I encountered the above question in the paper. I have no idea how to obtain the right entry.
Any help would be greatly appreciated!
$$I-\hat x \hat x^T= (\hat x^T E)^T(\hat x^T E)$$
where $\hat x = \frac{x}{\vert\vert x \vert\vert}, x\in \mathbb R^2$, and $E=\left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right]$
$E$ is interesting, since for any vector $x=\pmatrix{a\\b}\;$ the product $E^Tx=\pmatrix{-b\\+a}\;$ and so $\,x\perp E^Tx$
${\mathbb R}^{2}$ is interesting, since any orthonormal vector pair $(x,y)$ forms a basis and so $\;I = xx^T + yy^T$
Combining these two interesting facts one can write $$\eqalign{ I - xx^T &= yy^T \\&= (E^Tx)(E^Tx)^T \\&= (x^TE)^T(x^TE) }$$
You now that $|\hat x|=1$, so we can write $$\hat x=\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}$$ Then $$\begin{align}I-xx^T&=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}-\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}\begin{pmatrix}\cos\theta &\sin\theta\end{pmatrix}\\&=\begin{pmatrix}1-\cos^2\theta&-\sin\theta\cos\theta\\-\sin\theta\cos\theta&1-\sin^2\theta\end{pmatrix}\\&=\begin{pmatrix}\sin^2\theta&-\sin\theta\cos\theta\\-\sin\theta\cos\theta&\cos^2\theta\end{pmatrix}\end{align}$$ For the other side of the equation $$x^TE=\begin{pmatrix}\cos\theta &\sin\theta\end{pmatrix}\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}=\begin{pmatrix}-\sin\theta &\cos\theta\end{pmatrix}$$ Transpose it, do the last multiplication, and compare to the left hand side
