by consider the column vector
find its diagonalization and characteristic polynomial.
The problem is too easy if all you have to find is the diagonal matrix. This is because we see that A^3 = I and thus the characteristic polynomial must divide x^3 - 1. But A is a linear transformation on a vec. space of dimension 3, thus the characteristic poly must be a monic of a degree 3. Thus x^3 - 1 is the char poly. The roots are cube roots of unity. Since the roots are distinct, the lin. transformation is diagonalizable. And thus is similar to diag(1, v , v^2) where v^3 = 1.
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.