ABSTRACT

In this article, we wish to expand on some of the results obtained from the first article entitled Projection Theory. We have already established that one-parameter projection operators can be constructed from the unit circle ${\mathbb{S}}^1$ . As discussed in the previous article these operators form a Lie group known as the Projection Group. In the first section, we will show that the concepts from my first article are consistent with existing theory [1] [2] . In the second section, it will be demonstrated that not only such operators are mutually congruent but also we can define a group action on $G_P\left(\left[\theta \right]\right)$ by using the rotation group $SO\left(\mathrm{2,}ℝ\right)$ [3] [4] . It will be proved that the group $SO\left(\mathrm{2,}ℝ\right)$ acts on elements of $G_P\left(\left[\theta \right]\right)$ in a non-faithful but ∞-transitive way consistent with both group operations. Finally, in the last section we define the group operation $\phi \left(P_{\left[\theta \right]}\mathrm{,}{P}_{\left[{\theta }^{\prime }\right]}\right)$ in terms of matrix operations using the $Vec$ operator and the Hadamard Product; this construction is consistent with the group operation defined in the first article.

Keywords:

Projection Theory, Projection Manifolds, Projectors, Congruent Projection Matrices

1. Notation System

In this article the notation has been kept the same as in the previous article.

• ${\Vert \cdot \Vert }_F$ is the Frobenius Norm.

• $P_{\left[\theta \right]}$ is some projection operator in $G_P\left(\left[\theta \right]\right)$ .

• $I_2$ is the $2\times 2$ identity matrix.

• $Rnk\left(P_{\left[\theta \right]}\right)$ is the Rank of a projection operator.

• $\sigma \left(P_{\left[\theta \right]}\right)$ is the spectrum of a projection operator.

• $Sta{b}_{SO\left(2\right)}\left(P_{\left[\theta \right]}\right)$ is the Stabilizer subgroup for some fixed element of $G_P\left(\left[\theta \right]\right)$ .

• $R\left(\alpha \right)$ is an element of $SO\left(2\right)$ for angle $\alpha$ .

• $Vec$ is the Vectorisation Operator.

• $\odot$ is the Hadamard Product.

• $\otimes$ is the Kronecker Product.

2. Some Important Theorems and Lemmas

From [1] [2] we can get the following results.

Lemma 1. The Frobenius morn ${\Vert \cdot \Vert }_F$ of a projection matrix is its trace. That is

${\Vert P_{\left[\theta \right]}\Vert }_F=\sqrt{Tr\left(P_{\left[\theta \right]}\right)}=1$

Proof. It is well known that

${\Vert P_{\left[\theta \right]}\Vert }_F^2\mathrm{<mo>\; :\; </mo>}={\displaystyle \underset{i\mathrm{,}j}{\sum }}{\left|p_{ij}\left(\theta \right)\right|}^2=Tr\left(P_{\left[\theta \right]}^{\text{T}}{P}_{\left[\theta \right]}\right)$

Given that $P_{\left[\theta \right]}$ is both symmetric and idempotent implies that

$Tr\left(P_{\left[\theta \right]}^{\text{T}}{P}_{\left[\theta \right]}\right)=Tr\left(P_{\left[\theta \right]}{P}_{\left[\theta \right]}\right)=Tr\left(P_{\left[\theta \right]}^2\right)=Tr\left(P_{\left[\theta \right]}\right)$

Hence,

${\Vert P_{\left[\theta \right]}\Vert }_F=\sqrt{Tr\left(P_{\left[\theta \right]}\right)}$

therefore,

$Tr\left(\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\mathrm{si}{n}^{\prime }\left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]\right)={\cos }^2\left[\theta \right]+{\sin }^2\left[\theta \right]=\mathrm{1,}\forall \left[\theta \right]\in \mathbb{T}$

Theorem 1 The necessary and sufficient condition for $P_{\left[\theta \right]}\in G_P\left(\left[\theta \right]\right)$ is given by

$P_{\left[\theta \right]}^2=P_{\left[\theta \right]},\forall \left[\theta \right]\in \mathbb{T}$

Proof. We know, from the previous article that $P_{\left[\theta \right]}$ is constructed as follows

$\Phi \left({\text{e}}^{i\left[\theta \right]}\right)\mathrm{<mo>\; :\; </mo>}={\text{e}}^{i\left[\theta \right]}\otimes {\left({\text{e}}^{i\left[\theta \right]}\right)}^{\text{T}}=\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]=P_{\left[\theta \right]}$

$\begin{array}{c}{P}_{\left[\theta \right]}^2={\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]}^2\\ =\left[\begin{array}{cc}{\cos }^4\left[\theta \right]+{\cos }^2\left[\theta \right]{\sin }^2\left[\theta \right]& {\cos }^3\left[\theta \right]\sin [\theta ]+\cos \left[\theta \right]{\sin }^3\left[\theta \right]\\ {\cos }^3\left[\theta \right]\sin \left[\theta \right]+\cos \left[\theta \right]{\sin }^3\left[\theta \right]& {\sin }^4\left[\theta \right]+{\cos }^2\left[\theta \right]{\sin }^2\left[\theta \right]\end{array}\right]\\ =\left[\begin{array}{cc}{{\displaystyle \cos }}^2\left[\theta \right]\left({\cos }^2\left[\theta \right]+{\sin }^2\left[\theta \right]\right)& \cos \left[\theta \right]\sin \left[\theta \right]\left({\cos }^2\left[\theta \right]+{\sin }^2\left[\theta \right]\right)\\ \cos \left[\theta \right]\sin \left[\theta \right]\left({\cos }^2\left[\theta \right]+{\sin }^2\left[\theta \right]\right)& {\sin }^2\left[\theta \right]\left({\sin }^2\left[\theta \right]+{\cos }^2\left[\theta \right]\right)\end{array}\right]\\ =\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]=P_{\left[\theta \right]}\end{array}$

