# orthogonal decomposition theorem

Let $X$ be an Hilbert space and $A\subseteq X$ a closed subspace. Then the orthogonal complement (http://planetmath.org/Complimentary) 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 (http://planetmath.org/Complete) and $A$ is closed, $A$ is a 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 (http://planetmath.org/BestApproximationInInnerProductSpaces)).

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

 $x=a_{0}+(x-a_{0})$

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

Title orthogonal decomposition theorem OrthogonalDecompositionTheorem 2013-03-22 17:32:34 2013-03-22 17:32:34 asteroid (17536) asteroid (17536) 4 asteroid (17536) Theorem msc 46A99 closed subspaces of Hilbert spaces are complemented