---- > [!definition] Definition. ([[complement of a linear subspace]]) > Let $X$ be a [[vector space]] and let $Y$ be a [[linear subspace]] of $X$. > We say that $Y$ is **(algebraically) complemented in $X$** if there exists a further [[linear subspace]] $Z$ of $X$ such that $X= Y\oplus Z$ as a [[direct sum of vector spaces|(algebraic) direct sum]]. In this case $Z$ is said to be an **algebraic complement of $Y$**. It follows from [[Zorn's lemma]] that every [[linear subspace]] is algebraically complemented ([[every vector space has a basis|complete a basis]]). > If $X$ is in fact a [[topological vector space]], we say $Y$ is **(topologically) complemented** in $X$ if there exists a further [[linear subspace]] $Z$ of $X$ such that $X=Y \oplus Z$ as a [[TVS direct sum|topological direct sum]]. We call $Z$ a **(topological) complement of $Y$**. > A complement $Z$ is in general not unique. > > > [!specialization] > If $X$ is a [[dimension|finite-dimensional]] [[topological vector space|TVS]] the two notions coincide.[^7] ^specialization [^7]: Since all the linear maps characterizing the [[TVS direct sum|topological direct sum]], being [[characterizing continuity of linear maps|being]] [[linear map|linear maps]] with [[dimension|finite-dimensional]] source, are automatically [[continuous]]. >[!equivalence] Equivalence for algebraic complements. > The space $Y$ is algebraically complemented if and only if $Y=\operatorname{im }P$ for some [[linear projector|linear]] [[projection]] $P \in \operatorname{End}X$; [[linear projector iff induces direct sum decomposition|see here]]. ^equivalence > [!equivalence] Equivalence for topological complements. > The space $Y$ is topologically complemented if and only if $Y=\operatorname{im }P$ for some [[operator norm|continuous]] [[projection]] $P \in \operatorname{End}X$. > > > [!proof]- Proof. > > This follows from the final equivalence in [[TVS direct sum|topological direct sum]]. To be (too) explicit: > > > > If $Y \leq X$ is topologically complemented, then there exists $Z$ such that $X=Y \oplus Z$ as an algebraic direct sum, such that the projections $p_{Y}$, $p_{Z}$ defined in [[TVS direct sum|topological direct sum]] are bounded. We see $Y=\operatorname{im }(y+z \xmapsto{p_{Y}} y)$ is indeed the image of a bounded projection $P=p_{Y}$. > > > > Conversely, if $Y=\operatorname{im }P$ for some bounded projection $P \in \operatorname{End}X$, then we already know there exists $Z$ such that $X=Y \oplus Z$; we just need to show that the map $(+):Y \oplus Z \to Y+Z$ given by $(+)(y,z)=y+z$ has continuous inverse. This follows from the formula $(+)^{-1} x=\big( Px, (I-P)x \big)$ > > and continuity of $P$ and $I-P$. > > >>(For Banach spaces lots of alternative proofs. E.g. $X=\operatorname{im }P \oplus \operatorname{ker }P$ has closed factors since $P$ is a [[projections are open maps|closed map]] and $\operatorname{ker }P=P ^{-1}(\text{closed set})$, then use the equivalence [[direct sum of Banach spaces|here]]. Or just directly check $(+)^{-1}$ is bounded.) > [!basicproperties] > > - It follows from [[Zorn's lemma]] that every every [[linear subspace]] is algebraically complemented. > - If $X$ is a [[T1 Axiom|T1]] (e.g. [[Hausdorff space|Hausdorff]]) [[topological vector space|TVS]] (e.g. [[Banach space]]), then any topologically complemented subspace $Y$ is [[closed set|closed]], as is any topological complement $Z$. > [!basicexample] > - Finite-dimensional subspaces of [[norm|normed spaces]] (more generally, [[locally convex space|locally convex spaces]] where [[Hahn-Banach Extension Theorem|HBE]] applies) are always topologically complemented. > > > [!proof]- Proof. > > > > Let $Y \leq X$, $\text{dim }Y=n<\infty$. > > > > Let $\{ y_{i} \}_{i=1}^{n}$ be a [[basis]] for $Y$ and consider the corresponding coordinate functionals $\{ y^{i}:Y \to \mathbb{F} \}_{i=1}^{n}$, which are [[continuous]] (being maps between finite dimensions). Using [[Hahn-Banach Extension Theorem|HBE]], extend the $y^{i}$ to continuous linear functionals $\tilde{y}^{i}:X \to \mathbb{F}$. The linear projection onto $Y$ given by $\begin{align} > > P: X & \to X \\ > > x & \mapsto \widetilde{y}^{i}(x) \, y_{i} , > > \end{align}$ > > which is continuous as the sum of continuous functions, witnesses the result. > > > > Closed finite-codimensional subspaces $Y \leq X$ are always topologically complemented. Indeed, obtain an algebraic direct sum $X=Y \oplus Z$; is the projection $P:X \twoheadrightarrow Z$ continuous? Yes: its kernel is $Y$ [[characterization of quotienting a (semi)normed vector space|and]] $P$ factors as $P: X \xrightarrow{ \pi}\frac{X}{Y} \xrightarrow{\widetilde{P}}Z$ where $\widetilde{P}$ is an isomorphism. Now $\pi$ is [[continuous]] and $\widetilde{P}$ is too: since $\text{dim }X / Y < \infty$, $\text{dim }Z<\infty$ and so $\widetilde{P}$ is continuous as a linear map between finite-dimensional spaces. > [!proof]- Proof. > > As a complemented subspace, $Y=\operatorname{im }P$ for some [[continuous]] $P:X \to X$ satisfying $P^{2}=P$. Then $Y=\operatorname{im }P=\operatorname{ker }(I-P)$. Since $I-P$ is [[continuous]] and $\{ 0 \}$ is [[closed set|closed]] (since $X$ is $T_{1}$), $\operatorname{ker }(I-P)=(I-P)^{-1}(\{ 0 \})$ is [[closed set|closed]] as a subset of $X$. Hence $\operatorname{im }P=Y$ is also. Similarly, the complement $Z$ of $Y$ equals $\operatorname{ker }P=P ^{-1}(\{ 0 \})$ and is therefore closed since $P$ is [[continuous]]. > > Simpler: assume $Y$ topologically complemented, so $X=Y \oplus Z$ (internal) with $(+):Y \times Z \to Y+Z$ a [[homeomorphism]], then $Y,Z$ are closed as images of closed sets $Y \times \{ 0 \}$, $\{ 0 \} \times Z$ (since $X$ is $T_{1}$) in $Y \times Z$ under the [[homeomorphism]] $(+)$. > (^ this is the same as first part of argument in the equivalence in [[direct sum of Banach spaces|Banach direct sum]]) - If $X$ is a [[Hilbert space]], the [[Hilbert projection theorem]] implies that *any* [[closed set|closed]] [[linear subspace]] is topologically complemented, and the witnessing [[projection]] can even be taken to be unit-[[norm]] ---- #### ---- #### 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 > ```