On Feb 19, 12:06 am, Robert Israel <isr...@math.MyUniversitysInitials.ca> wrote: > jude <jw12jw12j...@yahoo.com> writes: > > Suppose A and B are commuting nxn matrices. It does not follow that > > one of these can be expressed as a polynomial in the other. It also > > does not follow that A and B can both be expressed as polynomials in > > terms of a third matrix C.
> > You're right (assuming those are M & M's matrices: I don't have that > book). > > > (1) What is the example supposed to be? > > I don't know, but I have managed to come up with one of my own. > > Consider the 4 x 4 case > > [ 0 1 0 0 ] [ 1 0 1 0 ] > [ 0 0 0 0 ] [ 0 1 0 1 ] > A = [ 0 0 0 1 ], B = [-1 0 -1 0 ] > [ 0 0 0 0 ] [ 0 -1 0 -1 ] > > Suppose there is a matrix C with f(C) = A and g(C) = B. > If v is an eigenvector for C, it must also be an eigenvector for > A and for B. But the only common eigenvectors of A and B are > the multiples of > > [ 1 ] > [ 0 ] > [-1 ] > [ 0 ] > > Adding a multiple of I if necessary, we can assume wlog that > the eigenvalue is 0. > Now the Jordan canonical form of C must be > [ 0 1 0 0 ] > [ 0 0 1 0 ] > [ 0 0 0 1 ] > [ 0 0 0 0 ] > > If f is a polynomial such that f(C) = A, then we can take f to be of > degree at most 3 (because C^4 = 0), and the coefficients of x^0 and x^1 > must be 0 (otherwise we couldn't have A^2 = 0). Thus A = a C^2 + b C^3 > for some a and b, and the kernel of A is the same as the kernel > of C^2, which has dimension 2. The same must be true of the kernel > of B. But A and B do not have the same kernel, so we have a contradiction. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada
There's just one step in your explanation I don't follow (the "wlog" part). A,B, and C are commuting matrices and you want to see if A and B are expressible as polynomials in C. I see that adding a multiple of I to C will not affect the commutativity, but then A and B won't necessarily be polynomials in C+kI. Isn't that a problem?