Show that $\frac{1-\lambda}{1+\lambda}$ is the eigen value

Question

$\alpha$ is an eigen vector of the real skew symmetric matrix $S$ with eigen value $\lambda$. How can I show that $\alpha$ is eigen vector of $(I+S)^{-1}(I-S)$ with eigen value $\frac{1-\lambda}{1+\lambda}$ ?

My attempt:

$$S\alpha=\lambda\alpha$$ $$\lambda\alpha-S\alpha=0$$ $$det(\lambda\alpha-S\alpha)=0$$ $$det(I\alpha-S\alpha+\lambda\alpha-I\alpha)=0$$ $$det((I-S)\alpha-(1-\lambda)\alpha)=0$$

How to proceed now?

Answer 1

You have $S\alpha=\lambda\alpha$

So, $(I+S)^{-1}(I-S)\alpha=(I+S)^{-1}(1-\lambda)\alpha=(1-\lambda)(I+S)^{-1}\alpha\quad...(1)$

Also, $(I+S)\alpha=(1+\lambda)\alpha$

$\Rightarrow \alpha=(1+\lambda)(I+S)^{-1}\alpha$

$\Rightarrow \frac{1}{1+\lambda}\alpha=(I+S)^{-1}\alpha\quad...(2)$, $\quad($as $1+\lambda\ne0)$

Using $(1)$ and $(2)$ your result follows.

Answer 2

I would avoid determinants. You have a nonzero vector with $\newcommand{\al}{\alpha}\newcommand{\la}{\lambda}S\al=\la\al$. Then $(I+S)\al=(1+\la)\al$ and $(I-S)\al=(1-\la)\al$. As long as $\la\ne-1$ then $(I+S)^{-1}\al=(1+\la)^{-1}\al$ and so $$(I+S)^{-1}(1-S)\al=(1-\la)(I+S)^{-1}\al=\frac{1-\la}{1+\la}\al.$$