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Thread: Orthogonal complement of a Hilbert Space

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    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. $\displaystyle H^\bot=\left\{0 \right\}$
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    Quote Originally Posted by russel View Post
    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

    $\displaystyle \mathbf{0}$
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by russel View Post
    Could someone explain me why, given a hilbert space, its orthogonal complement is always the one-element set containing 0, i.e. $\displaystyle H^\bot=\left\{0 \right\}$
    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?
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    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 $\displaystyle S^\bot=\left\{0 \right\}$.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by russel View Post
    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 $\displaystyle x=0$.
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