On 1/29/08, Rutwig <rutwig@...> wrote:
>
> A quite interesting problem. The point as I see it is to ensure that
> the identity multiple is not zero.
> Actually, take a,b,c any nonzero scalars and consider the matrices
>
> X= a(I_12 + I_13) [I_{jk} having 1 in j-row and k-column, rest zero)
>
> Y= b(I_21 + I_23)
>
> Z= c(I_31+ I_32)
>
> Then
>
> XY= ab(I_11+I_13), YX=ab(I_22+I_23), XZ= ac(I_11+I_12),
>
> ZX= ac(I_32+I_33), YZ= bc(I_21+I_22), ZY= bc(I_31+I_33).
>
> Now the tri-products:
>
> XYZ= bXZ, XZY= cXY, YXZ= aYZ, YZX= cYX, ZXY= aZY, ZYX= bZX.
>
> Thus the six matrices are independent for any values of a,b,c. Now,
> these matrices have all the same trace, namely abc. Thus the
> diagonal operator vanishes.
>
> Now, if you ensure that the products XY,YX,... have different trace
> (at least a pair), then the seventh matrix will provide the kE matrix.
>
> This for the moemnt. I will have a closer look to this.


Many thanks. Actually, for my purposes the question is even simpler
than I formulated, and your example provides the answer. Namely, I
have the following situation: for any n x n (but it is obvious that it
is enough to consider the case 3x3) traceless matrices X,Y,Z, the
following equality holds:
XYZ \otimes a + ... + ZYX \otimes f + (Tr(XYZ) - Tr(XZY)E \otimes g = 0,
where dotted terms run through all triple products of X,Y,Z, and a,
... , f, g are elements of a certain ring. From this I want to infer
that all elements a, ..., f, g vanish separately. Surely that would
follow from the statement of linear independence I formulated
initially, but we may choose an easier tactic and chop out this
equality piece by piece. Indeed, substituting your example we see that
a, ..., f vanish separately, so we are left with (Tr(XYZ) - Tr(XZY)E
\otimes g = 0. Now, obviously, there are matrices X,Y,Z such that
Tr(XYZ) - Tr(XZY) not zero (irrespective of the first example), and we
have g = 0.

I don't know why this or similar natural examples didn't came to my
mind - probably because I immediatley rushed for computer and produced
some randomly looking cumbersome examples.


> --- In liealgebras@yahoogroups.com, "Pasha Zusmanovich"
>
> <justpasha@...> wrote:
> >
> > Dear people,
> >
> > I have a question which belongs more to the domain of linear algebra
> > then to those of Lie algebras. My excuse is that it appeared genuinely
> > in Lie-algebraic context - namely, in description of Poisson
> > structures on the Lie algebra sl_n(A).
> >
> > Consider the following statement: there are three 3x3 traceless
> > matrices X, Y, Z such that 7 matrices XYZ, XZY, YXZ, YZX, ZXY, ZYX,
> > and (Tr(XYZ) - Tr(XZY))E (E is the identity matrix) are linearly
> > independent. Play with computer suggests that apparently any three 3x3
> > traceless matrices in general position are such. I would like to have
> > an elegant (algebro-geometric, linear-algebraic, whatever) proof.