L-x said:
This looks interesting but I'm having quite a lot of trouble following your notation. Do you know the name of the theorem so i could look it up perhaps?
Thanks for the help.
The old books “Matrices”, by Gantmacher, or the book “Theory of Matrices”, by P. Lancaster (Academic Press, 1969) have these results. I am sure there are numerous more modern treatments, but those two are the ones I have on my bookshelf.
Anyway, below I will give a short proof for the case of a 3 x 3 matrix A having 3 distinct, non-zero real eigenvalues r1, r2 and r3. The same form of proof goes through for a general nxn matrix with n distinct, real, nonzero eigenvalues. The basic result is the same, but with a somewhat different proof, if some of the eigenvalues are complex or some are zero. Distinctness remains important; without it we need to invoke Jordan Canonical forms, and f(A) can involve not only f(r) for eigenvalue r, but also f’(r). f’’’(r), etc., depending on the multiplicity of r and the form of its Jordan block(s).
So, let ui and vi be the left and right-eigenvectors of A for eigenvalue ri, chosen so that ui*vi = 1. Note that for rj different from ri we have ui*vj = 0; see, eg,
http://answers.yahoo.com/question/index?qid=20100729070957AAlQuHF . Note that the vi form a basis for the whole space. Now let Ei = vi*ui for all i; we have that Ei is a 3x3 matrix because it is of the form column x row in that order. The orthogonality results imply that Ei*Ei = Ei and Ei*Ej = 0 if i is different from j. Letting B = sum ri*Ei, we have that B*vi = ri*vi for all i, hence (B-A)*vi = 0 for all vi, hence (B-A)w = 0 for all vectors w, because all such w have the form sum ci*vi . Thus, B = A, so we have shown that A = sum ri*Ei. Now look at C = E1+E2+E3. We have C*vi = vi for all i, so C*w = w for all vectors w. That is, C = Identity matrix I. Now we are almost done.
Look at any polynomial p(x) = c0 + c1*x + c2*x^2 + … + cm*x^m, and _define_ p(A) = c0*I + c1*A + c2*A^2 + … + cm*A^m. We have A^2 = sum ri^2 * Ei*Ei + sum_{i < j}ri*rj Ei*Ej + sum_{j < i} rj*ri*Ej*Ei = sum ri^2 * Ei + 0 + 0. Similarly, A^3 = sum ri^3*Ei, etc. Finally, we have I = sum Ei = sum ri^0 * Ei (this is where having all ri nonzero comes in!), so we have that p(A) = c0*A^0 + c1*A^1 + … + cm*A^m is of the form p(A) = sum p(ri)*Ei. If f(x) is an analytic function whose radius of convergence includes the largest |eigenvalue|, then the same type of argument goes through, by taking some limits. The result is f(A) = sum f(ri)*Ei.
This stuff is typically used in solving constant-coefficient couples ODEs, because if x is a vector and A is a constant matrix, the DE system dx/dt = Ax has solution x(t) = exp(A*t)*x(0), so we need to know how to compute matrix exponentials.
RGV
