I am trying to understand the proof in the Appendix of the paper "Conic Optimization via Operator Splitting and Homogeneous Self-Dual Embedding" by O'Donoghue et al. (link: https://web.stanford.edu/~boyd/papers/scs.html)
I am stuck on the following part, which I present in a self-contained way here. Define the mapping $L(u,v):= (I+Q)^{-1}(u+v)-v$, where $u,v\in\mathbb{R}^n$, and $Q\in\mathbb{R}^{n\times n}$ is a skew-symmetric matrix. The authors want to show that this mapping is nonexpansive, i.e. $\lVert L(u,v) - L(\hat{u},\hat{v}) \rVert_2 \leq \lVert (u-\hat{u},v-\hat{v}) \rVert_2$ for all $u,v,\hat{u},\hat{v}$. Here the parentheses $(u,v)$ denote the vector $\begin{bmatrix} u\\v \end{bmatrix}$.
In the proof, the authors rearrange terms to show that \begin{align*} \lVert L(u,v) - L(\hat{u},\hat{v}) \rVert_2 = \lVert [(I+Q)^{-1},-(I-(I+Q)^{-1})](u-\hat{u},v-\hat{v})\rVert_2, \end{align*} and then they proceed to conclude that \begin{align*} \lVert [(I+Q)^{-1},-(I-(I+Q)^{-1})](u-\hat{u},v-\hat{v})\rVert_2 \leq \lVert (u-\hat{u},v-\hat{v}) \rVert_2, \end{align*} on the basis of the observation that for a skew-symmetric $Q$, it holds that \begin{align*} [(I+Q)^{-1},-(I-(I+Q)^{-1})][(I+Q)^{-1},-(I-(I+Q)^{-1})]^\top = I. \end{align*}
I don't see how the result follows from this observation. Shouldn't the correct thing to prove be the following?
\begin{align*} [(I+Q)^{-1},-(I-(I+Q)^{-1})]^\top[(I+Q)^{-1},-(I-(I+Q)^{-1})] = I. \end{align*}
Any help would be appreciated.
Observe that $$ \begin{aligned} M&=\begin{bmatrix}(I+Q)^{-1}&(I+Q)^{-1}-I\end{bmatrix}\\ &=\begin{bmatrix}(I+Q)^{-1}&(I+Q)^{-1}-(I+Q)^{-1}(I+Q)\end{bmatrix}\\ &=(I+Q)^{-1}\begin{bmatrix}I&-Q\end{bmatrix}.\\ \end{aligned} $$ Therefore $$ \begin{aligned} MM^T &=(I+Q)^{-1} \begin{bmatrix}I&-Q\end{bmatrix} \begin{bmatrix}I\\ Q\end{bmatrix} (I-Q)^{-1}\\ &=(I+Q)^{-1}(I-Q^2)(I-Q)^{-1}\\ &=\left[(I+Q)^{-1}(I-Q)\right]\left[(I+Q)(I-Q)^{-1}\right]\\ &=UU^T \end{aligned} $$ where $U=(I+Q)^{-1}(I-Q)$ is an orthogonal matrix because it is the Cayley transform of the skew-symmetric matrix $Q$. Hence $MM^T=UU^T=I$, meaning that $M$ has orthonormal rows and $\|M\|_2=1$.