Linear algebra question , help!?

It is easy to show that the set of eigenvalues of A^k is the set of k-th degrees of eigenvalues of A.

tr(B) = trace(B) is the sum of all eigenvalues of B, so as tr(A) = tr(A^2) = tr(A^3) = 0 we have (denote the eigenvalues of A a, b, c - three complex numbers)

a + b + c = 0, a^2 + b^2 + c^2 = 0, a^3 + b^3 + c^3 = 0. Let's show that a = b = c.

Really, if a, b and c are all distinct numbers, then the matrix (a b c; a^2 b^2 c^2; a^3 b^3 c^3) is invertible because its determinant is not equal to zero (similar to Vandermonde one), so the corresponding homogeneous system has a single solution and we know this solution: a = b = c = 0. If all a, b and c are equal then they are of course equal to zero. If for example a = b, then c = -2a, c^2 = -2a^2 and hence a = 0 and c = 0. So a = b = c = 0 in any case.

So the Jordan normal form of A is an upper triangular matrix with all the numbers on the diagonal equal to zero. It is trivial that then its third degree is equal to zero and so the JNF of A^3 is equal to zero. Then, of course, A^3 = 0.

I can now honestly say that brute force, while seemingly effective, is not necessarily the best idea on a problem like this. The link provided gives a word document that gives the exact approach, start to finish, that I used.

Again, while not ideal, it was effective, and did eventually give me a result, albeit not quite the one I had originally expected.

http://www.sendspace.com/file/rnzodl

Critiques are encouraged, and feel free to point out any errors, as arithmetic is often a fickle beast...

Another approach would be to say that the trace of A is the sum of the eigenvalues of A, and the trace of A^2 is the sum of eigenvalues of A^2, trace of A^3 is the sum of the eigenvalues of A^3...

You might be able to use the relations you get out of that to tackle this problem in a much shorter, more elegant way, but as I just thought of it and haven't tried it yet, I can't be sure.