Theorem 2 The orthogonal complement of $P_{\left[\theta \right]}$ is given by

$P_{\left[\theta \pm \pi /2\right]}=I_2-P_{\left[\theta \right]}$

where $I_2$ is the $2\times 2$ identity matrix. This is known as a Vector Rejection.

Proof. We first show that

$P_{\left[\theta \pm \pi /2\right]}=I_2-P_{\left[\theta \right]}$

the RHS of the equation gives us

$\begin{array}{c}{I}_2-P\left(\left[\theta \right]\right)=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]\\ =\left[\begin{array}{cc}1-{\cos }^2\left[\theta \right]& -\cos \left[\theta \right]\sin \left[\theta \right]\\ -\sin \left[\theta \right]\cos \left[\theta \right]& 1-{\sin }^2\left[\theta \right]\end{array}\right]\\ =\left[\begin{array}{cc}{\sin }^2\left[\theta \right]& -\cos \left[\theta \right]\sin \left[\theta \right]\\ -\sin \left[\theta \right]\cos \left[\theta \right]& {\cos }^2\left[\theta \right]\end{array}\right]\end{array}$

Now, the LHS of th equation gives us the following result

$\begin{array}{c}{P}_{\left[\theta \pm \pi /2\right]}=\left[\begin{array}{cc}{\cos }^2\left(\left[\theta \right]\pm \pi /2\right)& \cos \left(\left[\theta \right]\pm \pi /2\right)\sin \left(\left[\theta \right]\pm \pi /2\right)\\ \sin \left(\left[\theta \right]\pm \pi /2\right)\cos \left(\left[\theta \right]\pm \pi /2\right)& {\sin }^2\left(\left[\theta \right]\pm \pi /2\right)\end{array}\right]\\ =\left[\begin{array}{cc}{\sin }^2\left[\theta \right]& -\cos \left[\theta \right]\sin \left[\theta \right]\\ -\sin \left[\theta \right]\cos \left[\theta \right]& {\cos }^2\left[\theta \right]\end{array}\right]\end{array}$

since

${\cos }^2\left(\left[\theta \right]\pm \pi /2\right)={\left(\cos \left[\theta \right]\cos \left(\pi /2\right)\mp \sin \left[\theta \right]\sin \left(\pi /2\right)\right)}^2={\left(\mp \sin \left[\theta \right]\right)}^2={\sin }^2\left[\theta \right]$

${\sin }^2\left(\left[\theta \right]\pm \pi /2\right)={\left(\sin \left[\theta \right]\cos \left(\pi /2\right)\pm \cos \left[\theta \right]\sin \left(\pi /2\right)\right)}^2={\left(\pm \cos \left[\theta \right]\right)}^2={\cos }^2\left[\theta \right]$

Secondly, we show the effect these operators have on some vector $x\in {ℝ}^2$ . We easily show this in the following way

$\langle P_{\left[\theta \right]}x\mathrm{,}{P}_{\left[\theta \pm \pi /2\right]}x\rangle =0$

We proceed as follows

$\begin{array}{c}\langle P_{\left[\theta \right]}x\mathrm{,}{P}_{\left[\theta \pm \pi /2\right]}x\rangle =\langle P_{\left[\theta \right]}x\mathrm{,}\left(I_2-P_{\left[\theta \right]}\right)x\rangle =\langle P_{\left[\theta \right]}x\mathrm{,}{I}_{2}x-P_{\left[\theta \right]}x\rangle \\ =\langle P_{\left[\theta \right]}x\mathrm{,}x-P_{\left[\theta \right]}x\rangle =\langle P_{\left[\theta \right]}x\mathrm{,}x\rangle -\langle P_{\left[\theta \right]}x\mathrm{,}{P}_{\left[\theta \right]}x\rangle \\ =\Vert P_{\left[\theta \right]}x\Vert \Vert x\Vert \cos \theta -{\Vert P_{\left[\theta \right]}x\Vert }^2\\ =\Vert P_{\left[\theta \right]}x\Vert \Vert P_{\left[\theta \right]}x\Vert -{\Vert P_{\left[\theta \right]}x\Vert }^2=0\end{array}$

Lemma 2 The product of the operators $P_{\left[\theta \right]}$ and $P_{\left[\theta \pm \pi /2\right]}$ is zero. We shall call $P_{\left[\theta \pm \pi /2\right]}$ the Orthogonal Complement Operator of $P_{\left[\theta \right]}$ which I shall denote as $P_{\left[\theta \right]}^{\perp }$ .

