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
In this article the notation has been kept the same as in the previous article.
• ‖ ⋅ ‖ F is the Frobenius Norm.
• P [ θ ] is some projection operator in G P ( [ θ ] ) .
• I 2 is the 2 × 2 identity matrix.
• R n k ( P [ θ ] ) is the Rank of a projection operator.
• σ ( P [ θ ] ) is the spectrum of a projection operator.
• S t a b S O ( 2 ) ( P [ θ ] ) is the Stabilizer subgroup for some fixed element of G P ( [ θ ] ) .
• R ( α ) is an element of S O ( 2 ) for angle α .
• V e c is the Vectorisation Operator.
• ⊙ is the Hadamard Product.
• ⊗ is the Kronecker Product.
From [
Lemma 1. The Frobenius morn ‖ ⋅ ‖ F of a projection matrix is its trace. That is
‖ P [ θ ] ‖ F = T r ( P [ θ ] ) = 1
Proof. It is well known that
‖ P [ θ ] ‖ F 2 : = ∑ i , j | p i j ( θ ) | 2 = T r ( P [ θ ] T P [ θ ] )
Given that P [ θ ] is both symmetric and idempotent implies that
T r ( P [ θ ] T P [ θ ] ) = T r ( P [ θ ] P [ θ ] ) = T r ( P [ θ ] 2 ) = T r ( P [ θ ] )
Hence,
‖ P [ θ ] ‖ F = T r ( P [ θ ] )
therefore,
T r ( [ cos 2 [ θ ] cos [ θ ] si n ′ [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] ) = cos 2 [ θ ] + sin 2 [ θ ] = 1, ∀ [ θ ] ∈ T
Theorem 1 The necessary and sufficient condition for P [ θ ] ∈ G P ( [ θ ] ) is given by
P [ θ ] 2 = P [ θ ] , ∀ [ θ ] ∈ T
Proof. We know, from the previous article that P [ θ ] is constructed as follows
Φ ( e i [ θ ] ) : = e i [ θ ] ⊗ ( e i [ θ ] ) T = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = P [ θ ]
P [ θ ] 2 = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] 2 = [ cos 4 [ θ ] + cos 2 [ θ ] sin 2 [ θ ] cos 3 [ θ ] sin [ θ ] + cos [ θ ] sin 3 [ θ ] cos 3 [ θ ] sin [ θ ] + cos [ θ ] sin 3 [ θ ] sin 4 [ θ ] + cos 2 [ θ ] sin 2 [ θ ] ] = [ cos 2 [ θ ] ( cos 2 [ θ ] + sin 2 [ θ ] ) cos [ θ ] sin [ θ ] ( cos 2 [ θ ] + sin 2 [ θ ] ) cos [ θ ] sin [ θ ] ( cos 2 [ θ ] + sin 2 [ θ ] ) sin 2 [ θ ] ( sin 2 [ θ ] + cos 2 [ θ ] ) ] = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = P [ θ ]
Theorem 2 The orthogonal complement of P [ θ ] is given by
P [ θ ± π / 2 ] = I 2 − P [ θ ]
where I 2 is the 2 × 2 identity matrix. This is known as a Vector Rejection.
Proof. We first show that
P [ θ ± π / 2 ] = I 2 − P [ θ ]
the RHS of the equation gives us
I 2 − P ( [ θ ] ) = [ 1 0 0 1 ] − [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = [ 1 − cos 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] 1 − sin 2 [ θ ] ] = [ sin 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] cos 2 [ θ ] ]
Now, the LHS of th equation gives us the following result
P [ θ ± π / 2 ] = [ cos 2 ( [ θ ] ± π / 2 ) cos ( [ θ ] ± π / 2 ) sin ( [ θ ] ± π / 2 ) sin ( [ θ ] ± π / 2 ) cos ( [ θ ] ± π / 2 ) sin 2 ( [ θ ] ± π / 2 ) ] = [ sin 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] cos 2 [ θ ] ]
since
cos 2 ( [ θ ] ± π / 2 ) = ( cos [ θ ] cos ( π / 2 ) ∓ sin [ θ ] sin ( π / 2 ) ) 2 = ( ∓ sin [ θ ] ) 2 = sin 2 [ θ ]
sin 2 ( [ θ ] ± π / 2 ) = ( sin [ θ ] cos ( π / 2 ) ± cos [ θ ] sin ( π / 2 ) ) 2 = ( ± cos [ θ ] ) 2 = cos 2 [ θ ]
Secondly, we show the effect these operators have on some vector x ∈ ℝ 2 . We easily show this in the following way
〈 P [ θ ] x , P [ θ ± π / 2 ] x 〉 = 0
We proceed as follows
〈 P [ θ ] x , P [ θ ± π / 2 ] x 〉 = 〈 P [ θ ] x , ( I 2 − P [ θ ] ) x 〉 = 〈 P [ θ ] x , I 2 x − P [ θ ] x 〉 = 〈 P [ θ ] x , x − P [ θ ] x 〉 = 〈 P [ θ ] x , x 〉 − 〈 P [ θ ] x , P [ θ ] x 〉 = ‖ P [ θ ] x ‖ ‖ x ‖ cos θ − ‖ P [ θ ] x ‖ 2 = ‖ P [ θ ] x ‖ ‖ P [ θ ] x ‖ − ‖ P [ θ ] x ‖ 2 = 0
Lemma 2 The product of the operators P [ θ ] and P [ θ ± π / 2 ] is zero. We shall call P [ θ ± π / 2 ] the Orthogonal Complement Operator of P [ θ ] which I shall denote as P [ θ ] ⊥ .
P [ θ ] P [ θ ] ⊥ = 0, ∀ [ θ ] ∈ T
We say that P [ θ ] ⊥ P [ θ ] ⊥
Proof.
P [ θ ] P [ θ ] ⊥ = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ sin 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] cos 2 [ θ ] ] = [ cos 2 [ θ ] sin 2 [ θ ] − cos 2 [ θ ] sin 2 [ θ ] − cos 3 [ θ ] sin [ θ ] + cos 3 [ θ ] sin [ θ ] sin 3 [ θ ] cos [ θ ] − sin 3 [ θ ] cos [ θ ] − cos 2 [ θ ] sin 2 [ θ ] + sin 2 [ θ ] cos 2 [ θ ] ] = [ 0 0 0 0 ]
Lemma 3 P [ θ ] + P [ θ ] ⊥ = I 2
Proof.
