Why does (2z²-x²-y², x²-y²) have -1 trace in the character table of Td for C3 rotation?
I think I know now.

Choose rotation axis that makes equal angles with the $x$-, $y$-, and $z$-axes so that the $C_3$ rotation essentially permutes the axes. Without loss of generality, I could choose a $C_3$ rotation that permutes $x$ to $z$, $z$ to $y$, and $y$ to $x$. Rotation in either direction will give the same trace with this $C_3$ rotation. With this rotation (permutation), $(2z^2-x^2-y^2,x^2-y^2)$ becomes $(2y^2-z^2-x^2,z^2-x^2)$. Multiplication of basis function vector a by $2\times 2$ matrix in this representation yields

$$\begin{pmatrix}
  -1/2 &amp; -3/2\\
  1/2 &amp; -1/2 
\end{pmatrix}
\begin{pmatrix}
  2z^2-x^2-y^2 \\
  x^2-y^2
\end{pmatrix}
= 
\begin{pmatrix}
  2y^2-z^2-x^2\\
  z^2-x^2
\end{pmatrix}
$$

The trace of this matrix is $-1$ as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.

The tricky thing about this example was that I needed to use a rotation in a three-dimensional vector space and project it into a rotation in a two-dimensional vector space.