$P_{\left[\theta \right]}{P}_{\left[\theta \right]}^{\perp }=\mathrm{0,}\forall \left[\theta \right]\in \mathbb{T}$

We say that $P_{\left[\theta \right]}\perp P_{\left[\theta \right]}^{\perp }$

Proof.

$\begin{array}{l}{P}_{\left[\theta \right]}{P}_{\left[\theta \right]}^{\perp }=\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]\left[\begin{array}{cc}{\sin }^2\left[\theta \right]& -\cos \left[\theta \right]\sin \left[\theta \right]\\ -\sin \left[\theta \right]\cos \left[\theta \right]& {\cos }^2\left[\theta \right]\end{array}\right]\\ =\left[\begin{array}{cc}{{\displaystyle \cos }}^2\left[\theta \right]{\sin }^2\left[\theta \right]-{\cos }^2\left[\theta \right]{\sin }^2\left[\theta \right]& -{\cos }^3\left[\theta \right]\sin \left[\theta \right]+{\cos }^3\left[\theta \right]\sin \left[\theta \right]\\ {\sin }^3\left[\theta \right]\cos \left[\theta \right]-{\sin }^3\left[\theta \right]\cos \left[\theta \right]& -{\cos }^2\left[\theta \right]{\sin }^2\left[\theta \right]+{\sin }^2\left[\theta \right]{\cos }^2\left[\theta \right]\end{array}\right]\\ =\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Lemma 3 $P_{\left[\theta \right]}+P_{\left[\theta \right]}^{\perp }=I_2$

Proof.

$\begin{array}{l}{P}_{\left[\theta \right]}+P_{\left[\theta \right]}^{\perp }\\ =\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]+\left[\begin{array}{cc}{\sin }^2\left[\theta \right]& -\cos \left[\theta \right]\sin \left[\theta \right]\\ -\sin \left[\theta \right]\cos \left[\theta \right]& {\cos }^2\left[\theta \right]\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I_2\end{array}$

Theorem 3 A projection operator of dimension onto $V\subset {ℝ}^2$ where $\dim \left(V\right)=r$ then

$P_{\left[\theta \right]}=T{\Delta }_r{T}^{-1}$ (1)

where

$\Delta r:=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$ (2)

The matrix $T$ is formed by the columns of the basis vectors which span the subspaces V and W respectively.

Proof. We know that $V\mathrm{,}W\subset {ℝ}^2$ , furthermore we also know that $\dim \left(V\right)=\dim \left(W\right)=1\Rightarrow r=1$ .

Suppose the vector $a={\left(\cos \theta \mathrm{,}\sin \theta \right)}^{\text{T}}$ is a basis for subspaces V i.e. $Spn\left\{a\right\}=V$ then the matrix T can be computed in the following way.

Now, taking into account the fact that $W=V^{\perp }$ implies that $b$ is given by

$b=\left(\cos \left(\theta \pm \pi /2\right)\mathrm{,}\sin \left(\theta \pm \pi /2\right)\right)=\left(\mp \sin \theta \mathrm{,}\pm \cos \theta \right)$

Hence, $T$ is computed in the following way

$T=\left[\begin{array}{cc}\cos \left[\theta \right]& \mp \sin \left[\theta \right]\\ \sin \left[\theta \right]& \pm \cos \left[\theta \right]\end{array}\right]\therefore \det \left(T\right)=\pm {\cos }^2\left[\theta \right]\pm {\sin }^2\left[\theta \right]=\pm 1$

Hence, $T^{-1}$ is given by

$T^{-1}=\pm 1\left[\begin{array}{cc}\pm \cos \left[\theta \right]& \pm \sin \left[\theta \right]\\ -\sin \left[\theta \right]& \cos \left[\theta \right]\end{array}\right]=\left[\begin{array}{cc}\cos \left[\theta \right]& \sin \left[\theta \right]\\ \mp \sin \left[\theta \right]& \pm \cos \left[\theta \right]\end{array}\right]=T^{\text{T}}\Rightarrow T\mathrm{,}{T}^{\text{T}}\in O(2,ℝ)$

Hence, $P_{\left[\theta \right]}$ is

$\begin{array}{c}T{\Delta }_r{T}^{-1}=\left[\begin{array}{cc}\cos \left[\theta \right]& \mp \sin \left[\theta \right]\\ \sin \left[\theta \right]& \pm \cos \left[\theta \right]\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\pm 1\left[\begin{array}{cc}\pm \cos \left[\theta \right]& \pm \sin \left[\theta \right]\\ -\sin \left[\theta \right]& \cos \left[\theta \right]\end{array}\right]\\ =\pm 1\left[\begin{array}{cc}\cos \left[\theta \right]& \mp \sin \left[\theta \right]\\ \sin \left[\theta \right]& \pm \cos \left[{\theta }^{\prime }\right]\end{array}\right]\left[\begin{array}{cc}\pm \cos \left[\theta \right]& \pm \sin \left[\theta \right]\\ 0& 0\end{array}\right]\\ =\pm 1\left[\begin{array}{cc}\pm {\cos }^2\left[\theta \right]& \pm \cos \left[\theta \right]\sin \left[\theta \right]\\ \pm \sin \left[\theta \right]\cos \left[\theta \right]& \pm {\sin }^2\left[\theta \right]\end{array}\right]\\ =P_{\left[\theta \right]}\end{array}$

We can see that $T\mathrm{,}{T}^{-1}\in O\left(\mathrm{2,}ℝ\right)$ and are orthogonal reflection matrices.

