Are you familiar with Schur's theorem? A real matrix has a Schur decomposition iff all of its eigenvalues are real. So its characteristic polynomial must split.
I hope this helps.
You are correct that every complex matrix has a Schur decomposition
Hi I've got this question on an upcoming assignment and I can't think what to do except I think V can't be complex.
Give an example of a vector space V and a linear transformation T such that T cannot be represented by an upper triangular matrix with respect to any basis of V
Any help would be great