I was trying to evaluate the eigendecomposition of the matrix $$A=\left(\begin{matrix}0&b\\-b&0\end{matrix}\right)$$I easily found the eigenvalues and eigenvectors to be
$$ \lambda_1 = ib \longrightarrow v_{\lambda_1} = \left(\begin{matrix}1\\i\end{matrix}\right) \\[20pt] \lambda_2 = -ib \longrightarrow v_{\lambda_2} = \left(\begin{matrix}1\\-i\end{matrix}\right) $$
And from this is just a matter of evaluating the projection matrices on $v_1,v_2$ because for eigendecomposition we have that
$$ A = \sum_{j=1}^n \lambda_i\hat{\mathcal{P}}_{v_i}\tag 1 $$
The projection matrix can be found easily as
$$ \hat{\mathcal{P}}_{\mathbf{v}} = \mathbf{v}\mathbf{v}^T $$
so in my case
$$ \hat{\mathcal{P}}_{v_1} = \left(\begin{matrix}1\\i\end{matrix}\right)\left(\begin{matrix}1&i\end{matrix}\right) = \left(\begin{matrix}1&i\\i&-1\end{matrix}\right) \\ \hat{\mathcal{P}}_{v_1} = \left(\begin{matrix}1\\-i\end{matrix}\right)\left(\begin{matrix}1&-i\end{matrix}\right) = \left(\begin{matrix}1&-i\\-i&1\end{matrix}\right) $$
Then I tried, just for double-checking, that this were indeed projection matrices. I used the definition
$$ \hat{\mathcal{P}}_{\mathbf v}^2 = \hat{\mathcal{P}}_{\mathbf v} $$
but this is not clearly the case. As you can see, for $\hat{\mathcal{P}}_{v_1}$ we have
$$ \hat{\mathcal{P}}_{v_1}^2 = \left(\begin{matrix}1&i\\i&-1\end{matrix}\right)\left(\begin{matrix}1&i\\i&-1\end{matrix}\right) = \left(\begin{matrix}0&0\\0&0\end{matrix}\right) $$
Clearly a projection shouldn't be nilpotent! And not even only that! Formula $(1)$ doesn't hold, neither do
$$ \hat{\mathcal{P}}_{v_i}\hat{\mathcal{P}}_{v_j} = \delta_{ij}\hat{\mathcal{P}}_{v_j}\\ \sum_{j=1}^n \hat{\mathcal{P}}_{v_j} = \mathbb{I} $$
Question: where is the problem in this solution?
If you're doing it in $\Bbb C$, you have to use the conjugate transponse $v^H$, not $v^T$ (and be careful that you may need to normalize the vectors).