P [ θ ] + P [ θ ] ⊥ = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] + [ sin 2 [ θ ] − cos [ θ ] sin [ θ ] − sin [ θ ] cos [ θ ] cos 2 [ θ ] ] = [ 1 0 0 1 ] = I 2
Theorem 3 A projection operator of dimension onto V ⊂ ℝ 2 where dim ( V ) = r then
P [ θ ] = T Δ r T − 1 (1)
where
Δ r : = [ 1 0 0 0 ] (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 , W ⊂ ℝ 2 , furthermore we also know that dim ( V ) = dim ( W ) = 1 ⇒ r = 1 .
Suppose the vector a = ( cos θ , sin θ ) T is a basis for subspaces V i.e. S p n { a } = V then the matrix T can be computed in the following way.
Now, taking into account the fact that W = V ⊥ implies that b is given by
b = ( cos ( θ ± π / 2 ) , sin ( θ ± π / 2 ) ) = ( ∓ sin θ , ± cos θ )
Hence, T is computed in the following way
T = [ cos [ θ ] ∓ sin [ θ ] sin [ θ ] ± cos [ θ ] ] ∴ det ( T ) = ± cos 2 [ θ ] ± sin 2 [ θ ] = ± 1
Hence, T − 1 is given by
T − 1 = ± 1 [ ± cos [ θ ] ± sin [ θ ] − sin [ θ ] cos [ θ ] ] = [ cos [ θ ] sin [ θ ] ∓ sin [ θ ] ± cos [ θ ] ] = T T ⇒ T , T T ∈ O (2,ℝ)
Hence, P [ θ ] is
T Δ r T − 1 = [ cos [ θ ] ∓ sin [ θ ] sin [ θ ] ± cos [ θ ] ] [ 1 0 0 0 ] ± 1 [ ± cos [ θ ] ± sin [ θ ] − sin [ θ ] cos [ θ ] ] = ± 1 [ cos [ θ ] ∓ sin [ θ ] sin [ θ ] ± cos [ θ ′ ] ] [ ± cos [ θ ] ± sin [ θ ] 0 0 ] = ± 1 [ ± cos 2 [ θ ] ± cos [ θ ] sin [ θ ] ± sin [ θ ] cos [ θ ] ± sin 2 [ θ ] ] = P [ θ ]
We can see that T , T − 1 ∈ O ( 2, ℝ ) and are orthogonal reflection matrices.
Lemma 4 Let P [ θ ] be some projection matrix for some [ θ ] ∈ T then we have
R n k ( P [ θ ] ) = T r ( P θ ) (3)
Proof. We know that
P [ θ ] = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] ⇒ T r ( P [ θ ] ) = cos 2 [ θ ] + sin 2 [ θ ] = 1
We can now calculate its rank. The column space of P [ θ ] is given by
C l n ( P [ θ ] ) = { [ cos 2 [ θ ] sin [ θ ] cos [ θ ] ] , [ cos [ θ ] sin [ θ ] sin 2 [ θ ] ] }
where C l n ( P [ θ ] ) is the column space of P [ θ ] . R n k ( P [ θ ] ) = 1 iff ∃ α 1 , α 2 ≠ 0 ∈ ℝ such that
α 1 [ cos 2 [ θ ] sin [ θ ] cos [ θ ] ] + α 2 [ cos [ θ ] sin [ θ ] sin 2 [ θ ] ] = [ 0 0 ]
This gives us the following linear system
[ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ α 1 α 2 ] = [ 0 0 ]
Given that det ( P [ θ ] ) = 0 implies an infinite set of solutions exists.
Writing the homogeneous linear system we find the general set of vector solutions as follows
α 1 cos 2 [ θ ] + α 2 cos [ θ ] sin [ θ ] = 0
α 1 sin [ θ ] cos [ θ ] + α 2 sin 2 [ θ ] = 0
therefore
α 1 sin [ θ ] cos [ θ ] + α 2 sin 2 [ θ ] = 0
⇒ α 1 cos [ θ ] + α 2 sin [ θ ] = 0
∴ α 1 = − α 2 sin [ θ ] cos [ θ ] = − α 2 tan [ θ ]
Hence, the solution vector
U = { α 2 ∈ ℝ | α = α 2 ( − tan [ θ ] ,1 ) }
This implies that R n k ( P [ θ ] ) = T r ( P [ θ ] ) = 1 .
Theorem 4 Let A = [ a 1 ] such that V = S p n ( A ) . Then we have
P [ θ ] = A ( A T A ) − 1 A T (4)
Proof. a 1 = ( cos [ θ ] , sin [ θ ] ) T = A , then we have
A ( A T A ) − 1 A T = A ( a 1 T a 1 ) − 1 A = A A T = a 1 a 1 T = a 1 ⊗ a 1 = P [ θ ]
Lemma 5 The spectrum σ ( P [ θ ] ) of the P [ θ ] for some [ θ ] ∈ T is given by
σ ( P [ θ ] ) { λ 1 = 0, λ 2 = 1, ∀ P [ θ ] ∈ G P ( [ θ ] ) }
Proof. We simply perform the following calculation
det ( P [ θ ] − λ I 2 ) = | cos 2 θ − λ cos θ sin θ sin θ cos θ sin 2 θ − λ | = ( cos 2 θ − λ ) ( sin 2 θ − λ ) − cos 2 θ sin 2 θ = cos 2 θ sin 2 θ − λ cos 2 θ − λ sin 2 θ + λ 2 − cos 2 θ sin 2 θ = λ 2 − λ
therefore we see that
det ( P [ θ ] − λ I 2 ) = 0 ⇒ λ 2 − λ = 0 ⇒ λ ( λ − 1 ) = 0 ⇒ λ = 0 ∨ λ = 1
Hence, σ ( P [ θ ] ) = { 0 , 1 } as claimed.
We can calculate the associated Eigenspaces. For λ = 0 we get
P [ θ ] E 0 = λ E 0 = 0
Hence, it is clear that given some P [ θ ] its Eigenspace for λ = 0 is K e r ( [ P [ θ ] ] ) , that is, in this case, the Eigenspace is spanned by the following vectors
E 0 ( [ θ ] ) : = ( cos ( [ θ ] + π / 2 ) , sin ( [ θ ] + π / 2 ) ) T = [ − sin ( [ θ ] ) cos ( [ θ ] ) ]
Therefore, we can say that is I m ( P [ θ ] ) is V ⊂ ℝ 2 then S p n { E 0 ( [ θ ] ) } = V ⊥ .
