---- - [ ] there should be a way to unify with [[annihilator of a module]] > [!definition] Definition. ([[orthogonal complement|(left, right) kernel relative to a subspace]]) > Let $V$, $U$ be [[vector space|vector spaces]] [[field|over]] $k=\mathbb{R}$ or $\mathbb{C}$. Let $\langle -,- \rangle:V \times U \to k$ be a [[bilinear map|bilinear (or sesquilinear) form]] . Let $W \subset V$ and $Z \subset U$ be subsets. > > The **left kernel of $\langle -,- \rangle$ relative to $Z \subset U$** is the [[linear subspace]] $^{\perp}Z:=\operatorname{ker }_{L}(\langle -,- \rangle; U ):=\{ v \in V : \langle v, z \rangle=0 \text{ for all } z \in Z \}.$ > Note that if $Z$ is a [[linear subspace]] of $U$, then $^{\perp}Z=\operatorname{ker }(v \mapsto \langle v,- \rangle |_{Z})$ > The **right kernel of $\langle -,- \rangle$ relative to $W \subset V$** is the [[linear subspace]] $W^{\perp}:=\operatorname{ker }_{R}(\langle -,- \rangle; W ):=\{ w \in W: \langle u, w \rangle=0 \text{ for all } w \in W \}.$ > Note that if $W$ is a [[linear subspace|linear subspace]] of $V$, then $W^{\perp}= \operatorname{ker }(u \mapsto \langle -,u \rangle |_{W}).$ > The **left kernel of $\langle -,- \rangle$** is its kernel relative to $U$: $\operatorname{ker }_{L}(\langle -,- \rangle )= \operatorname{ker }(v \mapsto \langle v,- \rangle ) \subset V.$ > The **right kernel of $\langle -,- \rangle$** is its kernel relative to $V$: $\operatorname{ker }_{R}(\langle -,- \rangle )=\operatorname{ker }(u \mapsto \langle -,u \rangle ) \subset U.$ > We say that $\langle -,- \rangle$ **separates points** if its left and right kernels are both zero. In this case $(V, U, \langle -,- \rangle)$ is called a **dual pair**. > >When $\langle -,- \rangle$ is [[symmetric multilinear map|symmetric]] the left- and right- notions coincide and so the prefixes are dropped. > > [!basicproperties] > - Clearly, $W^{\perp}=\bigcap_{w \in W}^{}\operatorname{ker }(\langle w, - \rangle |_{W}\rangle)$ > - Hence $^{\perp}Z$ and $W^{\perp}$ are [[closed set|closed]] wrt the [[weak topology]] induced by $\langle -,- \rangle$.[^7] > - $^{\perp}(W^{\perp})=\overline{W}^{\sigma(V,U)}$ and similarly $(^{\perp}Z)^{\perp}=\overline{Z}^{\sigma(U,V)}$.[^8] > > > [!proof]- Proof. > > If $w \in W$ then of course $\langle w,u \rangle=0$ for all $u \in U$ satisfying $\langle W,u \rangle \equiv 0$, i.e., for all $u \in W^{\perp}$, hence $w \in ^{\perp}\!\!(W^{\top})$. Thus $W \subset^{\perp}\!\!(W ^{\perp})$. Since $^{\perp}(W^{\perp})$ is [[closed set|closed]] in $\sigma(V,U)$, we have $\overline{W}^{\sigma(V,U)} \subset \overline{^{\perp}(W ^{\perp})}^{\sigma(V,U)}=^{\perp}(W^{\perp})$. which completes one inclusion. > > > > The other inclusion relies on [[Hahn-Banach Extension Theorem|Hahn-Banach]] and is omitted until then. > [^7]: Indeed, by definition the maps $\{ \langle -,u \rangle |_{W}: u \in U \}$ are [[continuous]], hence their [[kernel of a linear map|kernels]] are $\sigma(U,V)$-[[closed set|closed]]. Now $W^{\perp}=\bigcap_{w \in W} \operatorname{ker }(\langle w,-\rangle |_{W})$ is closed [[closed sets behave complementarily to open sets|as an arbitrary intersection of closed sets]]. [^8]: This recovers e.g. the familiar result $(W^{\perp})^{\perp}=W$ for orthogonal complements in finite-dimensional inner product spaces. and $u \in U$ satsifies $\langle W,u \rangle \equiv 0$, then $\langle w, u \rangle$ Various special cases of this construction garner independent names and characterizations, some of which we illustrate below. > [!definition] Definition. ([[orthogonal complement|annihilator]]) > When $\langle -,- \rangle: V^{* } \times V \to \mathbb{F}$ pairs a [[vector space]] and its [[dual vector space|dual]] via evaluation $\langle \varphi, v \rangle:=\varphi(v)$, the (left-)kernel of $\langle -,- \rangle$ relative to a subset $W \subset V$ is called the **annihilator** of $W$ and denoted $W^{\circ}= \{ \varphi \in V^{*} : \varphi(W)=0 \}.