Krdytkyu

$p in mathbb{R}^2$ is a unit-length column vector. $E=begin{bmatrix}0 & 1 \ -1 & 0end{bmatrix}$ 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 $I-p p^T$?

The given unit vector $,p,$ spans a $1$-dimensional subspace,
and $:pp^T=p,langle, pmidcdot,rangle,$ 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 ${p,pm Ep}subsetmathbb R^2$ being an orthonormal basis.

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 $qmathbb{R}$" is the conjugate operation of "projection on $pmathbb{R}$" up to a $-pi/2$ rotation, which is a geometrical evidence.