Lemma 4 Let $P_{\left[\theta \right]}$ be some projection matrix for some $\left[\theta \right]\in \mathbb{T}$ then we have

$Rnk\left(P_{\left[\theta \right]}\right)=Tr\left(P_{\theta }\right)$ (3)

Proof. We know that

$P_{\left[\theta \right]}=\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]\Rightarrow Tr\left(P_{\left[\theta \right]}\right)={\cos }^2\left[\theta \right]+{\sin }^2\left[\theta \right]=1$

We can now calculate its rank. The column space of $P_{\left[\theta \right]}$ is given by

$Cln\left(P_{\left[\theta \right]}\right)=\left\{\left[\begin{array}{c}{\cos }^2\left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]\end{array}\right]\mathrm{,}\left[\begin{array}{c}\cos \left[\theta \right]\sin \left[\theta \right]\\ {\sin }^2\left[\theta \right]\end{array}\right]\right\}$

where $Cln\left(P_{\left[\theta \right]}\right)$ is the column space of $P_{\left[\theta \right]}$ . $Rnk\left(P_{\left[\theta \right]}\right)=1$ iff $\exists {\alpha }_1,{\alpha }_2\ne 0\in ℝ$ such that

${\alpha }_1\left[\begin{array}{c}{\cos }^2\left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]\end{array}\right]+{\alpha }_2\left[\begin{array}{c}\cos \left[\theta \right]\sin \left[\theta \right]\\ {\sin }^2\left[\theta \right]\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

This gives us the following linear system

$\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]\left[\begin{array}{c}{\alpha }_1\\ {\alpha }_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

Given that $\det \left(P_{\left[\theta \right]}\right)=0$ implies an infinite set of solutions exists.

Writing the homogeneous linear system we find the general set of vector solutions as follows

${\alpha }_1{\cos }^2\left[\theta \right]+{\alpha }_2\cos \left[\theta \right]\sin \left[\theta \right]=0$

${\alpha }_1\sin \left[\theta \right]\cos \left[\theta \right]+{\alpha }_2{\sin }^2\left[\theta \right]=0$

therefore

${\alpha }_1\sin \left[\theta \right]\cos \left[\theta \right]+{\alpha }_2{\sin }^2\left[\theta \right]=0$

$\Rightarrow {\alpha }_1\cos \left[\theta \right]+{\alpha }_2\sin \left[\theta \right]=0$

$\therefore {\alpha }_1=-{\alpha }_2\frac{\sin \left[\theta \right]}{\cos \left[\theta \right]}=-{\alpha }_2\tan \left[\theta \right]$

Hence, the solution vector

$\mathcal{U}=\left\{{\alpha }_2\in ℝ\mathrm{<mo>\; |\; </mo>}\alpha ={\alpha }_2\left(-\tan \left[\theta \right]\mathrm{,1}\right)\right\}$

This implies that $Rnk\left(P_{\left[\theta \right]}\right)=Tr\left(P_{\left[\theta \right]}\right)=1$ .

Theorem 4 Let $A=\left[a_1\right]$ such that $V=Spn\left(A\right)$ . Then we have

$P_{\left[\theta \right]}=A{\left(A^{\text{T}}A\right)}^{-1}{A}^{\text{T}}$ (4)

Proof. $a_1={\left(\cos \left[\theta \right]\mathrm{,}\sin \left[\theta \right]\right)}^{\text{T}}=A$ , then we have

$A{\left(A^{\text{T}}A\right)}^{-1}{A}^{\text{T}}=A{\left(a_1^{\text{T}}{a}_1\right)}^{-1}A=A{A}^{\text{T}}=a_1{a}_1^{\text{T}}=a_1\otimes a_1=P_{\left[\theta \right]}$

Lemma 5 The spectrum $\sigma \left(P_{\left[\theta \right]}\right)$ of the $P_{\left[\theta \right]}$ for some $\left[\theta \right]\in \mathbb{T}$ is given by

$\sigma \left(P_{\left[\theta \right]}\right)\left\{{\lambda }_1=\mathrm{0,}{\lambda }_2=\mathrm{1,}\forall P_{\left[\theta \right]}\in G_P\left(\left[\theta \right]\right)\right\}$

Proof. We simply perform the following calculation

$\begin{array}{c}\det \left(P_{\left[\theta \right]}-\lambda I_2\right)=\left|\begin{array}{cc}{\cos }^2\theta -\lambda & \cos \theta \sin \theta \\ \sin \theta \cos \theta & {\sin }^2\theta -\lambda \end{array}\right|\\ =\left({\cos }^2\theta -\lambda \right)\left({\sin }^2\theta -\lambda \right)-{\cos }^2\theta {\sin }^2\theta \\ ={\cos }^2\theta {\sin }^2\theta -\lambda {\cos }^2\theta -\lambda {\sin }^2\theta +{\lambda }^2-{\cos }^2\theta {\sin }^2\theta \\ ={\lambda }^2-\lambda \end{array}$

therefore we see that

$\det \left(P_{\left[\theta \right]}-\lambda I_2\right)=0\Rightarrow {\lambda }^2-\lambda =0\Rightarrow \lambda \left(\lambda -1\right)=0\Rightarrow \lambda =0\vee \lambda =1$

Hence, $\sigma \left(P_{\left[\theta \right]}\right)=\left\{0,1\right\}$ as claimed.

