Orthogonal complement explained
of a
vector space
equipped with a
bilinear form
is the set
of all vectors in
that are
orthogonal to every vector in
. Informally, it is called the
perp, short for
perpendicular complement. It is a subspace of
.
Example
Let
V=(\R5,\langle ⋅ , ⋅ \rangle)
be the vector space equipped with the usual
dot product
(thus making it an
inner product space), and let
with
then its orthogonal complement
can also be defined as
being
The fact that every column vector in
is orthogonal to every column vector in
} can be checked by direct computation. The fact that the spans of these vectors are orthogonal then follows by bilinearity of the dot product. Finally, the fact that these spaces are orthogonal complements follows from the dimension relationships given below.
General bilinear forms
Let
be a vector space over a
field
equipped with a
bilinear form
We define
to be left-orthogonal to
, and
to be right-orthogonal to
, when
For a subset
of
define the left-orthogonal complement
to be
There is a corresponding definition of the right-orthogonal complement. For a reflexive bilinear form, where
B(u,v)=0\impliesB(v,u)=0 \forall u,v\inV
, the left and right complements coincide. This will be the case if
is a
symmetric or an alternating form.
The definition extends to a bilinear form on a free module over a commutative ring, and to a sesquilinear form extended to include any free module over a commutative ring with conjugation.[1]
Properties
- An orthogonal complement is a subspace of
;
then
;
of
is a subspace of every orthogonal complement;
;
is
non-degenerate and
is finite-dimensional, then
\dim(W)+\dim(W\perp)=\dim(V)
.
are subspaces of a finite-dimensional space
and
then
.
Inner product spaces
.
[2] Two vectors
and
are called if
, which happens
if and only if
scalars
.
If
is any subset of an inner product space
then its is the vector subspace
which is always a closed subset (hence, a closed vector subspace) of
[3] that satisfies:
C\bot=\left(\operatorname{cl}H\left(\operatorname{span}C\right)\right)\bot
;
C\bot\cap\operatorname{cl}H\left(\operatorname{span}C\right)=\{0\}
;
C\bot\cap\left(\operatorname{span}C\right)=\{0\}
;
C\subseteq\left(C\bot\right)\bot
;
\operatorname{cl}H\left(\operatorname{span}C\right)=\left(C\bot\right)\bot
.
If
is a vector subspace of an inner product space
then
If
is a closed vector subspace of a Hilbert space
then
where
is called the of
into
and
and it indicates that
is a
complemented subspace of
with complement
Properties
The orthogonal complement is always closed in the metric topology. In finite-dimensional spaces, that is merely an instance of the fact that all subspaces of a vector space are closed. In infinite-dimensional Hilbert spaces, some subspaces are not closed, but all orthogonal complements are closed. If
is a vector subspace of an
inner product space the orthogonal complement of the orthogonal complement of
is the
closure of
that is,
Some other useful properties that always hold are the following. Let
be a Hilbert space and let
and
be linear subspaces. Then:
;
then
;
;
;
is a closed linear subspace of
then
;
is a closed linear subspace of
then
the (inner)
direct sum.
The orthogonal complement generalizes to the annihilator, and gives a Galois connection on subsets of the inner product space, with associated closure operator the topological closure of the span.
Finite dimensions
For a finite-dimensional inner product space of dimension
, the orthogonal complement of a
-dimensional subspace is an
-dimensional subspace, and the double orthogonal complement is the original subspace:
If
, where
,
, and
refer to the
row space,
column space, and
null space of
(respectively), then
[4] Banach spaces
There is a natural analog of this notion in general Banach spaces. In this case one defines the orthogonal complement of
to be a subspace of the
dual of
defined similarly as the annihilator
It is always a closed subspace of
. There is also an analog of the double complement property.
is now a subspace of
(which is not identical to
). However, the
reflexive spaces have a
natural isomorphism
between
and
. In this case we have
This is a rather straightforward consequence of the Hahn–Banach theorem.
Applications
In special relativity the orthogonal complement is used to determine the simultaneous hyperplane at a point of a world line. The bilinear form
used in
Minkowski space determines a
pseudo-Euclidean space of events.
[5] The origin and all events on the
light cone are self-orthogonal. When a
time event and a
space event evaluate to zero under the bilinear form, then they are
hyperbolic-orthogonal. This terminology stems from the use of
conjugate hyperbolas in the pseudo-Euclidean plane:
conjugate diameters of these hyperbolas are hyperbolic-orthogonal.
External links
Notes and References
- Adkins & Weintraub (1992) p.359
- Adkins&Weintraub (1992) p.272
- If
then
which is closed in
so assume
Let where
is the underlying scalar field of
and define
by
L(h):=\left(\langleh,c\rangle\right)c,
which is continuous because this is true of each of its coordinates h\mapsto\langleh,c\rangle.
Then C\bot=L-1(0)=L-1\left(\{0\}\right)
is closed in
because
is closed in
and
is continuous. If
is linear in its first (respectively, its second) coordinate then
is a linear map (resp. an antilinear map); either way, its kernel \operatorname{ker}L=L-1(0)=C\bot
is a vector subspace of
Q.E.D.
- https://www.mathwizurd.com/linalg/2018/12/10/orthogonal-complement "Orthogonal Complement"
- [G. D. Birkhoff]