On this question, someone was kind enough to suggest the following outline of a proof.
1. Since z is a simple eigenvalue A, dim Nullspace(A-zI) = 1, i.e. there is a single linearly independent eigenvector associated with A-zI.
Let's call this vector x and let's complete it with n-1 vector x1, x2, ..., xn-1, xn = x so that the n vectors form a basis for my vector space (Rn or Cn). (Notice that the eigenvector is placed in the last column of P).
2. Place the n vectors as column in the matrix P: P = (x1|x2|...|xn). Now rank(P) = n so P is non-singular.
3. Show the form of P^(-1)AP. I use block matrix multiplication, so:
P^(-1)AP = transpose((y1|y2|...|yn)) A (x1|x2|...|xn) =
[y1Ax1 y1Ax2 ... y1Axn
y2Ax1 y2Ax2 ... y2Axn
ynAx1 ynAx2 ... ynAxn]
I can see straight away that the last column is 0nx1 since Axn = 0, but I fail to see why the last row should be 0nx1 too.
I would really appreciate any help here.