1. Notation System
In this article the notation has been kept the same as in the previous article.
•
is the Frobenius Norm.
•
is some projection operator in
.
•
is the
identity matrix.
•
is the Rank of a projection operator.
•
is the spectrum of a projection operator.
•
is the Stabilizer subgroup for some fixed element of
.
•
is an element of
for angle
.
•
is the Vectorisation Operator.
•
is the Hadamard Product.
•
is the Kronecker Product.
2. Some Important Theorems and Lemmas
From [1] [2] we can get the following results.
Lemma 1. The Frobenius morn
of a projection matrix is its trace. That is
Proof. It is well known that
Given that
is both symmetric and idempotent implies that
Hence,
therefore,
Theorem 1 The necessary and sufficient condition for
is given by
Proof. We know, from the previous article that
is constructed as follows
Theorem 2 The orthogonal complement of
is given by
where
is the
identity matrix. This is known as a Vector Rejection.
Proof. We first show that
the RHS of the equation gives us
Now, the LHS of th equation gives us the following result
since
Secondly, we show the effect these operators have on some vector
. We easily show this in the following way
We proceed as follows
Lemma 2 The product of the operators
and
is zero. We shall call
the Orthogonal Complement Operator of
which I shall denote as
.
We say that
Proof.
Lemma 3
Proof.
Theorem 3 A projection operator of dimension onto
where
then
(1)
where
(2)
The matrix
is formed by the columns of the basis vectors which span the subspaces V and W respectively.
Proof. We know that
, furthermore we also know that
.
Suppose the vector
is a basis for subspaces V i.e.
then the matrix T can be computed in the following way.
Now, taking into account the fact that
implies that
is given by
Hence,
is computed in the following way
Hence,
is given by
Hence,
is
We can see that
and are orthogonal reflection matrices.
Lemma 4 Let
be some projection matrix for some
then we have
(3)
Proof. We know that
We can now calculate its rank. The column space of
is given by
where
is the column space of
.
iff
such that
This gives us the following linear system
Given that
implies an infinite set of solutions exists.
Writing the homogeneous linear system we find the general set of vector solutions as follows
therefore
Hence, the solution vector
This implies that
.
Theorem 4 Let
such that
. Then we have
(4)
Proof.
, then we have
Lemma 5 The spectrum
of the
for some
is given by
Proof. We simply perform the following calculation
therefore we see that
Hence,
as claimed.
We can calculate the associated Eigenspaces. For
we get
Hence, it is clear that given some
its Eigenspace for
is
, that is, in this case, the Eigenspace is spanned by the following vectors
Therefore, we can say that is
is
then
.
For
we get the following Eigenvectors
In matrix form, this gives us
This gives us the following system
therefore, from (24) we get
Therefore,
Therefore, the Eigenvector is given
Lemma 6 The Eigenvectors
and
are linearly independent. Moreover,
.
Proof. For
we have
Clearly,
are linearly independent iff
This can be written matrix form in the following way
Now,
We can clearly see that, the determinant is non-zero hence,
and
are linearly independent.
We can now talk about the Diagonalizability of Projection Matrices
. Given that
has distinct eigenvalues implies that it is diagonalizable i.e.
where
is a diagonal matrix
Lemma 7
where
Proof.
Theorem 5 The projection matrix
with spectrum
is diagonalizable, hence there exists matrices
such that
(5)
where
are projection matrices onto
along
. Furthermore,
1)
whenever
.
2)
.
Proof. We know that the Eigenvalues are
this obviously implies that
This implies that
.
Lemma 8 Let
and
be the spectral projectors for the eigenvalues
and
respectively. Then we show that
1)
2)
Proof.
As claimed.
For the second part of the proof, we simply add the matrices
3. Lie Group Action of
on
From [3] [4] we have
Definition 1 Two squares matrices
and
are said to be Congruent if there exists an invertible matrix
such that
(6)
The form of the equation would tend to suggest that
such that
.