We can calculate the associated Eigenspaces. For $\lambda =0$ we get

$P_{\left[\theta \right]}{E}_0=\lambda E_0=0$

Hence, it is clear that given some $P_{\left[\theta \right]}$ its Eigenspace for $\lambda =0$ is $Ker\left(\left[P_{\left[\theta \right]}\right]\right)$ , that is, in this case, the Eigenspace is spanned by the following vectors

$E_0\left(\left[\theta \right]\right)\mathrm{<mo>\; :\; </mo>}={\left(\cos \left(\left[\theta \right]+\pi /2\right)\mathrm{,}\sin \left(\left[\theta \right]+\pi /2\right)\right)}^{\text{T}}=\left[\begin{array}{c}-\sin \left(\left[\theta \right]\right)\\ \cos \left(\left[\theta \right]\right)\end{array}\right]$

Therefore, we can say that is $Im\left(P_{\left[\theta \right]}\right)$ is $V\subset {ℝ}^2$ then $Spn\left\{E_0\left(\left[\theta \right]\right)\right\}=V^{\perp }$ .

For $\lambda =1$ we get the following Eigenvectors

$P_{\left[\theta \right]}{E}_1=\lambda E_1=E_1$

$P_{\left[\theta \right]}{E}_1-E_1=\left(P_{\left[\theta \right]}-I_2\right)E_1=0$

In matrix form, this gives us

$\left[\begin{array}{cc}{\cos }^2\theta -1& \cos \theta \sin \theta \\ \sin \theta \cos \theta & {\sin }^2\theta -1\end{array}\right]\left[\begin{array}{c}{E}_1\\ E_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

$\left[\begin{array}{cc}-{\sin }^2\theta & \cos \theta \sin \theta \\ \sin \theta \cos \theta & -{\cos }^2\theta \end{array}\right]\left[\begin{array}{c}{E}_1\\ E_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

This gives us the following system

$-{\sin }^2\theta E_1+\cos \theta \sin \theta E_2=0$

$\sin \theta \cos \theta E_1-{\cos }^2\theta E_2=0$

therefore, from (24) we get

$\sin \theta \cos \theta E_1={\cos }^2\theta \Rightarrow E_2\tan \theta E_1=E_2$

Therefore,

$\frac{E_2}{E_1}=\tan \theta \Rightarrow E_1\left(\theta \right)=E_2\cot \theta$

Therefore, the Eigenvector is given

$E_1=E_2\left[\begin{array}{c}\cot \theta \\ 1\end{array}\right]$

Lemma 6 The Eigenvectors $E_0$ and $E_1$ are linearly independent. Moreover, $Spn\left\{E_0\mathrm{,}{E}_1\right\}={ℝ}^2$ .

Proof. For ${\gamma }_0\mathrm{,}{\gamma }_1\in ℝ$ we have

${\gamma }_0{E}_0+{\gamma }_1{E}_1=0$

Clearly, $E_0\mathrm{,}{E}_1$ are linearly independent iff ${\gamma }_0={\gamma }_1=0$

$-{\gamma }_0\sin \theta +{\gamma }_1\cot \theta =0$

${\gamma }_0\cos \theta +{\gamma }_1=0$

This can be written matrix form in the following way

$\left[\begin{array}{cc}-\sin \theta & \cot \theta \\ \cos \theta & 1\end{array}\right]\left[\begin{array}{c}{\gamma }_0\\ {\gamma }_1\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

Now,

$\left|\begin{array}{cc}-\sin \theta & \cot \theta \\ \cos \theta & 1\end{array}\right|=-\sin \theta -\cot \theta \cos \theta \ne 0,\forall \theta$

We can clearly see that, the determinant is non-zero hence, $E_0$ and $E_1$ are linearly independent.

We can now talk about the Diagonalizability of Projection Matrices $P_{\left[\theta \right]}$ . Given that $P_{\left[\theta \right]}$ has distinct eigenvalues implies that it is diagonalizable i.e. $P_{\left[\theta \right]}=PD{P}^{-1}$ where $D$ is a diagonal matrix

Lemma 7

$P_{\left[\theta \right]}=P\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]P^{-1}$

where

$P=\left[E_0\mathrm{,}{E}_1\right]=\left[\begin{array}{cc}-\sin \theta & \cot \theta \\ \cos \theta & 1\end{array}\right]\mathrm{,}{P}^{-1}=-\frac{1}{\sin \theta +\cot \theta \cos \theta }\left[\begin{array}{cc}1& -\cot \theta \\ -\cos \theta & -\sin \theta \end{array}\right]$

Proof.

