Conference rusure::math

    Look at it the same way as before, by changing to an appropriate
    basis.  For instance, if the matrices are diagonizable via the same
    similarity then nothing stops one or both of them from having zeroes
    on the diagonal, when their product will zeroes on its diagonal as
    well...  The same appplies to reasoning using Jordan normal forms.

    - Jim

>Also can AB - BA  ever equal  I  ??
                                                           
    I think so.  If you look at L-2 space and look at the operators
                               
    A = (x + d/dx) and B = (x - d/dx) you will find that AB - BA = kI 
    
    (I the identity operator, and k a constant).  Of course, this is for
    
    the infinite dimensional L-2, but it shouldn't be much problem
    
    Restricting it to make the whole thing finite dimensional (using
    
    step functions, and etc).
    
    Eugene

   Re .-1

	I don't know how you thought of that!  It does work for L-two
    space, but doesn't apply to the matrix case.

	In fact, [A,B] = AB-BA cannot ever equal the identity because
    the trace of [A,B] is always zero.  This is because the traces of
    AB and BA are equal.

	Actually, not only are the traces equal, but so is the determinant
    and in fact so are the characteristic polynomials of the matrices AB and BA.
    This is clear if A or B is nonsingular, since a similarity transform
    by A (say) reverses the order and the characteristic polynomial of similar
    matrices is the same, but is true in general.

	There is an exception - if the matrices are in base 2, then you
    could do it.

	Are the minimal polynomials of AB and BA the same?

	What is the "trace" of the linear operator in .-1?

    - Jim

    re .3,
    You are right, it doesn't work for the finite dimension cases.
    As to the "traces" of the operators in .2....  In order for an operator
    on L-2 to have a trace, it must belong to the trace class or
    equivalently the operator has to be a product of two Hilbert-Schmidt 
    operators.  The guys in .2 aren't in the trace class.  As a matter of 
    fact they aren't even bounded operators on L-2.
    
    Eugene

    Right, that would work... but Hilbert-Schmidt operators are really like
    infinite dimensional matrices (so you have to talk about convergent
    series), and in that case the idea of trace is fairly natural.  If you
    let the number of "entries" in the matrices become continuously many
    then there are problems - series go to integrals for example, and you
    have to use delta functions (distributions.)

    The practical importance of trace class operators has to do with quantum
    statistics.  If you compute an expectation from a density operator
    then the trace has to converge - that may not happen even though the
    sum of squares does converge (consider 1 + 1/2 + 1/3 + 1/4 + ...)

    - Jim

	Re .-3

	>Are the minimal polynomials of AB and BA the same?

	Let A = /0 1\ , B = /0 0\ .
                \0 0/       \0 1/
	Then AB = /0 1\ and BA = 0.
                  \0 0/
	Hence the minimal polinomials of AB and BA are different.

	-hiroshi