Eigenvector of $A$ and $(I-A)$

Question

$A\in \mathbb {R}^{n\times n}$, $\|A\|<1$, and $\vec b\in \mathbb {R}^{n}$. If $\vec b$ is the eigenvector of $A$ corresponding to the largest eigenvelue, which is $0<\lambda<1$. Is is possible for $\vec b$ to be the eigenvector of $(I-A)$ corresponding to the largest eigenvalue?

Answer 1

Yes, it is possible. Just take $A=\lambda\operatorname{Id}$.