$\begin{array}{c}{P}_{\left[\theta \right]}=-\frac{1}{\sin \theta +\cot \theta \cos \theta }\left[\begin{array}{cc}-\sin \theta & \cot \theta \\ \cos \theta & 1\end{array}\right]\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& -\cot \theta \\ -\cos \theta & -\sin \theta \end{array}\right]\\ =-\frac{1}{\sin \theta +\cot \theta \cos \theta }\left[\begin{array}{cc}-\sin \theta & \cot \theta \\ \cos \theta & 1\end{array}\right]\left[\begin{array}{cc}0& 0\\ -\cos \theta & -\sin \theta \end{array}\right]\\ =-\frac{1}{\sin \theta +\cot \theta \cos \theta }\left[\begin{array}{cc}-\cot \theta \cos \theta & -\cot \theta \sin \theta \\ -\cos \theta & -\sin \theta \end{array}\right]\\ =\left[\begin{array}{cc}{{\displaystyle \cos }}^2\theta & \cos \theta \sin \theta \\ \sin \theta \cos \theta & {{\displaystyle \sin }}^2\theta \end{array}\right]=P_{\left[\theta \right]}\end{array}$

Theorem 5 The projection matrix $P_{\left[\theta \right]}$ with spectrum $\sigma =\left\{{\lambda }_1,{\lambda }_2\right\}$ is diagonalizable, hence there exists matrices $\left\{G_1\mathrm{,}{G}_2\right\}$ such that

$P_{\left[\theta \right]}={\lambda }_1{G}_1+{\lambda }_2{G}_2$ (5)

where $G_1\mathrm{,}{G}_2$ are projection matrices onto $N\left(P_{\left[\theta \right]}-{\lambda }_i{I}_2\right)$ along $R\left(P_{\left[\theta \right]}-{\lambda }_i{I}_2\right)$ . Furthermore,

1) $G_i\mathrm{,}{G}_j=0$ whenever $i\ne j$ .

2) ${\displaystyle {\sum }_{i=1}^{mk}}{G}_i=I_2$ .

Proof. We know that the Eigenvalues are ${\lambda }_1=0\wedge {\lambda }_2=1$ this obviously implies that

$P_{\left[\theta \right]}={\lambda }_2{G}_2=G_2$

$\begin{array}{c}{P}_{\left[\theta \right]}=PD{P}^{-1}=\left(X_1\mathrm{<mo>\; |\; </mo>}{X}_2\right)\left(\begin{array}{cc}{\lambda }_{1}I& 0\\ 0& {\lambda }_{2}I\end{array}\right)\left(\begin{array}{c}{Y}_1^{\text{T}}\\ Y_2^{\text{T}}\end{array}\right)\\ ={\lambda }_1{X}_1{Y}_1^{\text{T}}+{\lambda }_2{X}_2{Y}_2^{\text{T}}={\lambda }_2{X}_2{Y}_2^{\text{T}}=X_2{Y}_2^{\text{T}}\\ =-\frac{1}{\sin \theta +\cot \theta \cos \theta }\left[\begin{array}{c}\cot \theta \\ 1\end{array}\right]\left[\begin{array}{cc}-\cos \theta & -\sin \theta \end{array}\right]\\ =-\frac{1}{\sin \theta +\cot \theta \cos \theta }\left[\begin{array}{cc}-\cot \theta \cos \theta & -\cot \theta \sin \theta \\ -\cos \theta & -\sin \theta \end{array}\right]\\ =P_{\left[\theta \right]}\end{array}$

This implies that $G_2=X_2{Y}_2^{\text{T}}$ .

Lemma 8 Let $G_0$ and $G_1$ be the spectral projectors for the eigenvalues ${\lambda }_1$ and ${\lambda }_2$ respectively. Then we show that

1)

$G_0{G}_1=0$

2)

$G_0+G_1=I_2$

Proof.

$\begin{array}{c}{G}_0{G}_1=X_1{Y}_1^{\text{T}}\\ =\left[\begin{array}{c}-\sin \theta \\ \cos \theta \end{array}\right]\left[\begin{array}{cc}1& -\cot \theta \end{array}\right]\left[\begin{array}{c}\cot \theta \\ 1\end{array}\right]\left[\begin{array}{cc}-\cos \theta & -\sin \theta \end{array}\right]\\ =\left[\begin{array}{cc}-\sin \theta & \cot \theta \sin \theta \\ \cos \theta & -\cos \theta \cot \theta \end{array}\right]\left[\begin{array}{cc}-\cot \theta \cos \theta & -\sin \theta \cot \theta \\ -\cos \theta & -\sin \theta \end{array}\right]\\ =\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

As claimed.

For the second part of the proof, we simply add the matrices

$\begin{array}{l}{G}_0+G_1\\ =-\frac{1}{\sin \theta +\cot \theta \cos \theta }\left\{\left[\begin{array}{cc}-\sin \theta & \cot \theta \sin \theta \\ \cos \theta & -\cos \theta \cot \theta \end{array}\right]+\left[\begin{array}{cc}-\cot \theta \cos \theta & -\sin \theta \cot \theta \\ -\cos \theta & -\sin \theta \end{array}\right]\right\}\\ =-\frac{1}{\sin \theta +\cot \theta \cos \theta }\left[\begin{array}{cc}-\left(\sin \theta +\cot \theta \sin \theta \right)& 0\\ 0& -\left(\cos \theta \cot \theta +\sin \theta \right)\end{array}\right]\\ =I_2\end{array}$

3. Lie Group Action of $SO\left(2,ℝ\right)$ on $G_P\left[\theta \right]$

From [3] [4] we have

Definition 1 Two squares matrices $A$ and $B$ are said to be Congruent if there exists an invertible matrix $R$ such that

$B=R^{\text{T}}AR$ (6)

The form of the equation would tend to suggest that $R^{\text{T}}\mathrm{,}R\in SO\left(\mathrm{2,}ℝ\right)$ such that $R^{\text{T}}R=R{R}^{\text{T}}=I_2$ .

