Finding eigenpairs for Markov matrices

Question

If we're given a $ \displaystyle 2 \times 2 $ Markov matrix (so all entries are non-negative and columns add to 1) M$(a,b)$ such that $$M = M(a,b) := \begin{bmatrix}1 - a & b\\a & 1 - b \end{bmatrix}$$ where $a$ and $b$ are $ 0 ≤ a ≤ 1, 0 ≤ b ≤ 1$, how could you define $N := 1 - M$? I'm confused how to show that if $⟨µ,u⟩$ is an eigenpair for M, why is $⟨1 − λ,u⟩$ in an eigenpair for N?

Answer 1

Note that if $\langle 1 - \lambda, \vec u \rangle$ is an eigenpair for $N$, then: \begin{align*} N\vec u &= (1 - \lambda)\vec u \\ &= \vec u - \lambda \vec u \\ &= I_2\vec u - M \vec u &\text{since $\langle \lambda, \vec u \rangle$ is an eigenpair for $M$}\\ &= (I_2 - M) \vec u \end{align*} This suggests that $N$ should be the $2 \times 2$ identity matrix minus $M$.