Could someone explain me why, given a hilbert space, its orthogonal complement is always the one-element set containing 0, i.e. $\displaystyle H^\bot=\left\{0 \right\}$
A Hilbert Space is a vector space and all vector spaces have a basis.
Let
$\displaystyle e_\lambda , \quad \lambda \in \Lambda$
$\displaystyle \Lambda$
Can be an uncountable set.
So you are looking for a vector y such that
$\displaystyle <y,e_\lambda>=0, \quad \forall \lambda$
This last statement says that y is orthogonal to every basis vector and the only vector with that property is
$\displaystyle \mathbf{0}$
This question could be interpreted two ways,
a) Something that is orthogonal to everything is zero (this is clear since if $\displaystyle V$ is any inner product space and $\displaystyle x\in V^\perp$ then $\displaystyle x\perp x$ or $\displaystyle x=0$
b) Or why given a complete orthonormal set it is orthogonal to everything.
Which is it?
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 $\displaystyle S^\bot=\left\{0 \right\}$.
So, a more general statement is true (ask if you're curious). But, as for your problem: Let $\displaystyle x\in S^\perp$. Since $\displaystyle S$ is dense there exists a sequence $\displaystyle \{x_n\}$ in $\displaystyle \mathcal{H}$ with $\displaystyle \lim x_n=x$. Now, we know that the inner product on $\displaystyle \mathcal{H}$ (by definition) is continuous in the first entry and so $\displaystyle \left\langle x,x\right\rangle=\left\langle\lim x_n,x\right\rangle=\lim\left\langle x_n,x\right\rangle=\lim 0=0$ and so $\displaystyle x=0$.