Here' s the matrix in question

MatrixForm[A={{2,2,0},{-1,2,-1},{0,2,2}}]

First I find the characteristic polynomial of A and its roots

charpoly = Det[A-t IdentityMatrix[3]]

Solve[charpoly==0,t]

Then I find eigenvectors for the first two of these roots (I can ignore the third, because it's the complex conjugate of the second)

Solve[A.{x1,x2,x3}==2{x1,x2,x3}]

Solve[A.{x1,x2,x3}==(2-2I){x1,x2,x3}]

So (-1, 0, 1) is a basis for the eigenspace of the eigenvalue 2, and (1, -i, 1) is a basis for the eigenspace of the eigenvalue 2 - 2 i.  The real and imaginary parts of the latter vector are (1,0,1) and (0,1,0).  With these two vectors and the eigenvector for 2, I get a basis for R^3.

basis = {{-1,0,1},{1,0,1},{0,-1,0}}

And the change of coordinates matrix from this basis to the standard basis is just the matrix whose columns are the basis vectors.  Since I' ve entered the basis vectors as rows, I take the transpose

MatrixForm[chg = Transpose[basis]]

According to the theorem in class, the matrix for my linear transformation should be nicer if I use this basis :

MatrixForm[newA = Inverse[chg].A.chg]

So newA has ' block diagonal' form, as I hoped.  Moreover, the bottom block can be rewritten as

MatrixForm[block = r{{Cos[Theta],-Sin[Theta]},{Sin[Theta],Cos[Theta]}}]

where r and Theta are determined by the eigenvalue 2-2I = r e^(-i Theta).  That is, 
   r = | 2 - 2 i | = 2^(3/2)
is the magnitude of 2-2I, and 
   Theta = -tan^{-1}( 2/(-2)) = Pi/4
is minus the argument of 2-2I.  Just to be sure, I check it:

MatrixForm[block//.{r->2Sqrt[2],Theta->Pi/4}]

... which agrees with the bottom - right block in the matrix above