a simple way to prove this is as follows:
Doesn't Mrs Unreliable suffice for this ?
> suppose it was not true and then derive contradiction, which would
> prove that the converse is true. so here we go...
>
> suppose tr(A*B) != tr(B*A). Now let A=B=I where I is the identity
> matrix so tr(A*B)=tr(I*I)=tr(B*A) so we get a contradiction.
> therefore our assumption that tr(A*B)!= tr(B*A) must be false and
> hense tr(A*B)=tr(B*A) must be true.
>
> nizar
>
>
> On 19 Sep 2002, Sikari wrote:
>
>> Date: 19 Sep 2002 10:02:45 -0700
>> From: Sikari <sika...@spray.se>
>> Newsgroups: sci.math
>> Subject: Proof of tr(AB)=tr(BA)
Randy Poe
>
> a simple way to prove this is as follows:
> prove that the converse is true. so here we go...
>
> suppose tr(A*B) != tr(B*A). Now let A=B=I where I is the identity matrix
> so tr(A*B)=tr(I*I)=tr(B*A) so we get a contradiction.
> therefore our assumption that tr(A*B)!= tr(B*A) must be false and
> hense tr(A*B)=tr(B*A) must be true.
You have a small logical fallacy here.
The claim tr(A*B) = tr(B*A) means "for all A, B, this statement
is true". The converse tr(A*B) != tr(B*A) means "these are not
equal in general, or there exists at least one choice of A and
B for which the equality does not hold".
You can't go from there to letting A=B=I because the
converse is not claiming equality NEVER holds. Only that
there are A and B for which it fails.
Aside: In every logical fallacy is a proof there's usually
a proof that 0 = 1. Here's mine:
Theorem: 2x = x
Proof: Suppose 2x != x. Let x = 0, from which we see
2x = 0 = x. Contradiction. Therefore 2x = x.
Corollary 1: 1 = 2
Proof: Let x = 1 in the above theorem.
Corollary 2: 0 = 1
Proof: Since 1 = 2, subtract 1 from both sides.
Conjecture:
There is also an elementary proof of FLT in every
logical fallacy. Paging JSH, paging JSH...
- Randy
David C. Ullrich
<ni...@finn.cds.caltech.edu> wrote:
>a simple way to prove this is as follows:
>suppose it was not true and then derive contradiction, which would
>prove that the converse is true. so here we go...
>
>suppose tr(A*B) != tr(B*A). Now let A=B=I where I is the identity matrix
>so tr(A*B)=tr(I*I)=tr(B*A) so we get a contradiction.
>therefore our assumption that tr(A*B)!= tr(B*A) must be false and
>hense tr(A*B)=tr(B*A) must be true.
Excellent. One can give a similar proof that (a + b)^2 = a^2 + b^2:
Assume that (a + b)^2 != a^2 + b^2, now let a = 0 and b = 0 and
you see that the assumption that (a + b)^2 != a^2 + b^2 must be false.
Hint: The problem is to show that Tr(AB) = Tr(BA) for all A, B. The
negation of this is not "Tr(AB) != Tr(BA) for all A, B."
>nizar
>
>
>On 19 Sep 2002, Sikari wrote:
>
>> Date: 19 Sep 2002 10:02:45 -0700
>> From: Sikari <sika...@spray.se>
>> Newsgroups: sci.math
>> Subject: Proof of tr(AB)=tr(BA)
>>
>--
David C. Ullrich
Oscar Lanzi III
quantum mechanics. Cases where the equality may break down are those in
which the trace is not absolutely convergent.
So any proof requires finite matrices or, if for infinite matrices,
absolute convergence of the trace as given by the sum of the diagonal
elements.
--OL