How to get from $I-p p^T$ to $-E p p^T E$? $p \in \mathbb{R}^2$ is a unit-length column vector. $E=\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$ is the "$-\frac{\pi}{2}$" rotation matrix. So
how to prove that 
\begin{equation}
        I-p p^T = -E p p^T E \;?
\end{equation}
Of course you can let $p=[x,y]^T$ and prove this. But that is not the answer I want. I want to know how one can do some calculations and get $-E p p^T E$ just starting from $I-p p^T$?
 A: Objective : prove that $$(I - pp^T) = (-E pp^TE) \ \ \ \ \ \ (*)$$
Preliminary settings :
Let $q$ be the unit vector such that $(p,q)$ is a directly oriented base of the plane. Thus we have in particular $p^Tq=0$.
$E$ is clearly the $-\dfrac{\pi}{2}$ rotation matrix, with the important property that $-E = E^{-1}.$

I propose 2 proofs :
1) An algebraic proof : 
As $(p,q)$ is a basis, it suffices to prove that the action of the LHS operator and of the RHS operator of (*) on $p$ and on $q$ are the same, i.e., establish that :
$$(a) \ \ (I - pp^T)p = (-E pp^TE)p \ \ \ \ \text{and} \ \ \ \ (b) \ \ (I - pp^T)q = (-E pp^TE)q$$
which amounts to prove that :
$$(a) \ \ p - p(p^Tp) = -E pp^T(Ep) \ \ \ \ \text{and} \ \ \ \ (b) \ \ q - p(p^Tq) = E^{-1} pp^T(Eq)$$
or, equivalently :
$$(a) \ \ p - p(1) = -E pp^T(-q) \ \ \ \ \text{and} \ \ \ \ (b) \ \ q - p(0) = E^{-1} pp^T(p)$$
both (a) and (b) are true because the first one is $0=0$ and the second one is $q=q$.
2) A geometrical proof :
Let us give a geometrical meaning to the "characters" that appear on the scene of (*) : 


*

*$P_p := (p p^T)$ is the projection matrix onto subspace $\mathbb{R}p$ (proof : $(p p^T)p=p$ whereas $(p p^T)q=0).$

*$P_q :=(I - p p^T)$ is the projection matrix onto subspace $\mathbb{R}q$ (same kind of proof as before).


Thus (*) can be interpreted, in terms of transformations, as :
$$P_q = E^{-1} \circ P_p \circ E,$$
otherwise said, "projection on $q\mathbb{R}$" is the conjugate operation of "projection on $p\mathbb{R}$" up to a $-\pi/2$ rotation, which is a geometrical evidence.
A: The given unit vector $\,p\,$ spans a $1$-dimensional subspace,
and $\:pp^T=p\,\langle\, p\mid\cdot\,\rangle\,$ is the associated orthogonal projector onto that subspace.
The orthogonal complement is also $1$-dimensional as we are considering
$\mathbb R^2$, and it is (spanned by) "everything perpendicular to $p$", e.g. the unit vector $\,Ep\,$ or $\,-Ep$, the sign doesn't matter. The corresponding orthogonal projection is 
$$(\pm Ep)(\pm Ep)^T = Epp^T E^T = -Epp^TE\,.$$
Because subspace and orthogonal complement exhaust the whole vector space, the projections sum up to the identity:
$$-Epp^TE\,+\,p p^T\;=\;I$$
This equality is equivalent to $\{p,\pm Ep\}\subset\mathbb R^2$ being an orthonormal basis.
