Characterising a rotation matrix We have the operator $T: \mathbb{R}^3 \to \mathbb{R}^3$ given by $T(\vec{x}) = A\vec{x}$
$$A = \frac{1}{3}\begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix}$$
I have orthogonally diagonalized A and found that 
$$Q = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{2}} & \frac{-1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}} \end{pmatrix}$$
I now have to characterize the matrix by seeing what it does to the standard basis. The problem is, I find it very difficult to visualize such matters. 
I somehow see that the y-axis is rotated by 45 degrees counterclockwise (in one plane), but I'm rather uncertain about the other two axes, since they are rotations in 3D, rather than the simple 2D.
As a note, this is not a homework problem, but a practice problem for my exam. I have not done anything in particular to the Rodrigue's formula or about the trace. 
EDIT: The operator is indeed from $\mathbb{R}^3 \to \mathbb{R}^3$.

After orthogonally diagonalizing your matrix, you should find that $A = QDQ^T$, where $Q$ is as you've given it and
$$
D = \pmatrix{
0&0&0\\
0&-1&0\\
0&0&-1}
$$
We can easily see what happens to a vector in the eigenvector basis.  That is, if we define
$$
(a,b,c)_B = Q \pmatrix{a\\b\\c}
$$
Then you can see that
$$
A(a,b,c)_B = (0,-b,-c)
$$
That is: with respect to this new basis, we are projecting down to the "$yz$ plane" (with respect to the new basis) and then rotating in that plane by $180^\circ$.
If you want to see what $A$ does to the a vector in the standard basis, then there's no point in diagonalizing in the first place.
