TITLE: Showing $A-I$ is invertible, when $A$ is a skew-symmetric matrix
QUESTION [8 upvotes]: Let $A\in M_{n\times n}(\mathbb{R})$ be a skew-symmetric  matrix. show that $A-I$ is invertible and $(A-I)^{-1}(A+I)$ is an orthogonal matrix. 
$|A-I|=|(A-I)^T|=|-(A+I)|=(-1)^n|A+I|$
I have no clue to solve this problem, I appreciate any help.

REPLY [3 votes]: Here is a slightly different approach:
Note that $A$ has a basis of orthonormal eigenvectors (in the same manner as
Hermitian matrices). Suppose $Av= \lambda v$ for a unit vector $v$, then $v^*Av = \lambda = -v^* A^* v = - \overline{\lambda} $, hence all eigenvalues are
purely imaginary. In particular, they are not equal to one, hence
$A-I$ is invertible.
To show that $(A-I)^{-1}(A+I)$ is orthogonal, it is sufficient to show
that $\|(A-I)^{-1}(A+I) v \| = 1$ for all unit eigenvectors $v$. Since
$(A-I)^{-1}(A+I) v =  {\lambda +1 \over \lambda -1 }v$ and
$|{\lambda +1 \over \lambda -1 }| = 1$, we have the desired result.