| Oops, actually I should have thought before replying...
It is not necessry for the matrices to be simultaneously
diagonizable - another example that works is
A = I + Na, B = I + Nb, Na and Nb nilpotent such that Na*Nb = 0.
They must be transformable to an appropriate Jordan normal form
with a similarity is the real requirement.
The expression AB - BA is called a commutator and is often written
[A,B]. Commutators form an algebra (a Lie algebra) and satisfy something
called Jacobi's identity, which I have forgotten the exact form of, but
it relates 3 matrices something like this
[A,[B,C]] + [B,[C,A]] + [C,[A,B]] = 0.
I think this is right - it's a cyclic permutation of the matrices.
Offhand I'm not certain if [A,B] can be the identity or not.
- Jim
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Tr(AB)=Tr(BA). --- (1)
Tr() is a linear function. --- (2)
From (1) and (2), we have
Tr([A,B])=Tr(AB)-Tr(BA)=0. --- (3)
But we know that Tr(I)=n.
Hence [A,B] cannot be the identity.
-hiroshi
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| By using the formula Tr(XY)=Tr(YX),
you can show
Tr(ABAB)=Tr(A(BAB))=Tr((BAB)A)=Tr(BABA).
Tr(ABABAB)=Tr(BABABA).
...
Tr((AB)**i)=TR((BA)**i).
This shows the following.
For any i, sum of the i-th powers of the eigenvalues of AB is
equal to that of BA.
Now let x1,x2,.. xn be the eigenvalues of AB, y1,y2,... yn be
those of BA. Then we know that for any i,
(x1)**i + ... + (xn)**i = (y1)**i + ... (yn)**i.
And we can conclude that the values of the i-th symmetric
polynomials of x1,x2,... and y1,y2,... are equal. (*1)
Hence characteristic polynomials of AB and BA coinside.
Therefore their eigenvalues are equal.
(*1)
Symmetric polynomial is a polynomial of "sums of i-th powers."
e.g. (case n=3)
Let S1 = x1 + x2 + x3,
S2 = x1*x2 + x2*x3 + x3*x1,
S3 = x1*x2*x3 and
P1 = x1 + x2 + x3,
P2 = x1**2 + x2**2 + x3**2,
P3 = x1**3 + x3**3 + x3**3,
then
S1 = P1,
S2 = (P1**2 - P2)/2,
S3 = (P1**3 - 3*P1*P2 + P3).
General formula is known as Newton's formula.
-hiroshi
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