About the eigenvalues of $A = (B + I)(B-I)^{-1}$

Question

Let $B$ s.t. $spectrum(B)=(\lambda_i(B))_i$ does not contain $1$.

1. Prove that the eigenvalues of $A = (B + I)(B-I)^{-1}$ are $\left\{\lambda_i(A) = \frac{\lambda_i(B) + 1}{\lambda_i(B) - 1};i\le n\right\}$.

2. Prove that, for every $i$, $Re(\lambda_i(A))<0$ if and only if the spectral radius of $B$ satisfies $\rho(B) < 1$.

Thank you so much in advance, especially if you can correct my statement.

Answer 1

If the spectral radius is less than $1$, the result is pretty immediate.

Let $(v, \lambda)$ be an eigenpair for $B$. Then $(v, \lambda + 1)$ is an eigenpair for $B+I$ and $(v, \lambda-1)$ is an eigenpair for $B-I$. If $\lambda \not= 1$ (which is guaranteed if $\rho(B) < 1$) then $(v, \frac{1}{\lambda - 1})$ is an eigenpair for $(B-I)^{-1}$, and putting it together, $(v, \frac{\lambda +1}{\lambda-1})$ is an eigenpair for $A$.

The other direction is absolutely not true. We can just look at $1 \times 1$ matrices for a counterexample. For instance, take $B = 3$. Then $\rho(B)= 3$, and yet still $A = 2$ has eigenvalue $2 = \frac{3 + 1}{3 - 1}$.