$ The (right-)kernel of $\langle -,- \rangle$ relative to a subset $F \subset V^{*}$ is called the **preannihilator** of $F$ and denoted $F_{\circ}= \{ v \in V: \varphi(v)=0 \text{ for all } \varphi \in F \}.$ > >Note that if $F$ is [[dimension|finite-dimensional]] with [[basis]] $(f^{k})_{k=1}^{n}$, $F_{\circ}= \bigcap_{k=1}^{n}\operatorname{ker }f^{k}$.[^9] > Note that $W^{\circ}=\iota^{*}(\{ 0 \})$ is a closed subspace. [^9]: Feels a bit like [[nilradical equals intersection of all prime ideals]]. > [!basicproperties] > - If $F \leq V$ is a [[dimension|finite-dimensional]] [[linear subspace]], one has $(F_{\circ})^{\circ}=F$. More generally, see [[linear functional characterization of linear subspace closure|the result here]] for [[seminorm|seminormed spaces]] [^2] (though probably it holds more generally). > > > > [!proof]- Proof. > > ![[CleanShot 2025-10-23 at [email protected]]] > > #### (a) > > > > $\implies$. Suppose $T:X \to Y$ factors through $X \xrightarrow{S}Z$ via some [[linear map|linear map]] $\pi$. Then if $x \in \operatorname{ker }S$, we have $T(x)=\pi \circ S(x)=\pi(0)=0$. > > > > $\impliedby$. Suppose $\operatorname{ker }S \subset \operatorname{ker }T$. First assume $S$ is [[surjection|surjective]]. In this case, given $z \in Z$, choose any $x \in S ^{-1} (z) \subset X$. Define $\pi(z):=T(x)$. $\pi$ is [[well-defined]] because for any other element $x' \in S^{-1}(z)$, $S(x)=S(x')$ implies $S(x-x')=0$, hence $x-x' \in \operatorname{ker }S \subset \operatorname{ker }T$, so that $T(x)-T(x')=T(x-x')=0$. Hence $T(x)=T(x')$. In the case that $S$ is not [[surjective sheaf morphism|surjective]], choose an [[complement of a linear subspace|algebraic complement]] $W$ of $\operatorname{im }S$ and define $\pi: \underbrace{ \operatorname{im }S \oplus W }_{ Z } \to Y$ as before on $\operatorname{im }S$ and (e.g.) zero on $W$. $\pi$ is [[linear map|linear]] because it equals a direct sum $\hat{\pi} \oplus 0_{W \to Y}$ of linear maps. > > > > > > #### (b) > > Possibly related: [[linear subspaces as intersections of kernels of 1-forms]] > > > > $\implies$. Let $f \in \operatorname{span}\{ f^{1},\dots,f^{n} \}$, $f=\sum_{k=1}^{n}a_{k}f^{k}$ for some $a_{1},\dots,a_{n} \in \mathbb{F}$. Suppose $x \in \bigcap_{k=1}^{n} \operatorname{ker }f^{k}$, that is, $f^{k}(x)=0$ for all $k \in [n]$. Then clearly $f(x)=\sum_{k=1}^{n}a_{k}\cancel{ f^{k}(x) }^{=0}=0$. > > > > $\impliedby$. Suppose that $f,(f^{k})_{k=1}^{n}$ are [[dual vector space|linear functionals]] on $X$ satisfying $\bigcap_{k=1}^{n} \operatorname{ker }f^{k}\subset \operatorname{ker }f$, i.e., satisfying $\big(f^{k}(x)=0 \text{ for all }k=1,\dots,n\big) \implies f(x)=0.