We have a diagonal traceless matrix like the 3rd Pauli matrix, or Gell-Mann's matrices $\lambda_8$ and $\lambda_3$ in su(3). Now, we want to apply a unitary transformation on them and make them have a zero-diagonal and non zero-off diagonal elements somehow, such that the resulting matrix is orthogonal to the initial matrix. Is there such a unitary matrix?
$$U\lambda U^\dagger =\lambda^\prime , \qquad \operatorname{tr} \lambda^\prime \lambda=0 ~.$$
Well, you have been doing this for breakfast for SU(2), for Pauli matrices, Rotating by $\pi/2$ around the x-axis, $$ U=\frac{1}{\sqrt{2}} ( 1\!\!1 + i\sigma_1), $$ so that $$ \frac{1}{2} ( 1\!\!1 + i\sigma_1) \sigma_3 ( 1\!\!1 - i\sigma_1) =\sigma_2, $$ with tr$\sigma_2\sigma_3 $=0, of course.
In fact, any group rotation not around a Cartan-subalgebra member in general helps to get you there. (That is, avoid rotating $\lambda_3$ around $\lambda_8$.) Special care in the rotation accords to projecting the original direction out.