Let
be the rotation by some angle
such that
.
implies the rotation is counter-clockwise,
implies the rotation is clockwise.
Theorem 6 The action of
on
defines a congruence between two elements
and
in
i.e.
such that
(7)
Proof. A point on
can be represented as
.
Hence, we can choose two points on
,
and
.
It should be clear this action corresponds to the projector in the direction of
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
. Moreover, we know that I is the identity of
hence we get the following result
Finally, given some
the mapping
is bijective on
which implies that the mapping
is invertible and can be defined as
(8)
We can see that this is equal to
such that
Note that
is an counter-clockwise transformation of the projection operator.
Lemma 9 Let
and let
, then
(9)
Proof.
Lemma 10 Let
and
be elements of
and
respectively. Then we have
(10)
Proof. Beginning with the case where
we get the following result
Now choosing
for some
leads to
This shows us that projection matrices have a period of
.
We can now calculate the Stabilizer of
as follows.
Definition 2 The rotation group
acts on
then the Stabilizer of some element
denoted
is defined as
From the above definition it is clear that
Lemma 11
Proof. Choosing
implies
. We know that for every
s.t.
.
hence, for some
we have
This implies that
Theorem 7. Let
. Let
then
Proof.
Hence, we can conclude that
Lemma 12 The group of action of
on
not faithful but it is ∞-transitive.
Proof. for a group action to be faithful implies that for every pair of distinct elements of
there is some
such that
,
.
Consider the set
and let us choose some arbitrary projector in
. Specifically, let us consider the pair
such that
. Then we demonstrate that it is impossible to find an element of
such that the above definition is satisfied.
(11)
Since
and
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
each sequence is pairwise distinct, that is,
and
. Suppose that each sequence is chosen so that
and form two arithmetic sequences such that the quotient metric satisfies
. Then we can have the following result
we can define a refinement of the sequence in the following way
As
we have
.
where, it was shown that
is the topology
. Hence, this group action is ∞-transitive.
Lemma 13 The Kernel of
is given by
for some
.
Proof. Let
be some element of
and choose
the we have
Lemma 14 The Lie Group
has exactly one orbit.
Proof. We know that the action is transitive since for all pairs on
of the form
there exists
such that
.
We just need to show that for some fixed
we get
. It is clear that choosing
we get
Due to the fact that projector repeat every
we get see that
.
Definition 3 The Vec operator is a linear transformation which converts a matrix into a column vector. That is
defined as
Theorem 8 The group
defines a homeomorphism group on
. That is
For some
and for some translation angle
this mapping can be defined as follows
Proof.
Let
then we have
Knowing the size of the original matrix i.e. a
matrix and using the properties of the
Operator i.e. the isomorphism
we find that
4. The Group Operation of
as Matrix Products
We have already seen that the group operation on
is a follows
(12)
Hence,
.
Definition 4
is the Hadamard Product defined as follows.
Let
and
be two matrices the Hadamard product is
Theorem 9
(13)
where
where
and
. The matrices
and
are defined as
Proof.
For the anti-diagonal elements, it is clear that they are equal due to symmetry.
for the element
we get the following result
Now choosing
implies that we have
, that is we have
To be thorough let us choose
, we should expect to calculate
, that is
Last but least we will check that operation is associative that we want to make sure that
Proof. Clearly the operation is associative if all component functions are also associative. Hence we shave to show that
It is known that the Hadamard Product is known to be associative we can conclude that
Next, we deal with the anti-diagonal elements. We need to show that
We now use the following the following results(without proof)
with some algebra, we can show that
multiplication being commutative implies associativity for the anti-diagonal elements.
Finally, for element
we have to show that
We can see, from the previous section, that
it is also worthwhile noting since the Hadamard Product is commutative confirms that
is a commutative group operation which implies that the centre of
is itself.
5. In Conclusion
We have shown that the Lie Group
acts on the on the manifold
to generate new elements in
where many interesting properties of this group action have been demonstrated. The group operation
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.
Acknowledgements
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.