Let $\alpha ={\theta }^{\prime }-\theta$ be the rotation by some angle $\alpha$ such that $\alpha \in \left[\mathrm{0,}\text{π}\right)$ . $\alpha >0$ implies the rotation is counter-clockwise, $\alpha <0$ implies the rotation is clockwise.

Theorem 6 The action of $SO\left(2\right)$ on $G_P\left(\left[\theta \right]\right)$ defines a congruence between two elements $P_{\left[\theta \right]}$ and $P_{\left[{\theta }^{\prime }\right]}$ in $G_P\left(\left[\theta \right]\right)$ i.e.

$\xi \mathrm{<mo>\; :\; </mo>}SO\left(2\right)\times G_P\left(\left[\theta \right]\right)\to G_P\left(\left[\theta \right]\right)$

such that

$\xi \left(R\left(\alpha \right)\mathrm{,}{P}_{\left[\theta \right]}\right)=R^{\text{T}}\left(\alpha \right)P_{\left[\theta \right]}R\left(\alpha \right)=P_{\left[{\theta }^{\prime }\right]}$ (7)

Proof. A point on ${\mathbb{S}}^1$ can be represented as ${\text{e}}^{i\theta }={\left(\cos \theta \mathrm{,}\sin \theta \right)}^{\text{T}}$ .

Hence, we can choose two points on ${\mathbb{S}}^1$ , ${\text{e}}^{i\left[\theta \right]}={\left(\cos \left[\theta \right]\mathrm{,}\sin \left[\theta \right]\right)}^{\text{T}}$ and ${\text{e}}^{i\left[{\theta }^{\prime }\right]}={\left(\cos \left[{\theta }^{\prime }\right]\mathrm{,}\sin \left[{\theta }^{\prime }\right]\right)}^{\text{T}}$ .

$\begin{array}{c}\xi \left(R\left(\alpha \right)\mathrm{,}{P}_{\left[\theta \right]}\right)=R^{\text{T}}\left(\alpha \right)P_{\left[\theta \right]}R\left(\alpha \right)\\ =\left(R^{\text{T}}{\text{e}}^{i\left[\theta \right]}\right)\otimes \left({\left({\text{e}}^{i\left[\theta \right]}\right)}^{\text{T}}R\right)\\ ={\left({\left({\text{e}}^{i\left[\theta \right]}\right)}^{\text{T}}R\right)}^{\text{T}}\otimes \left({\left({\text{e}}^{i\left[\theta \right]}\right)}^{\text{T}}R\right)\\ ={\text{e}}^{\left[i{\theta }^{\prime }\right]}\otimes {\left({\text{e}}^{\left[i{\theta }^{\prime }\right]}\right)}^{\text{T}}=P_{\left[{\theta }^{\prime }\right]}\end{array}$

It should be clear this action corresponds to the projector in the direction of $\left[\theta \right]+\alpha ,\alpha <\left|\text{π}\right|$ which, as described in the first article, consistent with the topological structure. This is a clockwise transformation of the projection operator.

This, of course, implies that all projection operator are congruent matrices since it is always possible to find some $R\left(\alpha \right)\in SO\left(2\right)$ . Moreover, we know that I is the identity of $SO\left(2\right)$ hence we get the following result

$\xi \left(I\mathrm{,}{P}_{\left[\theta \right]}\right)=I^{\text{T}}{P}_{\left[\theta \right]}I=I{P}_{\left[\theta \right]}I=P_{\left[\theta \right]}I=P_{\left[\theta \right]}$

Finally, given some $P_{\left[\theta \right]}\in G_P\left(\left[\theta \right]\right)$ the mapping $\xi$ is bijective on $\left[\theta \right]+\alpha <\left|\text{π}\right|$ which implies that the mapping $\xi$ is invertible and can be defined as

${\xi }^{-1}\left(R\left(\alpha \right)\mathrm{,}{P}_{\left[\theta \right]}\right)\mathrm{<mo>\; :\; </mo>}=\xi \left(R\left(-\alpha \right)\mathrm{,}{P}_{\left[\theta \right]}\right)$ (8)

We can see that this is equal to

${\xi }^{-1}\left(R\left(-\alpha \right)\mathrm{,}{P}_{\left[\theta \right]}\right)=R^{\text{T}}\left(-\alpha \right)P_{\left[\theta \right]}R\left(-\alpha \right)=R\left(\alpha \right)P_{\left[\theta \right]}{R}^T(\alpha )$

such that

${\xi }^{-1}\circ \xi =R\left(\alpha \right)\left(R^{\text{T}}\left(\alpha \right)P_{\left[\theta \right]}R\left(\alpha \right)\right)R^{\text{T}}=I{P}_{\left[\theta \right]}I=P_{\left[\theta \right]}$

Note that ${\xi }^{-1}$ is an counter-clockwise transformation of the projection operator.

