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.
Linear Algebra – Showing A-I is Invertible for Skew-Symmetric Matrix A
linear algebramatricesorthogonality
Best Answer
If $A - I$ is not invertible, we can find a non-zero vector $x \in \ker(A - I)$ which will satisfy $Ax = x$. Calculating, we have
$$ \left< x, x \right>_{\mathbb{R}^n} = \left< Ax, x \right>_{\mathbb{R}^n} = (Ax)^T x = x^T A^T x = -x^T A x = -x^T x = - \left< x, x \right>_{\mathbb{R}^n}, $$
a contradiction. To see that $B := (A - I)^{-1}(A + I)$ is orthogonal, we can calculate
$$ B B^T = (A - I)^{-1}(A + I) \left( (A - I)^{-1}(A + I) \right)^T = (A - I)^{-1} (A + I) (A^T + I)(A^T - I)^{-1} \\ = (A - I)^{-1}(A + I)(-A + I)(-A - I)^{-1} = (-1)(I - A)^{-1}(I - A)(I + A)(-1)(A + I)^{-1} \\ = (-1)(-1)I = I.$$