Every orthogonal operator is diagonalizable? Answer is false and the rotation is a counterexample.
But I can't understand well.
Let $A=\begin {pmatrix}0&amp;-1\\1&amp;0\end{pmatrix}$ then it is rotation and also orthogonal operator. I think it can be diagonalizable by eigenvalue $i,-i$.
They (implicitly) meant "diagonalizable over the reals", and making a diagonal matrix whose entries are $i$ and $-i$ are not allowed when we are requiring all of our matrices' entries be real numbers. From the relevant Wikipedia article:


  In general, a rotation matrix is not diagonalizable over the reals, but all rotation matrices are diagonalizable over the complex field.
You should have been right that it is diagonalizable because there exists $X$ s.t. $A = X\Lambda X^{-1}$. There are also non-diagonalizable matrices that always have the form $A = XJX^{-1}$, where $J$ is the Jordan matrix of $A$.