Orthogonal complement

Let $V\; =\; (\backslash R^5,\; \backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle)$ be the vector space equipped with the usual dot product $\backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle$ (thus making it an inner product space), and let $$W\; =\; \backslash \{\backslash mathbf\{u\}\; \backslash in\; V:\; \backslash mathbf\{A\}x\; =\; \backslash mathbf\{u\},\backslash \; x\backslash in\; \backslash R^2\backslash \},$$ with

$$\backslash mathbf\{A\}\; =\; \backslash begin\{pmatrix\}$$

1 & 0\\

0 & 1\\

2 & 6\\

3 & 9\\

5 & 3\\

\end{pmatrix}.

then its orthogonal complement $$W^\backslash perp\; =\; \backslash \{\backslash mathbf\{v\}\backslash in\; V:\backslash langle\; \backslash mathbf\{u\},\backslash mathbf\{v\}\backslash rangle\; =\; 0\; \backslash \; \backslash \; \backslash forall\; \backslash \; \backslash mathbf\{u\}\; \backslash in\; W\backslash \}$$ can also be defined as $$W^\backslash perp\; =\; \backslash \{\backslash mathbf\{v\}\; \backslash in\; V:\; \backslash mathbf\{\backslash tilde\{A\}\}y\; =\; \backslash mathbf\{v\},\backslash \; y\; \backslash in\; \backslash R^3\backslash \},$$ being

$$\backslash mathbf\{\backslash tilde\{A\}\}\; =$$

\begin{pmatrix}

-2 & -3 & -5 \\

-6 & -9 & -3 \\

1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{pmatrix}.

The fact that every column vector in $\backslash mathbf\{A\}$ is orthogonal to every column vector in $\backslash mathbf\{\backslash tilde\{A\}\}$ 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.

Let $V$ be a vector space over a field $\backslash mathbb\{F\}$ equipped with a bilinear form $B.$ We define $\backslash mathbf\{u\}$ to be left-orthogonal to $\backslash mathbf\{v\}$, and $\backslash mathbf\{v\}$ to be right-orthogonal to $\backslash mathbf\{u\}$, when $B(\backslash mathbf\{u\},\backslash mathbf\{v\})\; =\; 0.$ For a subset $W$ of $V,$ define the left-orthogonal complement $W^\backslash perp$ to be

$$W^\backslash perp\; =\; \backslash left\backslash \{\; \backslash mathbf\{x\}\; \backslash in\; V\; :\; B(\backslash mathbf\{x\},\; \backslash mathbf\{y\})\; =\; 0\; \backslash \; \backslash \; \backslash forall\; \backslash \; \backslash mathbf\{y\}\; \backslash in\; W\; \backslash right\backslash \}.$$

There is a corresponding definition of the right-orthogonal complement. For a reflexive bilinear form, where $B(\backslash mathbf\{u\},\backslash mathbf\{v\})\; =\; 0\; \backslash implies\; B(\backslash mathbf\{v\},\backslash mathbf\{u\})\; =\; 0\; \backslash \; \backslash \; \backslash forall\; \backslash \; \backslash mathbf\{u\}\; ,\; \backslash mathbf\{v\}\; \backslash in\; V$, the left and right complements coincide. This will be the case if $B$ 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.Adkins & Weintraub (1992) p.359

• An orthogonal complement is a subspace of $V$;
• If $X\; \backslash subseteq\; Y$ then $X^\backslash perp\; \backslash supseteq\; Y^\backslash perp$;
• The radical $V^\backslash perp$ of $V$ is a subspace of every orthogonal complement;
• $W\; \backslash subseteq\; (W^\backslash perp)^\backslash perp$;
• If $B$ is non-degenerate and $V$ is finite-dimensional, then $\backslash dim(W)+\backslash dim\; (W^\backslash perp)=\backslash dim\; (V)$.
• If $L\_1,\; \backslash ldots,\; L\_r$ are subspaces of a finite-dimensional space $V$ and $L\_*\; =\; L\_1\; \backslash cap\; \backslash cdots\; \backslash cap\; L\_r,$ then $L\_*^\backslash perp\; =\; L\_1^\backslash perp\; +\; \backslash cdots\; +\; L\_r^\backslash perp$.

{{See also|Orthogonal projection}}

This section considers orthogonal complements in an inner product space $H$.Adkins&Weintraub (1992) p.272

Two vectors $\backslash mathbf\{x\}$ and $\backslash mathbf\{y\}$ are called {{em|orthogonal}} if $\backslash langle\; \backslash mathbf\{x\},\; \backslash mathbf\{y\}\; \backslash rangle\; =\; 0$, which happens if and only if $\backslash |\; \backslash mathbf\{x\}\; \backslash |\; \backslash le\; \backslash |\; \backslash mathbf\{x\}\; +\; s\backslash mathbf\{y\}\; \backslash |\; \backslash \; \backslash forall$ scalars $s$.{{sfn|Rudin|1991|pp=306-312}}

