by consider the column vector
find its diagonalization and characteristic polynomial.
However if you have to find the similarity matrices, you have to do find the eigen vectors and you have to do the following:
Call , v is a cube root of unity.
Then observe that . This means x is an eigenvector associated with v^2. But v^2 is also a cube root of unity,so replacing that in the place of v,we get the associated eigen vector Thus . Also 1 is a cube root of unity. Thus associated with that is and .
Thus x,y,z are the required eigenvectors. As a side note they are clearly linearly independent since the matrix formed by x,z and y will be Vandermonde.