For λ = 1 we get the following Eigenvectors
P [ θ ] E 1 = λ E 1 = E 1
P [ θ ] E 1 − E 1 = ( P [ θ ] − I 2 ) E 1 = 0
In matrix form, this gives us
[ cos 2 θ − 1 cos θ sin θ sin θ cos θ sin 2 θ − 1 ] [ E 1 E 2 ] = [ 0 0 ]
[ − sin 2 θ cos θ sin θ sin θ cos θ − cos 2 θ ] [ E 1 E 2 ] = [ 0 0 ]
This gives us the following system
− sin 2 θ E 1 + cos θ sin θ E 2 = 0
sin θ cos θ E 1 − cos 2 θ E 2 = 0
therefore, from (24) we get
sin θ cos θ E 1 = cos 2 θ ⇒ E 2 tan θ E 1 = E 2
Therefore,
E 2 E 1 = tan θ ⇒ E 1 ( θ ) = E 2 cot θ
Therefore, the Eigenvector is given
E 1 = E 2 [ cot θ 1 ]
Lemma 6 The Eigenvectors E 0 and E 1 are linearly independent. Moreover, S p n { E 0 , E 1 } = ℝ 2 .
Proof. For γ 0 , γ 1 ∈ ℝ we have
γ 0 E 0 + γ 1 E 1 = 0
Clearly, E 0 , E 1 are linearly independent iff γ 0 = γ 1 = 0
− γ 0 sin θ + γ 1 cot θ = 0
γ 0 cos θ + γ 1 = 0
This can be written matrix form in the following way
[ − sin θ cot θ cos θ 1 ] [ γ 0 γ 1 ] = [ 0 0 ]
Now,
| − sin θ cot θ cos θ 1 | = − sin θ − cot θ cos θ ≠ 0 , ∀ θ
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 [ θ ] . Given that P [ θ ] has distinct eigenvalues implies that it is diagonalizable i.e. P [ θ ] = P D P − 1 where D is a diagonal matrix
Lemma 7
P [ θ ] = P [ 0 0 0 1 ] P − 1
where
P = [ E 0 , E 1 ] = [ − sin θ cot θ cos θ 1 ] , P − 1 = − 1 sin θ + cot θ cos θ [ 1 − cot θ − cos θ − sin θ ]
Proof.
P [ θ ] = − 1 sin θ + cot θ cos θ [ − sin θ cot θ cos θ 1 ] [ 0 0 0 1 ] [ 1 − cot θ − cos θ − sin θ ] = − 1 sin θ + cot θ cos θ [ − sin θ cot θ cos θ 1 ] [ 0 0 − cos θ − sin θ ] = − 1 sin θ + cot θ cos θ [ − cot θ cos θ − cot θ sin θ − cos θ − sin θ ] = [ cos 2 θ cos θ sin θ sin θ cos θ sin 2 θ ] = P [ θ ]
Theorem 5 The projection matrix P [ θ ] with spectrum σ = { λ 1 , λ 2 } is diagonalizable, hence there exists matrices { G 1 , G 2 } such that
P [ θ ] = λ 1 G 1 + λ 2 G 2 (5)
where G 1 , G 2 are projection matrices onto N ( P [ θ ] − λ i I 2 ) along R ( P [ θ ] − λ i I 2 ) . Furthermore,
1) G i , G j = 0 whenever i ≠ j .
2) ∑ i = 1 m k G i = I 2 .
Proof. We know that the Eigenvalues are λ 1 = 0 ∧ λ 2 = 1 this obviously implies that
P [ θ ] = λ 2 G 2 = G 2
P [ θ ] = P D P − 1 = ( X 1 | X 2 ) ( λ 1 I 0 0 λ 2 I ) ( Y 1 T Y 2 T ) = λ 1 X 1 Y 1 T + λ 2 X 2 Y 2 T = λ 2 X 2 Y 2 T = X 2 Y 2 T = − 1 sin θ + cot θ cos θ [ cot θ 1 ] [ − cos θ − sin θ ] = − 1 sin θ + cot θ cos θ [ − cot θ cos θ − cot θ sin θ − cos θ − sin θ ] = P [ θ ]
This implies that G 2 = X 2 Y 2 T .
Lemma 8 Let G 0 and G 1 be the spectral projectors for the eigenvalues λ 1 and λ 2 respectively. Then we show that
1)
G 0 G 1 = 0
2)
G 0 + G 1 = I 2
Proof.
G 0 G 1 = X 1 Y 1 T = [ − sin θ cos θ ] [ 1 − cot θ ] [ cot θ 1 ] [ − cos θ − sin θ ] = [ − sin θ cot θ sin θ cos θ − cos θ cot θ ] [ − cot θ cos θ − sin θ cot θ − cos θ − sin θ ] = [ 0 0 0 0 ]
As claimed.
For the second part of the proof, we simply add the matrices
G 0 + G 1 = − 1 sin θ + cot θ cos θ { [ − sin θ cot θ sin θ cos θ − cos θ cot θ ] + [ − cot θ cos θ − sin θ cot θ − cos θ − sin θ ] } = − 1 sin θ + cot θ cos θ [ − ( sin θ + cot θ sin θ ) 0 0 − ( cos θ cot θ + sin θ ) ] = I 2
From [
Definition 1 Two squares matrices A and B are said to be Congruent if there exists an invertible matrix R such that
B = R T A R (6)
The form of the equation would tend to suggest that R T , R ∈ S O ( 2, ℝ ) such that R T R = R R T = I 2 .
Let α = θ ′ − θ be the rotation by some angle α such that α ∈ [ 0, π ) . α > 0 implies the rotation is counter-clockwise, α < 0 implies the rotation is clockwise.
Theorem 6 The action of S O ( 2 ) on G P ( [ θ ] ) defines a congruence between two elements P [ θ ] and P [ θ ′ ] in G P ( [ θ ] ) i.e.
