PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (194)
Orphanage
Unclass'd
Unproven (498)
Corrections (25)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: High Entry average rating: No information on entry rating
[parent] orthogonal decomposition theorem (Theorem)

Theorem - Let $ X$ be an Hilbert space and $ A \subseteq X$ a closed subspace. Then the orthogonal complement of $ A$, denoted $ A^{\perp}$, is a topological complement of $ A$. That means $ A^{\perp}$ is closed and

$\displaystyle X =A \oplus A^{\perp} \;. $

Proof :

  • $ A^{\perp}$ is closed :

    This follows easily from the continuity of the inner product. If a sequence $ (x_n)$ of elements in $ A^{\perp}$ converges to an element $ x_0 \in X$, then

    $\displaystyle \langle x_0, a \rangle = \langle \lim_{n \rightarrow \infty} x_n, a \rangle = \lim_{n \rightarrow \infty}\langle x_n, a \rangle = 0\;\;\;$   for every$\displaystyle a \in A $
    which implies that $ x_0 \in A^{\perp}$.
  • $ X=A \oplus A^{\perp}$ :

    Since $ X$ is complete and $ A$ is closed, $ A$ is a complete subspace of $ X$. Therefore, for every $ x \in X$, there exists a best approximation of $ x$ in $ A$, which we denote by $ a_0 \in A$, that satisfies $ x-a_0 \in A^{\perp}$ (see this entry).

    This allows one to write $ x$ as a sum of elements in $ A$ and $ A^{\perp}$

    $\displaystyle x= a_0 + (x-a_0) $
    which proves that
    $\displaystyle X= A + A^{\perp} \; . $

    Moreover, it is easy to see that

    $\displaystyle A \cap A^{\perp} = \{0\} $
    since if $ y \in A \cap A^{\perp}$ then $ \langle y, y \rangle = 0$, which means $ y=0$.

    We conclude that $ X=A \oplus A^{\perp}$. $ \square$



Anyone with an account can edit this entry. Please help improve it!

"orthogonal decomposition theorem" is owned by asteroid.
(view preamble)

View style:

Other names:  closed subspaces of Hilbert spaces are complemented

This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: easy to see, sum, best approximation, implies, converges, sequence, inner product, proof, topological complement, subspace, closed, Hilbert space
There are 3 references to this entry.

This is version 1 of orthogonal decomposition theorem, born on 2007-09-15.
Object id is 9942, canonical name is OrthogonalDecompositionTheorem.
Accessed 988 times total.

Classification:
AMS MSC: 46A99 (Functional analysis :: Topological linear spaces and related structures :: Miscellaneous)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | prove | add result | add corollary | add example | add (any)