Thread: 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.

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.

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

Let





Can be an uncountable set.

So you are looking for a vector y such that



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

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.

This question could be interpreted two ways,

a) Something that is orthogonal to everything is zero (this is clear since if is any inner product space and then or


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 .

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 .

So, a more general statement is true (ask if you're curious). But, as for your problem: Let . Since is dense there exists a sequence in with . Now, we know that the inner product on (by definition) is continuous in the first entry and so and so .