ξ : S O ( 2 ) × G P ( [ θ ] ) → G P ( [ θ ] )
such that
ξ ( R ( α ) , P [ θ ] ) = R T ( α ) P [ θ ] R ( α ) = P [ θ ′ ] (7)
Proof. A point on S 1 can be represented as e i θ = ( cos θ , sin θ ) T .
Hence, we can choose two points on S 1 , e i [ θ ] = ( cos [ θ ] , sin [ θ ] ) T and e i [ θ ′ ] = ( cos [ θ ′ ] , sin [ θ ′ ] ) T .
ξ ( R ( α ) , P [ θ ] ) = R T ( α ) P [ θ ] R ( α ) = ( R T e i [ θ ] ) ⊗ ( ( e i [ θ ] ) T R ) = ( ( e i [ θ ] ) T R ) T ⊗ ( ( e i [ θ ] ) T R ) = e [ i θ ′ ] ⊗ ( e [ i θ ′ ] ) T = P [ θ ′ ]
It should be clear this action corresponds to the projector in the direction of [ θ ] + α , α < | π | 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 ( α ) ∈ S O ( 2 ) . Moreover, we know that I is the identity of S O ( 2 ) hence we get the following result
ξ ( I , P [ θ ] ) = I T P [ θ ] I = I P [ θ ] I = P [ θ ] I = P [ θ ]
Finally, given some P [ θ ] ∈ G P ( [ θ ] ) the mapping ξ is bijective on [ θ ] + α < | π | which implies that the mapping ξ is invertible and can be defined as
ξ − 1 ( R ( α ) , P [ θ ] ) : = ξ ( R ( − α ) , P [ θ ] ) (8)
We can see that this is equal to
ξ − 1 ( R ( − α ) , P [ θ ] ) = R T ( − α ) P [ θ ] R ( − α ) = R ( α ) P [ θ ] R T (α)
such that
ξ − 1 ∘ ξ = R ( α ) ( R T ( α ) P [ θ ] R ( α ) ) R T = I P [ θ ] I = P [ θ ]
Note that ξ − 1 is an counter-clockwise transformation of the projection operator.
Lemma 9 Let R ( α ) , R ( α ′ ) ∈ S O ( 2 ) and let P [ θ ] ∈ G P ( [ θ ] ) , then
ξ ( R ( α ′ ) , ξ ( R ( α ) , P [ θ ] ) ) = R T ( α + α ′ ) P [ θ ] R ( α + α ′ ) = ξ ( R ( α ) R ( α ′ ) , P [ θ ] ) (9)
Proof.
ξ ( R ( α ′ ) , ξ ( R ( α ) , P [ θ ] ) ) = ξ ( R ( α ′ ) , R T ( α ) P [ θ ] R ( α ) ) = R T ( α ′ ) ( R T ( α ) P [ θ ] R ( α ) ) R ( α ′ ) = R T ( α ′ ) R T ( α ) P [ θ ] R ( α ) R ( α ′ ) = R T ( α + α ′ ) P [ θ ] R ( α + α ′ ) = ξ ( R ( α ) R ( α ′ ) , P [ θ ] )
Lemma 10 Let R ( α ) and P [ θ ] be elements of S O ( 2 ) and G P ( [ θ ] ) respectively. Then we have
ξ ( R ( 0 ) , P [ θ ] ) = ξ ( R ( k π ) , P [ θ ] ) = P [ θ ] , ∀ k ∈ ℤ (10)
Proof. Beginning with the case where φ = 0 we get the following result
ξ ( R ( 0 ) , P [ θ ] ) = R T ( 0 ) P [ θ ] R ( 0 ) = [ 1 0 0 1 ] [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ 1 0 0 1 ] = P [ θ ]
Now choosing φ = k π for some k ∈ ℤ leads to
ξ ( R ( k π ) , P [ θ ] ) = R T ( k π ) P [ θ ] R ( k π ) = [ cos ( k π ) sin ( k π ) − sin ( k π ) cos ( k π ) ] [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ cos ( k π ) − sin ( k π ) sin ( k π ) cos ( k π ) ] = [ ( − 1 ) k 0 0 ( − 1 ) k ] [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] [ ( − 1 ) k 0 0 ( − 1 ) k ] = [ ( − 1 ) k cos 2 [ θ ] ( − 1 ) k cos [ θ ] sin [ θ ] ( − 1 ) k sin [ θ ] cos [ θ ] ( − 1 ) k sin 2 [ θ ] ] [ ( − 1 ) k 0 0 ( − 1 ) k ] = [ ( − 1 ) 2 k cos 2 [ θ ] ( − 1 ) 2 k cos [ θ ] sin [ θ ] ( − 1 ) 2 k sin [ θ ] cos [ θ ] ( − 1 ) 2 k sin 2 [ θ ] ] = P [ θ ]
This shows us that projection matrices have a period of π .
We can now calculate the Stabilizer of G P ( [ θ ] ) as follows.
Definition 2 The rotation group S O ( 2 ) acts on G P ( [ θ ] ) then the Stabilizer of some element P [ θ ] ∈ G P ( [ θ ] ) denoted S t a b S O ( 2 ) is defined as
S t a b S O ( 2 ) ( P [ θ ] ) : = { R ( α ) ∈ S O ( 2 ) : ξ ( R ( α ) , P [ θ ] ) = P [ θ ] }
From the above definition it is clear that
S t a b S O ( 2 ) ( P [ θ ] ) : = { R [ k π ] : α = k π , k ∈ ℤ }
Lemma 11
S t a b S O ( 2 ) ( P [ θ ] ) ≤ S O (2)
Proof. Choosing k = 0 implies R ( 0 ) = I 2 ⇒ R ( 0 ) P [ θ ] = I 2 P [ θ ] = P [ θ ] . We know that for every R ( k π ) ∃ R z − 1 ( k π ) = R T ( k π ) s.t. R ( k π ) R T ( k π ) = R ( k π ) T R ( k π ) = I 2 .
hence, for some k , k ′ ∈ ℤ we have
R ( k π ) ( R T ( k π ) P [ θ ] R ( k π ) ) R T ( k π ) = R ( k π ) ( P [ θ ] ) R T ( k π ) = P [ θ ]
This implies that
R ( k π ) ( R T ( k π ) R ( k π ) ) R T ( k π ) = R ( k π ) I 2 R T ( k π ) = I 2 ∈ S t a b S O ( 2 ) ( P [ θ ] )
Theorem 7. Let P [ θ ] , P [ θ ′ ] ∈ G P ( [ θ ] ) . Let θ ′ = α + α ′ then
P [ θ + θ ′ ] = R T ( α + α ′ ) P [ θ ] R ( α + α ′ )
Proof.
