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In Wikipedia, bilinear mapping to the underlying field $f \in \operatorname{Hom}(V \otimes V, k)$ is defined as non-degenerate iff $X \mapsto f(-,X)$ is an isomorphism to the continuous (?) dual.

I tried to prove the proposition that $X \mapsto f(-,X)$ is an iso iff $X \mapsto f(X,-)$ is an iso, but only managed to prove that this proposition is equivalent to the proposition that $\varphi := \lbrace(f(-,X), f(X,-))\rbrace$ is an automorphism of $V^*$ (and a mapping, of course).

Is the proposition true? I assume it's likely because there is no mention of left vs. right degeneracy on the wiki.

I'm asking because I'm trying to study orthogonal complements in the most general setting, so I wanted to see what conditions on $f$ are required for it to define complement, when $V \cong U \oplus U^\perp$ and when left and right complements coincide.

EDIT: I am specifically concerned about the infinite-dimensional case. For the finite-dimensional case, the proposition is easy to prove using a variety of strategies (e.g. via determinants).

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    Wikipedia says "If either of B1 or B2 is an isomorphism, then both are, and the bilinear form B is said to be nondegenerate." It gives a proof only in the finite-dimensional case though.2011-02-21
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    Yeah, finite case is easy, I proved it yesterday without any trouble!2011-02-21
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    In the finite-dimensional case, the property that $f(-,X) = 0$ iff $X = 0$ is a criterion, and using $\varphi$ one can easily transport it to $f(X,-)$. In infinite-dimensional case, however, this is not a sufficient condition :(2011-02-21
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    @Alexei: is this question about topological vector spaces or just about vector spaces? If you are only studying the algebraic dual, then an infinite-dimensional vector space is _never_ isomorphic to its algebraic dual (given the axiom of choice).2011-02-24
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    @Qiaochu : I'm aware of that, and to be honest I'm not sure about it at all. Non-degeneracy has to make some sense in infinite-dimensional case, right? So it seems like a natural concession to only talk about topological vector spaces and continuous $f$.2011-02-24
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    @Alexei: non-degeneracy still makes sense as a property of a bilinear map on $V \times W$ for $V, W$ two vector spaces, topological or otherwise. In this generality the statement you're asking about is false.2011-02-24
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    @QiaochuYuan: can you please make your last comment an answer so that I could accept it?2011-10-02

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