If $C$ is any subset of an inner product space $H$ then its {{em|{{visible anchor|orthogonal complement}} in $H$}} is the vector subspace

$$\backslash begin\{align\}$$

C^\perp

:&= \{\mathbf{x} \in H : \langle \mathbf{x}, \mathbf{c} \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}

which is always a closed subset (hence, a closed vector subspace) of $H${{sfn|Rudin|1991|pp=306-312}}If $C\; =\; \backslash varnothing$ then $C^\{\backslash bot\}\; =\; H,$ which is closed in $H$ so assume $C\; \backslash neq\; \backslash varnothing.$ Let $P\; :=\; \backslash prod\_\{c\; \backslash in\; C\}\; \backslash mathbb\{F\}$ where $\backslash mathbb\{F\}$ is the underlying scalar field of $H$ and define $L\; :\; H\; \backslash to\; P$ by $L(h)\; :=\; \backslash left(\backslash langle\; h,\; c\; \backslash rangle\backslash right)\_\{c\; \backslash in\; C\},$ which is continuous because this is true of each of its coordinates $h\; \backslash mapsto\; \backslash langle\; h,\; c\; \backslash rangle.$ Then $C^\{\backslash bot\}\; =\; L^\{-1\}(0)\; =\; L^\{-1\}\backslash left(\backslash \{\; 0\; \backslash \}\backslash right)$ is closed in $H$ because $\backslash \{\; 0\; \backslash \}$ is closed in $P$ and $L\; :\; H\; \backslash to\; P$ is continuous. If $\backslash langle\; \backslash ,\backslash cdot\backslash ,,\; \backslash ,\backslash cdot\backslash ,\; \backslash rangle$ is linear in its first (respectively, its second) coordinate then $L\; :\; H\; \backslash to\; P$ is a linear map (resp. an antilinear map); either way, its kernel $\backslash operatorname\{ker\}\; L\; =\; L^\{-1\}(0)\; =\; C^\{\backslash bot\}$ is a vector subspace of $H.$ Q.E.D. that satisfies:

• $C^\{\backslash bot\}\; =\; \backslash left(\backslash operatorname\{cl\}\_H\; \backslash left(\backslash operatorname\{span\}\; C\backslash right)\backslash right)^\{\backslash bot\}$;
• $C^\{\backslash bot\}\; \backslash cap\; \backslash operatorname\{cl\}\_H\; \backslash left(\backslash operatorname\{span\}\; C\backslash right)\; =\; \backslash \{\; 0\; \backslash \}$;
• $C^\{\backslash bot\}\; \backslash cap\; \backslash left(\backslash operatorname\{span\}\; C\backslash right)\; =\; \backslash \{\; 0\; \backslash \}$;
• $C\; \backslash subseteq\; \backslash left(C^\{\backslash bot\}\backslash right)^\{\backslash bot\}$;
• $\backslash operatorname\{cl\}\_H\; \backslash left(\backslash operatorname\{span\}\; C\backslash right)\; \backslash subseteq\; \backslash left(C^\{\backslash bot\}\backslash right)^\{\backslash bot\}$.

If $C$ is a vector subspace of an inner product space $H$ then

$$C^\{\backslash bot\}\; =\; \backslash left\backslash \{\backslash mathbf\{x\}\; \backslash in\; H\; :\; \backslash |\backslash mathbf\{x\}\backslash |\; \backslash leq\; \backslash |\backslash mathbf\{x\}\; +\; \backslash mathbf\{c\}\backslash |\; \backslash \; \backslash \; \backslash forall\; \backslash \; \backslash mathbf\{c\}\; \backslash in\; C\; \backslash right\backslash \}.$$

If $C$ is a closed vector subspace of a Hilbert space $H$ then{{sfn|Rudin|1991|pp=306-312}}

$$H\; =\; C\; \backslash oplus\; C^\{\backslash bot\}\; \backslash qquad\; \backslash text\{\; and\; \}\; \backslash qquad\; \backslash left(C^\{\backslash bot\}\backslash right)^\{\backslash bot\}\; =\; C$$

where $H\; =\; C\; \backslash oplus\; C^\{\backslash bot\}$ is called the {{em|{{visible anchor|orthogonal decomposition}}}} of $H$ into $C$ and $C^\{\backslash bot\}$ and it indicates that $C$ is a complemented subspace of $H$ with complement $C^\{\backslash bot\}.$

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 $W$ is a vector subspace of a Hilbert space the orthogonal complement of the orthogonal complement of $W$ is the closure of $W,$ that is,

$$\backslash left(W^\backslash bot\backslash right)^\backslash bot\; =\; \backslash overline\; W.$$

Some other useful properties that always hold are the following. Let $H$ be a Hilbert space and let $X$ and $Y$ be linear subspaces. Then:

• $X^\backslash bot\; =\; \backslash overline\{X\}^\{\backslash bot\}$;
• if $Y\; \backslash subseteq\; X$ then $X^\backslash bot\; \backslash subseteq\; Y^\backslash bot$;
• $X\; \backslash cap\; X^\backslash bot\; =\; \backslash \{\; 0\; \backslash \}$;
• $X\; \backslash subseteq\; (X^\backslash bot)^\backslash bot$;
• if $X$ is a closed linear subspace of $H$ then $(X^\backslash bot)^\backslash bot\; =\; X$;
• if $X$ is a closed linear subspace of $H$ then $H\; =\; X\; \backslash oplus\; X^\backslash bot,$ 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.