R T ( α + α ′ ) P [ θ ] R ( α + α ′ ) = R T ( α ) R T ( α ′ ) P [ θ ] R ( α ′ ) R ( α ) = R T ( α ) P [ θ + α ′ ] R ( α ) = P [ θ + ( α ′ + α ) ] = P [ θ + θ ′ ]
Hence, we can conclude that
R T ( α + α ′ ) P [ θ ] R ( α + α ′ ) = P [ θ + θ ′ ]
Lemma 12 The group of action of S O ( 2, ℝ ) on G P ( [ θ ] ) not faithful but it is ∞-transitive.
Proof. for a group action to be faithful implies that for every pair of distinct elements of S O ( 2, ℝ ) there is some P [ θ ] such that R ( α ) P [ θ ] R T ( α ) ≠ R ( α ′ ) P [ θ ] R T ( α ′ ) , ∀ P [ θ ] ∈ G P ( [ θ ] ) .
Consider the set S O ( 2, ℝ ) × S O ( 2, ℝ ) and let us choose some arbitrary projector in G P ( [ θ ] ) . Specifically, let us consider the pair ( R ( α ) , R ( α ′ ) ) ∈ S O ( 2, ℝ ) × S O ( 2, ℝ ) such that α ′ = α ± π . Then we demonstrate that it is impossible to find an element of G P ( [ θ ] ) such that the above definition is satisfied.
R T ( α ′ ) P [ θ ] R ( α ′ ) = R T ( α ± π ) P [ θ ] R ( α ± π ) = − R T ( α ) P [ θ ] − R ( α ) = R T ( α ) P [ θ ] R ( α ) (11)
Since α ′ ≠ α and P [ θ ] is arbitrary, we conclude that this action is not faithful.
However, we are going to demonstrate that it is n-transitive. To show this we can consider two pairwise distinct projector sequences of form
( { P [ θ i ] } , { P [ θ r ] } ) , i , r = 1 , ⋯ , n
each sequence is pairwise distinct, that is, P [ θ i ] ≠ P [ θ j ] , ∀ i , j = 1 , ⋯ , n and P [ θ r ] ≠ P [ θ s ] , ∀ r , s = 1 , ⋯ , n . Suppose that each sequence is chosen so that ϕ ( P [ θ i ] ) , ϕ ( P [ θ r ] ) ∈ T and form two arithmetic sequences such that the quotient metric satisfies d T ( [ θ i ] , [ θ r ] ) = inf k ∈ ℤ { | θ i − θ r + 2 k π | } = β < π , ∀ i , r = 1, ⋯ , n . Then we can have the following result
R T ( β ) P [ θ i ] R ( β ) = P [ θ r ] , ∀ i , r = 1, ⋯ , n
we can define a refinement of the sequence in the following way
σ ( β ) = β ′ , β ′ < β ⇒ σ ( { P [ θ i ] } i = 1 n ) = { P [ θ i ] } i = 1 n ′ , n < n ′
As β → 0 ⇒ n → ∞ we have lim n → ∞ ( { P [ θ i ] } i = 1 n ∪ { P [ θ r ] } r = 1 n ) = P α ⊆ G P ( [ θ ] ) ∈ τ G P ( [ θ ] ) .
where, it was shown that τ G P ( [ θ ] ) is the topology G P [ θ ] . Hence, this group action is ∞-transitive.
Lemma 13 The Kernel of ξ is given by
K e r ξ : = { R ( α ) ∈ S O ( 2, ℝ ) : R T ( α ) P [ θ ] R ( α ) = P [ 0 ] } = { R ( α ) ∈ S O ( 2, ℝ ) : α : = − θ }
for some P [ θ ] ∈ G P ( [ θ ] ) .
Proof. Let P [ θ ] be some element of G P ( [ θ ] ) and choose α = − θ the we have
ξ ( R ( − θ ) P [ θ ] ) = R T ( − θ ) P [ θ ] R ( − θ ) = R ( θ ) P [ θ ] R T ( θ ) = ξ − 1 ( R ( θ ) , P [ θ ] ) = P [ 0 ] = I d G P ( [ θ ] )
Lemma 14 The Lie Group G P ( [ θ ] ) has exactly one orbit.
Proof. We know that the action is transitive since for all pairs on G P ( [ θ ] ) × G P ( [ θ ] ) of the form ( P [ θ ′ ] , P [ θ ] ) there exists R ( α + k π ) , k ∈ ℤ such that P [ θ ′ ] = R T ( α + k π ) P [ θ ] R ( α + k π ) , ∀ k ∈ ℤ .
We just need to show that for some fixed P [ θ ] ∈ G P ( [ θ ] ) we get R T ( α ) P [ θ ] R ( α ) = G P ( [ θ ] ) . It is clear that choosing − π 2 < α ≤ π 2 we get
{ R T ( π 2 ) P [ θ ] R ( π 2 ) = P [ θ + π 2 ] = P [ θ ] ⊥ R T ( − π 2 ) P [ θ ] R ( − π 2 ) = P [ θ − π 2 ] = R ( π 2 ) P [ θ ] R T ( π 2 ) = P [ θ ] ⊥
Due to the fact that projector repeat every π we get see that R T ( α ) P [ θ ] R ( α ) = G P ( [ θ ] ) .
Definition 3 The Vec operator is a linear transformation which converts a matrix into a column vector. That is
V e c : ℝ n × m → ℝ n m
defined as
V e c ( A ) = [ a 1,1 , ⋯ , a m ,1 , ⋯ , a 1,2 , ⋯ , a m ,2 , ⋯ , a m , n ] T , A ∈ ℝ n × m
Theorem 8 The group S O ( 2, ℝ ) defines a homeomorphism group on G P ( [ θ ] ) . That is
S O ( 2, ℝ ) : G P ( [ θ ] ) → G P ( [ θ ] )
For some P [ θ ] ∈ G P ( [ θ ] ) and for some translation angle α ∈ [ 0, π ) this mapping can be defined as follows
V e c − 1 { ( R z ( α ) ⊗ R z ( α ) ) V e c ( P [ θ ] ) }
Proof.
