You are here
Homeorthogonal decomposition theorem
Primary tabs
orthogonal decomposition 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
$X=A\oplus A^{{\perp}}\;.$ 
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
$\langle x_{0},a\rangle=\langle\lim_{{n\rightarrow\infty}}x_{n},a\rangle=\lim_{% {n\rightarrow\infty}}\langle x_{n},a\rangle=0\;\;\;\text{for every}a\in A$ 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 $xa_{0}\in A^{{\perp}}$ (see this entry).
This allows one to write $x$ as a sum of elements in $A$ and $A^{{\perp}}$
$x=a_{0}+(xa_{0})$ which proves that
$X=A+A^{{\perp}}\;.$ Moreover, it is easy to see that
$A\cap A^{{\perp}}=\{0\}$ since if $y\in A\cap A^{{\perp}}$ then $\langle y,y\rangle=0$, which means $y=0$.
We conclude that $X=A\oplus A^{{\perp}}$. $\square$
Mathematics Subject Classification
46A99 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
Recent Activity
new correction: Error in proof of Proposition 2 by alex2907
Jun 24
new question: A good question by Ron Castillo
Jun 23
new question: A trascendental number. by Ron Castillo
Jun 19
new question: Banach lattice valued Bochner integrals by math ias