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Let $V$ be an infinite dimensional vector space and $A$ be a subspace of $V$. Is there always an orthogonal complement of $A$ in $V$? If not, is there a counter-example? Thank you very much.

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    What is an orthogonal complement? To have orthogonal complements, you need to have the notion of orthogonality of vectors, i.e. you need some kind of additional structure. An inner product, for example. Otherwise the notion is simply undefined.2012-11-19
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    Perhaps there is a linear transformation you forgot to tell us about?2012-11-19
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    First of all, you need to fix a bilinear form $B$ to get a notion of orthogonality. If this is done, the set of vectors $x$ in $V$ such that $B(x,a)=0$ for all $a$ in $A$ is a subspace of $V$ and is _defined_ to be the orthogonal complement of $A$ (w.r.t. $B$). I am confused about what your question is exactly.2012-11-19
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    Maybe you are actually interested in complements as defined [here](http://en.citizendium.org/wiki/Complement_%28linear_algebra%29), i.e. in the existence of a subspace $B$ such that $V=A \oplus B$?2012-11-19
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    If A=V, then the orthogonal complement is just the 0 vector.2012-11-19
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    @Dan, thank you very much. A vector space always has a Hamel basis. Could we define orthogonal complement for every vector space?2012-11-19
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    @Adeel, thank you very much. Could we define orthogonal complement for every vector space?2012-11-19
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    @user9791 The question as it is sounds too vague. If you want a standard widely accepted notion - then to the best of my knowledge it doesn't exist. All the notions of orthogonality that I've encountered are defined for vector spaces with additional structure, not for general vector spaces. If you want to experiment and try to invent a new kind of orthogonal complements, then I strongly suspect that it will be impossible to make them "natural". This can probably be proved rigorously by [abstract nonsense](http://en.wikipedia.org/wiki/Category_theory#Natural_transformation).2012-11-19
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    So you are interested in the existence of a (non degenerate) bilinear form on $V$? The set of bilinear forms forms a vector space which can be identified with the space of linear maps from $V$ to $V^*$; under this identification, non degenerate bilinear forms correspond to isomorphisms, so if $V \cong V^*$ then there is a nondegenerate form. If we don't require nondegenerateness then the map $A \mapsto A^\perp$ is not necessarily a bijection.2012-11-19
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    Once you fix a basis $(b_i)$ (like a Hammel basis), you can get a bilinear form, hence an orthogonality concept, simply define $\langle b_i,b_j\rangle := \delta_{ij}$.2012-11-19

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