Let $\mathcal{E} = \lbrace v^1 ,v^2, \dotsm, v^m \rbrace$ be the set of right eigenvectors of $P$ and let $\mathcal{E^*} = \lbrace \omega^1 ,\omega^2, \dotsm, \omega^m \rbrace$ be the set of left eigenvectors of $P.$ Given any two vectors $v \in \mathcal{E}$ and $ \omega \in \mathcal{E^*}$ which correspond to the eigenvalues $\lambda_1$ and $\lambda_2$ respectively. If $\lambda_1 \neq \lambda_2$ then $\langle v, \omega^\tau\rangle = 0.$
Proof. For any eigenvector $v\in \mathcal{E}$ and $ \omega \in \mathcal{E^*}$ which correspond to the eigenvalues $\lambda_1$ and $\lambda_2$ where $\lambda_1 \neq \lambda_2$ we have, \begin{equation*} \begin{split}\langle\omega,v\rangle = \frac{1}{\lambda_2}\langle \lambda_2 \omega, v\rangle = \frac{1}{\lambda_2} \langle P^ \tau \omega ,v\rangle = \frac{1}{\lambda_2}\langle\omega,P v\rangle = \frac{1}{\lambda_2} \langle \omega,\lambda_1v\rangle = \frac{\lambda_1}{\lambda_2}\langle \omega,v\rangle .\end{split} \end{equation*} This implies $(\frac{\lambda_1}{\lambda_2} - 1)\langle\omega,v\rangle = 0.$ If $\lambda_1 \neq \lambda_2$ then $\langle \omega,v\rangle = 0.$
My question: what if $ \lambda_2 = 0 \neq \lambda_1,$ how can I include this case in my proof.