Orthogonal complement: Difference between revisions
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\end{pmatrix}.</math> | \end{pmatrix}.</math> | ||
The fact that every column vector in <math>\mathbf{A}</math> is orthogonal to every column vector in <math>\mathbf{\tilde{A}}</math> 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. | The fact that every [[Row and column vectors|column vector]] in <math>\mathbf{A}</math> is orthogonal to every column vector in <math>\mathbf{\tilde{A}}</math> 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 == | == General bilinear forms == | ||
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{{See also|Orthogonal projection}} | {{See also|Orthogonal projection}} | ||
This section considers orthogonal complements in an [[inner product space]] <math>H</math>.<ref>Adkins&Weintraub (1992) p.272</ref> | This section considers orthogonal complements in an [[inner product space]] <math>H</math>.<ref>Adkins & Weintraub (1992) p.272</ref> | ||
Two vectors <math>\mathbf{x}</math> and <math>\mathbf{y}</math> are called {{em|[[Orthogonal vectors (inner product space)|orthogonal]]}} if <math>\langle \mathbf{x}, \mathbf{y} \rangle = 0</math>, which happens [[if and only if]] <math>\| \mathbf{x} \| \le \| \mathbf{x} + s\mathbf{y} \| \ \forall</math> scalars <math>s</math>.{{sfn|Rudin|1991|pp=306-312}} | Two vectors <math>\mathbf{x}</math> and <math>\mathbf{y}</math> are called {{em|[[Orthogonal vectors (inner product space)|orthogonal]]}} if <math>\langle \mathbf{x}, \mathbf{y} \rangle = 0</math>, which happens [[if and only if]] <math>\| \mathbf{x} \| \le \| \mathbf{x} + s\mathbf{y} \| \ \forall</math> scalars <math>s</math>.{{sfn|Rudin|1991|pp=306-312}} | ||
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&= \{\mathbf{x} \in H : \langle \mathbf{c}, \mathbf{x} \rangle = 0 \ \ \forall \ \mathbf{c} \in C\} | &= \{\mathbf{x} \in H : \langle \mathbf{c}, \mathbf{x} \rangle = 0 \ \ \forall \ \mathbf{c} \in C\} | ||
\end{align}</math> | \end{align}</math> | ||
which is always a closed subset (hence, a closed vector subspace) of <math>H</math>{{sfn|Rudin|1991|pp=306-312}}<ref group="proof">If <math>C = \varnothing</math> then <math>C^{\bot} = H,</math> which is closed in <math>H</math> so assume <math>C \neq \varnothing.</math> Let <math display="inline">P := \prod_{c \in C} \mathbb{F}</math> where <math>\mathbb{F}</math> is the underlying scalar field of <math>H</math> and define <math>L : H \to P</math> by <math>L(h) := \left(\langle h, c \rangle\right)_{c \in C},</math> which is continuous because this is true of each of its coordinates <math>h \mapsto \langle h, c \rangle.</math> Then <math>C^{\bot} = L^{-1}(0) = L^{-1}\left(\{ 0 \}\right)</math> is closed in <math>H</math> because <math>\{ 0 \}</math> is closed in <math>P</math> and <math>L : H \to P</math> is continuous. If <math>\langle \,\cdot\,, \,\cdot\, \rangle</math> is linear in its first (respectively, its second) coordinate then <math>L : H \to P</math> is a [[linear map]] (resp. an [[antilinear map]]); either way, its kernel <math>\operatorname{ker} L = L^{-1}(0) = C^{\bot}</math> is a vector subspace of <math>H.</math> [[Q.E.D.]]</ref> that satisfies: | which is always a [[Closed set|closed subset]] (hence, a closed vector subspace) of <math>H</math>{{sfn|Rudin|1991|pp=306-312}}<ref group="proof">If <math>C = \varnothing</math> then <math>C^{\bot} = H,</math> which is closed in <math>H</math> so assume <math>C \neq \varnothing.</math> Let <math display="inline">P := \prod_{c \in C} \mathbb{F}</math> where <math>\mathbb{F}</math> is the underlying scalar field of <math>H</math> and define <math>L : H \to P</math> by <math>L(h) := \left(\langle h, c \rangle\right)_{c \in C},</math> which is continuous because this is true of each of its coordinates <math>h \mapsto \langle h, c \rangle.</math> Then <math>C^{\bot} = L^{-1}(0) = L^{-1}\left(\{ 0 \}\right)</math> is closed in <math>H</math> because <math>\{ 0 \}</math> is closed in <math>P</math> and <math>L : H \to P</math> is continuous. If <math>\langle \,\cdot\,, \,\cdot\, \rangle</math> is linear in its first (respectively, its second) coordinate then <math>L : H \to P</math> is a [[linear map]] (resp. an [[antilinear map]]); either way, its kernel <math>\operatorname{ker} L = L^{-1}(0) = C^{\bot}</math> is a vector subspace of <math>H.</math> [[Q.E.D.]]</ref> that satisfies: | ||
