# Proof for Cayley Transformations

I was given the following 4 facts to verify:

1) if $A$ is skew, then $A-1$ is invertible.

2) if $U=(A+1)(A-1)^{-1}$, then U is isometric.

3) $U-1$ is invertible.

4) If $U$ is isometric and $U-1$ is invertible, and if $A=(U+1)(U-1)^{-1}$, then A is skew.

1) A proof provided for me is that if $A$ is skew, then its eigenvalues are either $0$ or purely imaginary, which implies the eigenvalues of $A-1$ are of the form $\lambda - 1$, with $\lambda$ as the eigenvalue of $A$. It was implied in the proof the matrices of the form $A\pm 1$ have eigenvalues $\lambda \pm 1$, which I don't entirely understand as to why.

2) I assume this proof also assumes $A$ is skew from part 1), for which using the definitions of skew and isometry gives the rest of the proof.

3) and 4) is where things start to get blurry. I'm not sure how to proceed.

• If $\mathbf v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $(A+\beta I)\mathbf v=(\lambda+\beta)\mathbf v$. – amd Dec 14 '17 at 23:25
• What does "skew" mean? Skew-symmetric? In this case, for 1), notice that $\left(1+A\right)\left(1-A\right) = 1 - A^2 = 1 + A^T A$ is positive definite (since $1$ is positive definite while $A^T A$ is nonnegative semidefinite) and thus invertible, and therefore $1-A$ is invertible as well. – darij grinberg Dec 15 '17 at 3:30
• See also math.stackexchange.com/a/2409050 for part 2). – darij grinberg Dec 17 '17 at 4:07