( R ( α ) ⊗ R ( α ) ) V e c ( P [ θ ] ) = [ cos 2 α − cos α sin α − sin α cos α − sin 2 α cos α sin α cos 2 α − sin 2 α − sin α cos α sin α cos α − sin 2 α cos 2 α − cos α sin α sin 2 α sin α cos α cos α sin α cos 2 α ] [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = [ cos 2 α cos 2 [ θ ] − 2 cos α sin α cos [ θ ] sin [ θ ] + sin 2 α sin 2 [ θ ] cos α sin α cos 2 [ θ ] + cos 2 α cos [ θ ] sin [ θ ] − sin 2 α sin [ θ ] cos [ θ ] − sin α cos α sin 2 [ θ ] cos α sin α cos 2 [ θ ] + cos 2 α cos [ θ ] sin [ θ ] − sin 2 α sin [ θ ] cos [ θ ] − sin α cos α sin 2 [ θ ] sin 2 α cos 2 [ θ ] + 2 sin α cos α cos θ sin [ θ ] + cos 2 α sin 2 [ θ ] ]
= [ ( cos α cos [ θ ] − sin α sin [ θ ] ) 2 ( cos [ θ ] sin α ) ( cos [ θ ] cos α − sin α sin [ θ ] ) + cos α sin θ ( cos α cos [ θ ] − sin α sin [ θ ] ) cos α sin θ ( cos α cos [ θ ] − sin α sin [ θ ] ) ( sin α cos [ θ ] + cos α sin [ θ ] ) 2 ] = [ cos 2 ( α + [ θ ] ) sin ( α + [ θ ] ) cos ( α + [ θ ] ) cos ( α + [ θ ] ) sin ( α + [ θ ] ) sin 2 ( α + [ θ ] ) ]
Let θ ′ = α + [ θ ] then we have
( R ( α ) ⊗ R ( α ) ) V e c ( P [ θ ] ) = [ cos 2 [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ′ ] ]
Knowing the size of the original matrix i.e. a 2 × 2 matrix and using the properties of the V e c Operator i.e. the isomorphism ℝ 2 × 2 ≅ ℝ 4 we find that
V e c − 1 ( R ( α ) ⊗ R ( α ) ) V e c ( P [ θ ] ) = V e c − 1 ( [ cos 2 [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ′ ] ] ) = [ cos 2 [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ′ ] ] = P [ θ ′ ]
We have already seen that the group operation on G ( [ θ ] ) is a follows
φ ( P [ θ ] , P [ θ ′ ] ) = P [ θ + θ ′ ] , ∀ P [ θ ] , P [ θ ′ ] ∈ G P ( [ θ ] ) (12)
P [ θ ] = e i [ θ ] ⊗ ( e i [ θ ] ) T , P [ θ ′ ] = e i [ θ ′ ] ⊗ ( e i [ θ ′ ] ) T
Hence, P [ θ + θ ′ ] = e [ θ + θ ′ ] ⊗ ( e i [ θ + θ ′ ] ) T .
Definition 4 ⊙ is the Hadamard Product defined as follows.
Let A and B be two matrices the Hadamard product is
( A ⊙ B ) i , j : = ( A ) i , j ( B ) i , j
Theorem 9
P [ θ + θ ′ ] = φ ( P [ θ ] , P [ θ ′ ] ) = [ φ 11 ( P [ θ ] , P [ θ ′ ] ) φ 21 ( P [ θ ] , P [ θ ′ ] ) φ 21 ( P [ θ ] , P [ θ ′ ] ) φ 22 ( P [ θ ] , P [ θ ′ ] ) ] (13)
where
φ ( P [ θ ] , P [ θ ′ ] ) : = { φ 11 ( P [ θ ] , P [ θ ′ ] ) : = 1 − T V e c ( P [ θ ] ⊙ P [ θ ′ ] ) φ 12 ( P [ θ ] , P [ θ ′ ] ) = φ 21 ( P [ θ ] , P [ θ ′ ] ) : = 1 + T V e c ( P [ θ ] ⊙ Q P [ θ ′ ] Q * ) φ 22 ( P [ θ ] , P [ θ ′ ] ) : = 1 − T V e c ( P [ θ ] ⊙ P [ θ ′ ] ⊥ )
where 1 − T = [ 1, − 1, − 1,1 ] T and 1 + T = [ 1,1,1,1 ] T . The matrices Q and Q * are defined as
Q : = [ 0 1 1 0 ] , Q * = [ 1 0 0 − 1 ]
Proof.
1 − T V e c ( P [ θ ] ⊙ P [ θ ′ ] ) = [ 1, − 1, − 1,1 ] v e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] ⊙ [ cos 2 [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ′ ] ] ) = [ 1, − 1, − 1,1 ] v e c ( [ cos 2 [ θ ] cos 2 [ θ ′ ] cos [ θ ] sin [ θ ] cos [ θ ′ ] sin [ θ ′ ] sin [ θ ] cos [ θ ] sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ] sin 2 [ θ ′ ] ] ) = [ 1, − 1, − 1,1 ] [ cos 2 [ θ ] cos 2 [ θ ′ ] sin [ θ ] cos [ θ ] sin [ θ ′ ] cos [ θ ′ ] cos [ θ ] sin [ θ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ] sin 2 [ θ ′ ] ] = cos 2 [ θ ] cos 2 [ θ ′ ] − 2 sin [ θ ] cos [ θ ] sin [ θ ′ ] cos [ θ ′ ] + sin 2 [ θ ] sin 2 [ θ ′ ] = cos 2 ( [ θ + θ ′ ] )
For the anti-diagonal elements, it is clear that they are equal due to symmetry.
