Received 4 April 2016; accepted 16 May 2016; published 19 May 2016

1. Introduction
Two lines
and
in
are considered. When is the reflection over
followed by the reflection over
the same as the reflection over
followed by the reflection over
? It is easy to see that it is the case if and only if
or
.
When considering subspaces of
, we can ask similar questions for lines, for planes or for the mixed case of one line and one plane. Instead of addressing those cases one by one, we generalize the situation of arbitrary two linear subspaces of a vector space with an inner product.
2. Projection
Supposing that U is a vector space equipped with an inner product,
is a linear subspace of U. Given a vector
, we know from linear algebra [1] [2] that u can be decomposed uniquely as
where
is the projection of the vector u onto V and
, i.e.
.
Here are some elementary properties of the projection
:
1)
is linear.
2)
if and only if
.
3)
if and only if ![]()
4)
.
5) If V1 and V2 are subspaces of U, then
, for all
.
6) If V1, V2 and W are subspaces of U, then
.
7) If V1, V2 and W are subspaces of U, then
.
Lemma 2.1. Supposing that U is a linear space and V, W are two linear subspaces of U, if
then
.
Proof. We first show that
. Since
and
, we have
. On the other hand, if
, then
, hence
and thus
. As a result,
. The proof of
is similar. +
Suppose U is a vector space and V, W are two subspaces of U. Intersecting the identity
with V and W, we get
and
. It is obvious that these two sums are orthogonal.
Denote
and
. Using these notations,
and
.
Lemma 2.2.
if and only if
.
Poorf.
![]()
(Þ) If
, then
. On the other hand, by the fourth property of projection above,
. Similarly,
. Thus,
.
(Ü) By Lemma 2.1,
. For
,
![]()
and
, but
, we must have
, i.e.
. Similarly,
. +
Theorem 2.3. Supposing that U is a vector space and V, W are two subspaces of U, then
if and only if
.
Proof. (Þ) Assume that
. In particular,
![]()
Thus,
. Similarly,
.
(Ü) Assume
. By Lemma 2.2,
.
![]()
Similarly,
. +
3. Reflection over a Subspace
Supposing that U is a vector space equipped with an inner product,
is a subspace of U. We define the refection of
with respect to V as
![]()
The above formula can be easily derived from the observation that
. Note that if
,
then
.
Lemma 3.1. Supposing that U is a vector space and V, W are two vector subspaces of U, then
if and only if
.
Proof.
![]()
Similarly,
. Hence,
![]()
Theorem 3.2. Supposing that U is a vector space and V, W are two subspaces of U, then
if and only if
.
Poor. By Lemma 3.1,
if and only if
. By Theorem 2.3,
if and only if
. +
4. Projection onto a Translated Subspace
Define the projection of
onto a translated subspace
as
![]()
is well defined: supposing
, then
. Hence
and thus
![]()
Theorem 4.1.
if and only if
and
.
Proof.
![]()
Similarly,
.
Thus,
if and only if
![]()
(Þ) By Theorem 2.3, the first equation implies
. The second equation simply means that
.
(Ü) By Theorem 2.3, the first equation is satisifed. To show the second equation, since
, we have
, for some
and
, or
:
![]()
which is the second equation.
5. Reflection over a Translated Subspace
We next discuss the reflection over a translated subspace. Let
be a subspace. A translated subspace is
for some
. We define the reflection of
over
as
![]()
is well-defined: supposing
, then
and hence
. As a result,
![]()
Supposing
for some
is another translated subspace.
![]()
Similarly,
.
Theorem 5.1.
if and only if
and
.
Proof.
if and only if
![]()
(Þ) By Theorem 3.2,
implies
. The second equation simply means
.
(Ü) We express
and
in terms of projections:
![]()
![]()
By Theorem 3.2,
implies
. By Lemma 3.1, we also have
. To show
, it suffices to verify the second equation
![]()
Since
, we must have
for some
and
, or
:
![]()
6. Mixed Transformations
Theorem 6.1.
if and only if
and
.
Theorem 6.2.
if and only if
and
.
Theorem 6.3.
if and only if
and
.
7. Generalizations
If we denote
, the permutation group of order n, then
Theorem 7.1.
![]()
if and only if
![]()
Theorem 7.2.
![]()
if and only if
![]()
Theorem 7.3.
![]()
if and only if
and
![]()
Theorem 7.4.
![]()
if and only if
and
![]()