$ > > With $e_{1},\dots,e_{n}$ the standard [[module is free iff admits basis|coordinate]] [[basis]] for $\mathbb{R}^{n}$, define $\begin{align} > > S: X &\to \mathbb{R}^{n} \\ > > x & \mapsto \sum_{k=1}^{n} f^{k}(x) { e_{k} }_{ }. > > \end{align}$ > > Note that $x \in \operatorname{ker }S$ implies ($f^{k}(x)=0$ for all $k \in [n]$), which in turn implies (by hypothesis) $x \in \operatorname{ker }f$. Thus $\operatorname{ker }S \subset \operatorname{ker }f$, yielding from part $(a)$ a factorization > > > > ```tikz > > \usepackage{tikz-cd} > > \usepackage{amsmath} > > \usepackage{amsfonts} > > \begin{document} > > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZARgBoAGAXVJADcBDAGwFcYkQANEAX1PU1z5CKAEwVqdJq3YBrHnxAZseAkTLEJDFm0QgAOnoC29HAAsARueAAlbgD1CvfsqFFy4mlum6AHvOeCqijuGp5SOvp6cMyGAPrAMgC8xPZgAAQAZnYyABQ+AJRpMLFy3BIwUADm8ESgGQBOEIZI7iA4EEhkktrsGf4gDU0tNO1IYt3eIADKIDSM9OYwjAAKAirCIPVYlaY4-YPNiADMIx2IACwj9FiM7MZocKNOA42H46OIXYxYYBFQ9HBTBVZhMIgYYD4sI84GkDGgsCD5osVmtXLotjs9mVuEA > > \begin{tikzcd} > > x \arrow[d, maps to] & X \arrow[r, "f"] \arrow[d, "S"'] & k \\ > > \sum_{k=1}^n f^k(x) e_k & \mathbb{R}^n \arrow[ru, "\exists \pi"', dashed] & > > \end{tikzcd} > > \end{document} > > ``` > > Defining $a_{k}:=\pi(e_{k})$, we see that $f=\sum_{k=1}^{n} a_{k}f^{k}$, witnessing that $f \in \operatorname{span}\{ f^{1},\dots,f^{n} \}$. > > > > #### (c) > > > > Note that if $F$ is [[dimension|finite-dimensional]] with [[basis]] $(f^{k})_{k=1}^{n}$, $F_{\circ}= \bigcap_{k=1}^{n}\operatorname{ker }f^{k}$. Fix a [[basis]] $f^{1},\dots,f^{n}$ for $F$. If $\varphi \in (F_{\circ})^{\circ}$, i.e., if $\varphi(F_{\circ})=0$, then $F_{\circ} \subset \operatorname{ker }\varphi$, which by $(b)$ implies $\varphi \in \operatorname{span}(f^{1},\dots,f^{n})=F$. Thus $(F_{\circ})^{\circ} \subset F$. For the reverse inclusion, take $\varphi \in F$, and observe that if $v \in V$ is such that $f(v)=0$ for all $f \in F$, then of course $\varphi(v)=0$. > > > > > > If $W \leq V$ is an inclusion of finite-dimensional vector spaces, then $W^{*}=\frac{V^{*}}{W^{\circ}} \text{ and } \left( \frac{V}{W} \right)^{*} \cong W^{\circ}.$ See [[dual of quotient is annihilator]]. [^2]: That result relies oh [[Hahn-Banach Extension Theorem|Hahn-Banach]], so it is still nice to have the proof for finite dimensions provided in this note (part b of that proof gets used all the time as well). > [!definition] Definition. ([[orthogonal complement]]) > Suppose we have a single [[vector space]] $V$ endowed with a [[symmetric multilinear map|symmetric]] [[bilinear map|bilinear form]] $\langle -,- \rangle$, and $W$ a subset of $V$. In this case the kernel of $\langle -,- \rangle$ relative to $W$ is called the **orthogonal complement of $W$** and denoted $W^{\perp}:=\{ v \in V: \langle v,w \rangle=0 \text{ for all } w \in W \}.$ > > [[nondegenerate bilinear form|Nondegeneracy]] of $\langle -,- \rangle$ is precisely the condition $V^{\perp}=\{ 0 \}$. > Upon assuming more structure on $(V, \langle -,- \rangle)$ and $W\subset V$ we obtain more characterizations; see the equivalences and properties below. > ^definition > [!equivalence] Equivalences in special cases. > - If $W$ is a [[linear subspace]] of $V$,[^1] then $W^{\perp}=\operatorname{ker }(v \mapsto \langle v,- \rangle |_{W} ).$ > - If $\langle -,- \rangle$ is further an [[inner product]], then this induces a [[complement of a linear subspace|algebraic]] [[direct sum of vector spaces|direct sum]] $V =\overline{W} \oplus W^{\perp}$, [[linear projector iff induces direct sum decomposition|whose witnessing]] [[linear projector]] is exactly the [[orthogonal projection]] $P_{\overline{W}}$. [[first isomorphism theorem|In turn]], [[closure|we]] obtain [[natural transformation|canonical]] [[linear isomorphism|linear isomorphisms]] $V / \overline{W} \xrightarrow{\sim} W^{\perp} \text{ and } V / W^{\perp} \xrightarrow{\sim} \overline{W}.