φ 12 = φ 21 = [ 1,1,1,1 ] V e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] ⊙ [ 0 1 1 0 ] [ cos 2 [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ′ ] ] [ 1 0 0 − 1 ] ) = [ 1,1,1,1 ] V e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] ⊙ [ sin [ θ ′ ] cos [ θ ′ ] sin 2 [ θ ′ ] cos 2 [ θ ′ ] cos [ θ ′ ] sin [ θ ′ ] ] [ 1 0 0 − 1 ] )
= [ 1,1,1,1 ] V e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] ⊙ [ sin [ θ ′ ] cos [ θ ′ ] − sin 2 [ θ ′ ] cos 2 [ θ ′ ] − cos [ θ ′ ] sin [ θ ′ ] ] ) = [ 1,1,1,1 ] V e c ( [ cos 2 [ θ ] sin [ θ ′ ] cos [ θ ′ ] − cos [ θ ] sin [ θ ] sin 2 [ θ ′ ] sin [ θ ] cos [ θ ] cos 2 [ θ ′ ] − sin 2 [ θ ] cos [ θ ′ ] sin [ θ ′ ] ] ) = [ 1,1,1,1 ] [ cos 2 [ θ ] sin [ θ ′ ] cos [ θ ′ ] sin [ θ ] cos [ θ ] cos 2 [ θ ′ ] − cos [ θ ] sin [ θ ] sin 2 [ θ ′ ] − sin 2 [ θ ] cos [ θ ′ ] sin [ θ ′ ] ]
= cos 2 [ θ ] sin [ θ ′ ] cos [ θ ′ ] + sin [ θ ] cos [ θ ] cos 2 [ θ ′ ] − cos [ θ ] sin [ θ ] sin 2 [ θ ′ ] − sin 2 [ θ ] cos [ θ ′ ] sin [ θ ′ ] = cos ( [ θ + θ ′ ] ) sin ( [ θ + θ ′ ] )
for the element p 22 we get the following result
1 − T V e c ( P [ θ ] ⊙ P [ θ ′ ] ⊥ ) = [ 1, − 1, − 1,1 ] V e c ( [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] ⊙ [ sin 2 [ θ ′ ] − cos [ θ ′ ] sin [ θ ′ ] − cos [ θ ′ ] sin [ θ ′ ] cos 2 [ θ ′ ] ] ) = [ 1, − 1, − 1,1 ] V e c ( [ cos 2 [ θ ] sin 2 [ θ ′ ] − cos [ θ ] sin [ θ ] cos [ θ ′ ] sin [ θ ′ ] − sin [ θ ] cos [ θ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ] cos 2 [ θ ′ ] ] ) = [ 1, − 1, − 1,1 ] [ cos 2 [ θ ] sin 2 [ θ ′ ] − sin [ θ ] cos [ θ ] cos [ θ ′ ] sin [ θ ′ ] − cos [ θ ] sin [ θ ] cos [ θ ′ ] sin [ θ ′ ] sin 2 [ θ ] cos 2 [ θ ′ ] ] = cos 2 [ θ ] sin 2 [ θ ′ ] + 2 sin [ θ ] cos [ θ ] cos [ θ ′ ] sin [ θ ′ ] + sin 2 [ θ ] cos 2 [ θ ′ ] = sin 2 ( [ θ + θ ′ ] )
Now choosing θ ′ = 0 implies that we have φ ( P [ θ ] , P [ 0 ] ) = P [ θ + 0 ] = P [ θ ] , that is we have
φ ( P [ θ ] , P [ 0 ] ) = [ φ 11 ( P [ θ ] , P [ 0 ] ) φ 21 ( P [ θ ] , P [ 0 ] ) φ 21 ( P [ θ ] , P [ 0 ] ) φ 22 ( P [ θ ] , P [ 0 ] ) ] = [ 1 − T V e c ( P [ θ ] ⊙ P [ 0 ] ) 1 + T V e c ( P [ θ ] ⊙ Q P [ 0 ] Q * ) 1 + T V e c ( P [ θ ] ⊙ Q P [ 0 ] Q * ) 1 − T V e c ( P [ θ ] ⊙ P [ 0 ] ⊥ ) ] = [ cos 2 [ θ ] cos [ θ ] sin [ θ ] sin [ θ ] cos [ θ ] sin 2 [ θ ] ] = P [ θ ]
To be thorough let us choose θ ′ = − θ , we should expect to calculate φ ( P [ θ ] , P [ − θ ] ) = P [ θ ] + [ − θ ] = P [ 0 ] , that is
φ ( P [ θ ] , P [ − θ ] ) = [ φ 11 ( P [ θ ] , P [ − θ ] ) φ 21 ( P [ θ ] , P [ − θ ] ) φ 21 ( P [ θ ] , P [ − θ ] ) φ 22 ( P [ θ ] , P [ − θ ] ) ] = [ cos 2 ( θ − θ ) cos ( θ − θ ) sin ( θ − θ ) sin ( θ − θ ) cos ( θ − θ ) sin 2 ( θ − θ ) ] = [ 1 0 0 0 ] = P [ 0 ] = P e
Last but least we will check that operation is associative that we want to make sure that
φ ( P [ θ ″ ] , φ ( P [ θ ] , P [ θ ′ ] ) ) = φ ( φ ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )
Proof. Clearly the operation is associative if all component functions are also associative. Hence we shave to show that
φ 11 ( P [ θ ″ ] , φ 11 ( P [ θ ] , P [ θ ′ ] ) ) = φ 11 ( φ 11 ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )
φ 11 ( P [ θ ″ ] , φ 11 ( P [ θ ] , P [ θ ′ ] ) ) = φ 11 ( P [ θ ″ ] , P [ θ + θ ′ ] ) = 1 − T V e c ( P [ θ ″ ] ⊙ P [ θ + θ ′ ] ) = [ 1, − 1, − 1,1 ] [ cos 2 θ ″ cos 2 ( θ + θ ′ ) cos θ ″ sin θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) sin θ ″ cos θ ″ sin ( θ + θ ′ ) cos ( θ + θ ′ ) sin 2 θ ″ sin 2 ( θ + θ ′ ) ] = cos 2 θ ″ cos 2 ( θ + θ ′ ) − 2 cos θ ″ sin θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) + sin 2 θ ″ sin 2 ( θ + θ ′ )
= ( cos θ ″ cos ( θ + θ ′ ) − sin θ ″ sin ( θ + θ ′ ) ) 2 = ( cos ( θ ″ + θ ′ + θ ) ) 2 = cos 2 ( θ ″ + θ ′ + θ )
It is known that the Hadamard Product is known to be associative we can conclude that
φ 11 ( P [ θ ″ ] , φ 11 ( P [ θ ] , P [ θ ′ ] ) ) = φ 11 ( P [ θ ″ ] , P [ θ + θ ′ ] ) = 1 − T V e c ( P [ θ ″ ] ⊙ P [ θ + θ ′ ] ) = 1 − T V e c ( P [ θ ″ ] ⊙ ( P [ θ ′ ] ⊙ P [ θ ] ) ) = 1 − T V e c ( ( P [ θ ″ ] ⊙ P [ θ ′ ] ) ⊙ P [ θ ] ) = φ 11 ( φ 11 ( P [ θ ″ ] , P [ θ ] ) , P [ θ ] )