For a finite-dimensional inner product space of dimension $n$, the orthogonal complement of a $k$-dimensional subspace is an $(n-k)$-dimensional subspace, and the double orthogonal complement is the original subspace:

$$\backslash left(W^\{\backslash bot\}\backslash right)^\{\backslash bot\}\; =\; W.$$

If $\backslash mathbf\{A\}\; \backslash in\; \backslash mathbb\{M\}\_\{mn\}$, where $\backslash mathcal\{R\}(\backslash mathbf\{A\})$, $\backslash mathcal\{C\}\; (\backslash mathbf\{A\})$, and $\backslash mathcal\{N\}\; (\backslash mathbf\{A\})$ refer to the row space, column space, and null space of $\backslash mathbf\{A\}$ (respectively), then[https://www.mathwizurd.com/linalg/2018/12/10/orthogonal-complement "Orthogonal Complement"]

$$\backslash left(\backslash mathcal\{R\}\; (\backslash mathbf\{A\})\; \backslash right)^\{\backslash bot\}\; =\; \backslash mathcal\{N\}\; (\backslash mathbf\{A\})\; \backslash qquad\; \backslash text\{\; and\; \}\; \backslash qquad\; \backslash left(\backslash mathcal\{C\}\; (\backslash mathbf\{A\})\; \backslash right)^\{\backslash bot\}\; =\; \backslash mathcal\{N\}\; (\backslash mathbf\{A\}^\{\backslash operatorname\{T\}\}).$$

There is a natural analog of this notion in general Banach spaces. In this case one defines the orthogonal complement of $W$ to be a subspace of the dual of $V$ defined similarly as the annihilator

$$W^\backslash bot\; =\; \backslash left\backslash \{\; x\backslash in\; V^*\; :\; \backslash forall\; y\backslash in\; W,\; x(y)\; =\; 0\; \backslash right\backslash \}.$$

It is always a closed subspace of $V^*$. There is also an analog of the double complement property. $W^\{\backslash perp\; \backslash perp\}$ is now a subspace of $V^\{**\}$(which is not identical to $V$). However, the reflexive spaces have a natural isomorphism $i$ between $V$ and $V^\{**\}$. In this case we have $$i\backslash overline\{W\}\; =\; W^\{\backslash perp\backslash perp\}.$$

This is a rather straightforward consequence of the Hahn–Banach theorem.

In special relativity the orthogonal complement is used to determine the simultaneous hyperplane at a point of a world line. The bilinear form $\backslash eta$ used in Minkowski space determines a pseudo-Euclidean space of events.G. D. Birkhoff (1923) Relativity and Modern Physics, pages 62,63, Harvard University Press 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.

• {{annotated link|Complemented lattice}}
• {{annotated link|Complemented subspace}}
• {{annotated link|Hilbert projection theorem}}
• {{annotated link|Orthogonal projection}}

{{reflist|group=proof}}

{{reflist}}

• {{citation|last1=Adkins|first1=William A.|last2=Weintraub|first2=Steven H.|title=Algebra: An Approach via Module Theory| series=Graduate Texts in Mathematics|volume=136|publisher=Springer-Verlag|year=1992|isbn=3-540-97839-9| zbl=0768.00003 }}
• {{Citation|last1=Halmos|first1=Paul R.|author1-link=Paul R. Halmos|title=Finite-dimensional vector spaces|series=Undergraduate Texts in Mathematics|publisher=Springer-Verlag|location=Berlin, New York|isbn=978-0-387-90093-3| year=1974|zbl=0288.15002 }}
• {{citation|first1=J.|last1=Milnor|author1-link=John Milnor| first2=D.|last2=Husemoller|title=Symmetric Bilinear Forms|series=Ergebnisse der Mathematik und ihrer Grenzgebiete|volume=73|publisher=Springer-Verlag|year=1973|isbn=3-540-06009-X| zbl=0292.10016 }}
• {{Rudin Walter Functional Analysis|edition=2}}

• Orthogonal [https://www.youtube.com/watch?v=rnNxUekd3B4&t=645s complement; Minute 9.00 in the Youtube Video]
• [http://www.khanacademy.org/math/linear-algebra/alternate-bases/othogonal-complements/v/linear-algebra-orthogonal-complements Instructional video describing orthogonal complements (Khan Academy)]

{{Functional analysis}}

{{Hilbert space}}

{{DEFAULTSORT:Orthogonal Complement}}

Category:Linear algebra

Category:Functional analysis