Deriving the irreps of the d-orbitals under C3 and S6 operations in octahedral symmetry
The $C_3$ operation rotates $x \rightarrow y$, $y \rightarrow z$ and $z \rightarrow x$.
Hence,
$$\phi_1=2z^2-(x^2+y^2)$$ 
rotates into 
$$-(1/2)\phi_1 +(3/2)\phi_2$$ and
$$\phi_2=x^2-y^2$$
rotates into 
$$-(1/2)\phi_1 -(1/2)\phi_2$$
so, in the basis
 $( \phi_1, \phi_2 )$ the matrix that represents $C_3$ is

$$\left [\begin{matrix} -1/2 &amp; -1/2  \\ 3/2 &amp; -1/2  \end{matrix}\right]$$

Denoting this 2 by 2 matrix by M what we have is
$C_3(\phi_1  \phi_2) = (\phi_1 \phi_2)M$.
The trace of this matrix is $-1$. 

This is what you find in the character table.

Since $C_3^3=I$ (identity) its eigenvalues are the cubic roots of unity:
$1, \lambda$ and $ \lambda^*$
where $\lambda = \exp(2\pi i/3)=-1/2+i\sqrt 3/2$,
and $\lambda^*$ is the complex conjugate of $\lambda $.

It is easy to find eigenvectors with the eigenvalue 1 [e.g., an s orbital or x+y+z].

You can find a linear combination of $\phi_1$ and $\phi_2$ that will be an eigenvector of $C_3$ with the eigenvalue $\lambda$ (transform into itself multiplied by $\lambda$)
and another linear combination that will have eigenvalue $\lambda^*$. Now, the matrix representing $C_3$ will be diagonal with $ \lambda$ and $\lambda^*$ along the diagonal. The trace remains $-1$.
Something similar should work for $S_6$.