If $A$ is skew-symmetric, then $I-A$ is invertible and $Q = (I-A)^{-1}(I+A)$ is orthogonal 
Prove that if $A^T = -A$ is any skew -symmetric matrix, then $Q =
(I-A)^{-1}(I+A)$ is an orthogonal matrix. Can you prove that $(I - A)$ is
always invertible?

How do I go on to prove this? Is it similar to proving that $\det(Q) = \pm 1$ or that $A^T = A^{-1}$?
 A: Um. No, not determinant. Individual eigenvalues. The ordinary dot product of two column vectors $v,w$ is given by the matrix product $v^T w = w^T v$ because the transpose of a 1 by 1 matrix is itself. So suppose your $A$ has a real eigenvalue $\lambda,$ with an eigenvector $v.$ We have $Av = \lambda v,$ and
$$ \lambda v^T v = v^T (\lambda v) = v^T (Av) = (Av)^T v = v^T A^T v = -v^T A v = - \lambda v^T v.  $$ Now $v \neq 0,$ so $v^T v \neq 0.$ Thus 
$$ \lambda v^T v = - \lambda v^T v  $$ means $\lambda = 0.$
So, the only possible real eigenvalue is $0.$ In particular, $1$ is never an eigenvalue, we always have $Av \neq v,$ and $(I-A)v \neq 0.$ Put more simply, $0$ is not an eigenvalue of $(I-A),$ which is thus nonsingular.  
A: I assume that $A\in M_{n}(\mathbb{R})$.
Since $A$ is real skew-symmetric we know that $\det(I+A)\neq 0$ (also, it is know in the case of real skew-symmetric matrices $det(A)\geq0$).
So, $\det(I-A)=\det(I+A^{T})=\det((I+A)^{T})=\det(I+A)\neq0$.
A: Prove that Q is an orthogonal matrix:
$$QQ^T = I$$
Note, because A is a skew-symmetric matrix $A^T = -A$
$$Q^T = ((I-A)^{-1}(I+A))^T$$
$$= (I+A)^T((I-A)^{-1})^T$$
$$= (I+A)^T((I-A)^T)^{-1}$$
$$= (I^T+A^T)(I^T-A^T)^{-1}$$
$$= (I+A^T)(I-A^T)^{-1}$$
$$= (I-A)(I+A)^{-1}$$
Now show $QQ^T = I$,
$$QQ^T = (I-A)^{-1}(I+A)(I-A)(I+A)^{-1}$$
Note: $$(I+A)(I-A) = (I-A)(I+A)$$
Proof:
$$(I+A)(I-A)=(I+A)I-(I+A)A=I^2+A-A-A^2=I-A^2$$
$$(I-A)(I+A)= (I-A)I+(I-A)A = I^2-A+A-A^2 = I-A^2$$
So, $$(I+A)(I-A) = (I-A)(I+A)$$
Continuing $QQ^T = I$,
$$QQ^T = (I-A)^{-1}(I+A)(I-A)(I+A)^{-1} = (I-A)^{-1}(I-A)(I+A)(I+A)^{-1}$$
$$= II = I$$
Thus:
$$QQ^T = I$$
Prove that $(I−A)$ is always invertible. Because A is a skew-symmetric matrix, $det(A) = 0$. Since $det(I) = 1$, $det(I-A) = 1 \neq 0$. Thus, $(I−A)$ is always invertible.
