1. ## Tricky matrix question

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

2. Originally Posted by steve_young
For a n*n skew-symmetric matrix K (K transpose = -K), show that K+I (identity) is invertible.
first of all you should've mentioned that the entries of $K$ come from real numbers. to prove that $K+I$ is invertible, we need to prove that if $K \bold{x} = -\bold{x},$ for some vector $\bold{x} \in \mathbb{R}^n,$ then $\bold{x}=\bold{0}.$

this is easy to prove: $-||\bold{x}||^2=K \bold{x} \cdot \bold{x}=\bold{x} \cdot K^T \bold{x} = - \bold{x} \cdot K \bold{x}=\bold{x} \cdot \bold{x}=|| \bold{x}||^2.$ thus $|| \bold{x} ||^2 = 0$ and hence $\bold{x} = \bold{0}.$ similarly we see that $K-I$ is invertible too.

And show that (K-I)(K+I)^-1 is orthogonal (is it's own inverse)
let $A=(K-I)(K+I)^{-1}.$ we must show that $A^T=A^{-1}.$ well, we have $K-I=A(K+I)$ and thus $-K-I=(K-I)^T=(K+I)^TA^T=(-K+I)A^T.$ hence $A^T=(K-I)^{-1}(K+I).$

therefore $A^{-1}=(K+I)(K-I)^{-1}=(K-I)^{-1}(K+I)=A^T.$

and has determinant 1

this is false! for example if $K=0$ is a $3 \times 3$ matrix, then $A=-I$ and thus $\det A = -1.$ in general we can only say that $\det A = \pm 1,$ which is obvious because $A$ is orthogonal.

3. Originally Posted by NonCommAlg
and has determinant 1
this is false! for example if $K=0$ is a $3 \times 3$ matrix, then $A=-I$ and thus $\det A = -1.$ in general we can only say that $\det A = \pm 1,$ which is obvious because $A$ 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 $\det(K+I) = \det(K^{\textsc t}+I) = \det(-K+I) = (-1)^n\det(K-I)$, from which $\det\bigl((K-I)(K+I)^{-1}\bigr) = \det(K-I)\bigl(\det(K+I)\bigr)^{-1} = (-1)^n$.

Alternatively, you could reword the question to say that $\det\bigl((I-K)(I+K)^{-1}\bigr) = 1$.