Lemma 9 Let $R\left(\alpha \right)\mathrm{,}R\left({\alpha }^{\prime }\right)\in SO\left(2\right)$ and let $P_{\left[\theta \right]}\in G_P\left(\left[\theta \right]\right)$ , then

$\xi \left(R\left({\alpha }^{\prime }\right)\mathrm{,}\xi \left(R\left(\alpha \right)\mathrm{,}{P}_{\left[\theta \right]}\right)\right)=R^{\text{T}}\left(\alpha +{\alpha }^{\prime }\right)P_{\left[\theta \right]}R\left(\alpha +{\alpha }^{\prime }\right)=\xi \left(R\left(\alpha \right)R\left({\alpha }^{\prime }\right)\mathrm{,}{P}_{\left[\theta \right]}\right)$ (9)

Proof.

$\begin{array}{c}\xi \left(R\left({\alpha }^{\prime }\right)\mathrm{,}\xi \left(R\left(\alpha \right)\mathrm{,}{P}_{\left[\theta \right]}\right)\right)=\xi \left(R\left({\alpha }^{\prime }\right)\mathrm{,}{R}^{\text{T}}\left(\alpha \right)P_{\left[\theta \right]}R\left(\alpha \right)\right)\\ =R^{\text{T}}\left({\alpha }^{\prime }\right)\left(R^{\text{T}}\left(\alpha \right)P_{\left[\theta \right]}R\left(\alpha \right)\right)R\left({\alpha }^{\prime }\right)\\ =R^{\text{T}}\left({\alpha }^{\prime }\right)R^{\text{T}}\left(\alpha \right)P_{\left[\theta \right]}R\left(\alpha \right)R\left({\alpha }^{\prime }\right)\\ =R^{\text{T}}\left(\alpha +{\alpha }^{\prime }\right)P_{\left[\theta \right]}R\left(\alpha +{\alpha }^{\prime }\right)\\ =\xi \left(R\left(\alpha \right)R\left({\alpha }^{\prime }\right)\mathrm{,}{P}_{\left[\theta \right]}\right)\end{array}$

Lemma 10 Let $R\left(\alpha \right)$ and $P_{\left[\theta \right]}$ be elements of $SO\left(2\right)$ and $G_P\left(\left[\theta \right]\right)$ respectively. Then we have

$\xi \left(R\left(0\right)\mathrm{,}{P}_{\left[\theta \right]}\right)=\xi \left(R\left(k\text{π}\right)\mathrm{,}{P}_{\left[\theta \right]}\right)=P_{\left[\theta \right]}\mathrm{,}\forall k\in \mathbb{Z}$ (10)

Proof. Beginning with the case where $\phi =0$ we get the following result

$\begin{array}{c}\xi \left(R\left(0\right)\mathrm{,}{P}_{\left[\theta \right]}\right)=R^{\text{T}}\left(0\right)P_{\left[\theta \right]}R\left(0\right)\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ =P_{\left[\theta \right]}\end{array}$

Now choosing $\phi =k\text{π}$ for some $k\in \mathbb{Z}$ leads to

$\begin{array}{l}\xi \left(R\left(k\text{π}\right)\mathrm{,}{P}_{\left[\theta \right]}\right)=R^{\text{T}}\left(k\text{π}\right)P_{\left[\theta \right]}R\left(k\text{π}\right)\\ =\left[\begin{array}{cc}\cos \left(k\text{π}\right)& \sin \left(k\text{π}\right)\\ -\sin \left(k\text{π}\right)& \cos \left(k\text{π}\right)\end{array}\right]\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]\left[\begin{array}{cc}\cos \left(k\text{π}\right)& -\sin \left(k\text{π}\right)\\ \sin \left(k\text{π}\right)& \cos \left(k\text{π}\right)\end{array}\right]\\ =\left[\begin{array}{cc}{\left(-1\right)}^k& 0\\ 0& {\left(-1\right)}^k\end{array}\right]\left[\begin{array}{cc}{\cos }^2\left[\theta \right]& \cos \left[\theta \right]\sin \left[\theta \right]\\ \sin \left[\theta \right]\cos \left[\theta \right]& {\sin }^2\left[\theta \right]\end{array}\right]\left[\begin{array}{cc}{\left(-1\right)}^k& 0\\ 0& {\left(-1\right)}^k\end{array}\right]\\ =\left[\begin{array}{cc}{\left(-1\right)}^k{\cos }^2\left[\theta \right]& {\left(-1\right)}^k\cos \left[\theta \right]\sin \left[\theta \right]\\ {\left(-1\right)}^k\sin \left[\theta \right]\cos \left[\theta \right]& {\left(-1\right)}^k{\sin }^2\left[\theta \right]\end{array}\right]\left[\begin{array}{cc}{\left(-1\right)}^k& 0\\ 0& {\left(-1\right)}^k\end{array}\right]\\ =\left[\begin{array}{cc}{\left(-1\right)}^{2k}{\cos }^2\left[\theta \right]& {\left(-1\right)}^{2k}\cos \left[\theta \right]\sin \left[\theta \right]\\ {\left(-1\right)}^{2k}\sin \left[\theta \right]\cos \left[\theta \right]& {\left(-1\right)}^{2k}{\sin }^2\left[\theta \right]\end{array}\right]=P_{\left[\theta \right]}\end{array}$