Next, we deal with the anti-diagonal elements. We need to show that
φ 12 ( P [ θ ″ ] , φ 12 ( P [ θ ] , P [ θ ′ ] ) ) = ( φ 12 ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )
φ 12 ( P [ θ ″ ] , φ 12 ( P [ θ ] , P [ θ ′ ] ) ) = 1 + T V e c ( P [ θ ] ⊙ Q P [ θ + θ ′ ] Q * ) = 1 + T V e c ( [ cos 2 θ ″ cos θ ″ sin θ ″ sin θ ″ cos θ ″ sin 2 θ ″ ] ⊙ [ sin ( θ + θ ′ ) cos ( θ + θ ′ ) − sin 2 ( θ + θ ′ ) cos 2 ( θ + θ ′ ) − cos ( θ + θ ′ ) sin ( θ + θ ′ ) ] ) = 1 + T V e c ( [ cos 2 θ ″ sin ( θ + θ ′ ) cos ( θ + θ ′ ) − cos θ ″ sin θ ″ sin 2 ( θ + θ ′ ) sin θ ″ cos θ ″ cos 2 ( θ + θ ′ ) − sin 2 θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) ] )
= [ 1,1,1,1 ] [ cos 2 θ ″ sin ( θ + θ ′ ) cos ( θ + θ ′ ) sin θ ″ cos θ ″ cos 2 ( θ + θ ′ ) − cos θ ″ sin θ ″ sin 2 ( θ + θ ′ ) − sin 2 θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) ] = cos 2 θ ″ sin ( θ + θ ′ ) cos ( θ + θ ′ ) + sin θ ″ cos θ ″ cos 2 ( θ + θ ′ ) − cos θ ″ sin θ ″ sin 2 ( θ + θ ′ ) − sin 2 θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ )
We now use the following the following results(without proof)
cos ( ∑ i = 1 3 ) = cos θ ′ cos θ ″ cos θ ‴ − cos θ ′ sin θ ″ sin θ ‴ − cos θ ″ sin θ ′ sin θ ‴ − cos θ ‴ sin θ ′ sin θ ″
sin ( ∑ i = 1 3 ) = sin θ ′ cos θ ″ cos θ ‴ + sin θ ″ cos θ ′ cos θ ‴ + sin θ ‴ cos θ ″ cos θ ′ − sin θ ′ sin θ ″ sin θ ‴
with some algebra, we can show that
cos ( ∑ i = 1 3 ) sin ( ∑ i = 1 3 ) = 1 + T V e c ( P [ θ ] ⊙ Q P [ θ + θ ′ ] Q * )
multiplication being commutative implies associativity for the anti-diagonal elements.
Finally, for element φ 22 we have to show that
φ 22 ( P [ θ ″ ] , φ 22 ( P [ θ ] , P [ θ ′ ] ) ) = φ 22 ( φ 22 ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )
φ 22 ( P [ θ ″ ] , φ 22 ( P [ θ ] , P [ θ ′ ] ) ) = φ 22 ( P [ θ ″ ] , P [ θ + θ ′ ] ) = 1 − T V e c ( P [ θ ″ ] ⊙ P [ θ + θ ′ ] ⊥ ) = [ 1, − 1, − 1,1 ] [ cos 2 θ ″ sin 2 ( θ + θ ′ ) − cos θ ″ sin θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) − sin θ ″ cos θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) sin 2 θ ″ cos 2 ( θ + θ ′ ) ] = cos 2 θ ″ sin 2 ( θ + θ ′ ) + 2 cos θ ″ sin θ ″ cos ( θ + θ ′ ) sin ( θ + θ ′ ) + sin 2 θ ″ cos 2 ( θ + θ ′ )
= ( cos θ ″ sin ( θ + θ ′ ) + sin θ ″ cos ( θ + θ ′ ) ) 2 = ( sin ( θ ″ + θ ′ + θ ) ) 2 = sin 2 ( θ ″ + θ ′ + θ ) = 1 − T V e c ( P [ θ ″ + θ ′ ] ⊙ P [ θ ] ⊥ ) = φ 22 ( φ 22 ( P [ θ ″ ] , P [ θ ′ ] ) , P [ θ ] )
We can see, from the previous section, that
R z T ( α + α ′ ) P [ θ ] R z ( α + α ′ ) = P [ θ + θ ′ ] = φ ( P [ θ ] , P [ θ ′ ] ) = [ φ 11 ( P [ θ ] , P [ θ ′ ] ) φ 21 ( P [ θ ] , P [ θ ′ ] ) φ 21 ( P [ θ ] , P [ θ ′ ] ) φ 22 ( P [ θ ] , P [ θ ′ ] ) ]
it is also worthwhile noting since the Hadamard Product is commutative confirms that φ ( P [ θ ] , P [ θ ′ ] ) is a commutative group operation which implies that the centre of G P ( [ θ ] ) is itself.
We have shown that the Lie Group S O ( 2, ℝ ) acts on the on the manifold G P ( [ θ ] ) to generate new elements in G P ( [ θ ] ) where many interesting properties of this group action have been demonstrated. The group operation φ ( P [ θ ] , P [ θ ′ ] ) in terms of matrix operations requires the use of the vectorisation operator and the Hadamard Product because it is not a traditional vector sum when the angles are added together. Adding the vectors in a traditional way would require the tensor product of the sum and a normalisation constant. I believe that projection matrices have more interesting structures that can be further studied. I hope that this article will raise some interest in what, I think, does deserve more investigation.
I wish to personally thank the Editor(s) and Mrs Eunice Du for all her help. I also wish to extend my gratitude to the referee for their comments.
The author declares no conflicts of interest regarding the publication of this paper.
Niglio, J.-F. (2019) A Follow-Up on Projection Theory: Theorems and Group Action. Advances in Linear Algebra & Matrix Theory, 9, 1-19. https://doi.org/10.4236/alamt.2019.91001