$ > - If $(V, \langle -,- \rangle)$ is further a [[Hilbert space]], then $\overline{W} \oplus W^{\perp}$ is in fact a *[[TVS direct sum|topological]]* [[direct sum of vector spaces|direct sum]], so that $P_{\overline{W}}$ is [[continuous]] and the displayed [[isomorphism|isomorphisms]] are also [[homeomorphism|homeomorphisms]] (indeed, [[Lipschitz continuous|isometries]]). > > [^1]: If $W$ is just a subset, then $v \mapsto \langle v,- \rangle_{W}$ is not a map between vector spaces, so doesn't have a [[kernel of a linear map|kernel]]. > [!basicproperties] Basic properties. > - $W^{\perp}$ is a [[linear subspace]] of $V$ (even when $W$ is not) > - If $U \subset W$ then $W^{\perp} \subset U^{\perp}$ > - $W \subset (W^{\perp})^{\perp}$ ^properties > [!basicproperties] Properties for when $V$ is an [[inner product space]]. > Assume $\langle -,- \rangle$ is an [[inner product]].[^2] > - $W^{\perp}$ is [[closed set|closed]] as a [[linear subspace]] of $V$ > - $\overline{W}^{\perp}=W^{\perp}$ ^properties [^2]: Can probably relax this a bit, mostly just need the inducing of a [[well-defined]] [[topological space|topology]]. > [!basicproperties] Properties for when $V$ is a [[Hilbert space]]. > Assume $V$ is a [[Hilbert space]]. > - $\overline{W}=(W^{\perp})^{\perp}$ > - *([[dense|on denseness]])* If $W$ is a [[linear subspace]] of $V$, then $\overline{W}=V \text{ if and only if }W^{\perp}=\{ 0 \}.$ ^properties ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` # Definition If $U \subset V$, then the **orthogonal complement** of $U$, denoted $U^\perp$, is the [[linear subspace]] of all [[vector]]s in $V$ that are [[orthogonal]] to every [[vector]] in $U$: $U^\perp = \{ v \in V : \langle v,u \rangle = 0 \text{ for every } u \in U \}.$ # Justification - We should show that $U^{\perp}$ is indeed a [[linear subspace]]. We'll show [[linear subspace#Equivalence]]. [[inner product#Basic Properties#2|Clearly]] $0 \in U^{ \perp}$. Suppose $v,w \in u^{\perp}$ and $u \in U$. We have $\begin{align} \langle v+w, u \rangle = \langle v,u \rangle + \langle w,u \rangle = 0 + 0 =0 \end{align}$ hence $v+w \in U^{\perp}$. Finally, let $\lambda \in \ff$. We have $\langle \lambda v,u \rangle =\lambda \langle v,u \rangle= \lambda 0=0$. $\qedin$ # Properties - [[orthogonal complement#Basic Properties of orthogonal complement]] - [[direct sum of a subspace and its orthogonal complement]] - [[dimension of the orthogonal complement]] - [[the orthogonal complement of the orthogonal complement]] ## Basic Properties of [[orthogonal complement]] ###### 1 - $\{ 0 \}^{\perp}=V$. ###### 2 - $V^{\perp}=\{ 0 \}$. ###### 3 $U \subset V \implies U \cap U^{\perp} \subset \{ 0 \}$. ###### 4 $U \subset W \subset V \implies W^{\perp} \subset U^{\perp}$. ## Proof of [[orthogonal complement#Basic Properties of orthogonal complement|Basic Properties]] ###### [[orthogonal complement#1|1]] $0$ is [[orthogonal]] to all $v \in V$ by [[inner product#Basic Properties]]. ###### [[orthogonal complement#2|2]] $\{ 0 \} \subset V^\perp$ because $V^{\perp}$ [[orthogonal complement#Justification|is a subspace]]. Now let $v \in V^{\perp}$. We have $\langle v,v \rangle=0$, implying $\|v\|=0$, which by [[norm]] requirements enforces $v=0$. ###### [[orthogonal complement#3|3]] Suppose $U \subset V$ and let $u \in U$, $v \in U^{\perp}$. We want to show $u=v \iff u=v=0$. Since we have $\langle u,v \rangle=0$, this result is immediate from requirements of the [[norm]]. ###### [[orthogonal complement#4|4]] Suppose $U \subset W \subset V$ and $U \subset W$. Let $v \in W^{\perp}$. Then $\langle v,u \rangle = 0$ for all $u \in U$ (since $u \in W$). Hence $v \in U^{\perp}$. $\qedin$