$\displaystyle A: C^{3}\rightarrow C^{3} with A(e_{1})=e_{2}, A(e_{2})=e_{3}, A(e_{3})=e_{1} $
by consider the column vector $\displaystyle (1,v,v^{2})^{T} s.t. v^{3}=1 $
find its diagonalization and characteristic polynomial.
$\displaystyle A: C^{3}\rightarrow C^{3} with A(e_{1})=e_{2}, A(e_{2})=e_{3}, A(e_{3})=e_{1} $
by consider the column vector $\displaystyle (1,v,v^{2})^{T} s.t. v^{3}=1 $
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 $\displaystyle x = (1,v,v^{2})^{T} s.t. v^{3}=1 $, v is a cube root of unity.
Then observe that $\displaystyle Ax = v^2 x$. 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 $\displaystyle z = (1,v^2,v)^{T}$ Thus $\displaystyle Az = vz$. Also 1 is a cube root of unity. Thus associated with that is $\displaystyle y = (1,1,1)^{T}$ and $\displaystyle Ay = y$.
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.