* <math>C^{\bot} = \left(\operatorname{cl}_H \left(\operatorname{span} C\right)\right)^{\bot}</math>; | * <math>C^{\bot} = \left(\operatorname{cl}_H \left(\operatorname{span} C\right)\right)^{\bot}</math>; | ||
* <math>C^{\bot} \cap \operatorname{cl}_H \left(\operatorname{span} C\right) = \{ 0 \}</math>; | * <math>C^{\bot} \cap \operatorname{cl}_H \left(\operatorname{span} C\right) = \{ 0 \}</math>; | ||
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=== Properties === | === 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 space]]s, some subspaces are not closed, but all orthogonal complements are closed. If <math>W</math> is a vector subspace of a [[Hilbert space]] the orthogonal complement of the orthogonal complement of <math>W</math> is the [[Closure (topology)|closure]] of <math>W,</math> that is, | The orthogonal complement is always closed in the [[Metric space|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 space]]s, some subspaces are not closed, but all orthogonal complements are closed. If <math>W</math> is a vector subspace of a [[Hilbert space]] the orthogonal complement of the orthogonal complement of <math>W</math> is the [[Closure (topology)|closure]] of <math>W,</math> that is, | ||
<math display="block">\left(W^\bot\right)^\bot = \overline W.</math> | <math display="block">\left(W^\bot\right)^\bot = \overline W.</math> | ||
Latest revision as of 04:07, 3 November 2025
Template:Short description In the mathematical fields of linear algebra and functional analysis, the orthogonal complement of a subspace 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 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 , 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 ;
- If then ;
- The radical of is a subspace of every orthogonal complement;
- ;
- If is non-degenerate and is finite-dimensional, then .
- If are subspaces of a finite-dimensional space and then .
Inner product spaces
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This section considers orthogonal complements in an inner product space .[2]
Two vectors and are called Template:Em if , which happens if and only if scalars .Template:Sfn
If is any subset of an inner product space then its Template:Em is the vector subspace which is always a closed subset (hence, a closed vector subspace) of Template:Sfn[proof 1] that satisfies:
- ;
- ;
- ;
- ;
- .
If is a vector subspace of an inner product space then If is a closed vector subspace of a Hilbert space thenTemplate:Sfn where is called the Template:Em 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 a Hilbert 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:
- ;
- if then ;
- ;
- ;
- if is a closed linear subspace of then ;
- if 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[3]
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.[4] 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.
See also
Notes
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- ↑ If then which is closed in so assume Let where is the underlying scalar field of and define by which is continuous because this is true of each of its coordinates Then 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 is a vector subspace of Q.E.D.
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References
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- ↑ Adkins & Weintraub (1992) p.359
- ↑ Adkins & Weintraub (1992) p.272
- ↑ "Orthogonal Complement"
- ↑ G. D. Birkhoff (1923) Relativity and Modern Physics, pages 62,63, Harvard University Press
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Bibliography
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- Template:Rudin Walter Functional Analysis