# Orthogonal complement of a Hilbert Space

• May 16th 2011, 03:14 PM
russel
Orthogonal complement of a Hilbert Space
Could someone explain me why, given a hilbert space, its orthogonal complement is always the one-element set containing 0, i.e. $H^\bot=\left\{0 \right\}$
• May 16th 2011, 03:28 PM
TheEmptySet
Quote:

Originally Posted by russel
Could someone explain me why, given a hilbert space, its orthogonal complement is always the one-element set containing 0, i.e. $H^\bot=\left\{0 \right\}$

A Hilbert Space is a vector space and all vector spaces have a basis.

Let

$e_\lambda , \quad \lambda \in \Lambda$

$\Lambda$

Can be an uncountable set.

So you are looking for a vector y such that

$=0, \quad \forall \lambda$

This last statement says that y is orthogonal to every basis vector and the only vector with that property is

$\mathbf{0}$
• May 16th 2011, 03:29 PM
Drexel28
Quote:

Originally Posted by russel
Could someone explain me why, given a hilbert space, its orthogonal complement is always the one-element set containing 0, i.e. $H^\bot=\left\{0 \right\}$

This question could be interpreted two ways,

a) Something that is orthogonal to everything is zero (this is clear since if $V$ is any inner product space and $x\in V^\perp$ then $x\perp x$ or $x=0$

b) Or why given a complete orthonormal set it is orthogonal to everything.

Which is it?
• May 17th 2011, 02:25 AM
russel
Drexel, I think the question (b) concerns me. I suppose I must use something that has to do with the completeness of the hilbert space. But I guess TheEmptySet's answer is ok.
For example, I need this property when I want to prove that if S is a nonempty set, dense in an inner-product space X, then $S^\bot=\left\{0 \right\}$.
• May 17th 2011, 11:18 AM
Drexel28
Quote:

Originally Posted by russel
Drexel, I think the question (b) concerns me. I suppose I must use something that has to do with the completeness of the hilbert space. But I guess TheEmptySet's answer is ok.
For example, I need this property when I want to prove that if S is a nonempty set, dense in an inner-product space X, then $S^\bot=\left\{0 \right\}$.

So, a more general statement is true (ask if you're curious). But, as for your problem: Let $x\in S^\perp$. Since $S$ is dense there exists a sequence $\{x_n\}$ in $\mathcal{H}$ with $\lim x_n=x$. Now, we know that the inner product on $\mathcal{H}$ (by definition) is continuous in the first entry and so $\left\langle x,x\right\rangle=\left\langle\lim x_n,x\right\rangle=\lim\left\langle x_n,x\right\rangle=\lim 0=0$ and so $x=0$.