For a n*n skew-symmetric matrix K (K transpose = -K), show that K+I (identity) is invertible
And show that (K-I)(K+I)^-1 is orthogonal (is it's own inverse) and has determinant 1
first of all you should've mentioned that the entries of $\displaystyle K$ come from real numbers. to prove that $\displaystyle K+I$ is invertible, we need to prove that if $\displaystyle K \bold{x} = -\bold{x},$ for some vector $\displaystyle \bold{x} \in \mathbb{R}^n,$ then $\displaystyle \bold{x}=\bold{0}.$
this is easy to prove: $\displaystyle -||\bold{x}||^2=K \bold{x} \cdot \bold{x}=\bold{x} \cdot K^T \bold{x} = - \bold{x} \cdot K \bold{x}=\bold{x} \cdot \bold{x}=|| \bold{x}||^2.$ thus $\displaystyle || \bold{x} ||^2 = 0$ and hence $\displaystyle \bold{x} = \bold{0}.$ similarly we see that $\displaystyle K-I$ is invertible too.
let $\displaystyle A=(K-I)(K+I)^{-1}.$ we must show that $\displaystyle A^T=A^{-1}.$ well, we have $\displaystyle K-I=A(K+I)$ and thus $\displaystyle -K-I=(K-I)^T=(K+I)^TA^T=(-K+I)A^T.$ hence $\displaystyle A^T=(K-I)^{-1}(K+I).$And show that (K-I)(K+I)^-1 is orthogonal (is it's own inverse)
therefore $\displaystyle A^{-1}=(K+I)(K-I)^{-1}=(K-I)^{-1}(K+I)=A^T.$
and has determinant 1
this is false! for example if $\displaystyle K=0$ is a $\displaystyle 3 \times 3$ matrix, then $\displaystyle A=-I$ and thus $\displaystyle \det A = -1.$ in general we can only say that $\displaystyle \det A = \pm 1,$ which is obvious because $\displaystyle A$ is orthogonal.
In fact, we can say a bit more than that. The determinant of a matrix is equal to the determinant of the transpose. So $\displaystyle \det(K+I) = \det(K^{\textsc t}+I) = \det(-K+I) = (-1)^n\det(K-I)$, from which $\displaystyle \det\bigl((K-I)(K+I)^{-1}\bigr) = \det(K-I)\bigl(\det(K+I)\bigr)^{-1} = (-1)^n$.
Alternatively, you could reword the question to say that $\displaystyle \det\bigl((I-K)(I+K)^{-1}\bigr) = 1$.