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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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    @QiaochuYuan: can you please make your last comment an answer so that I could accept it?2011-10-02

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Let $f:V\otimes V\to k$ be a bilinear form, and suppose that $k$ is a field and that $V$ is finite dimensional.

Define $f_l$, $f_r:V\to V^*$ by setting $f_l(v)(w)=f(v\otimes w)$ and $f_r(v)(w)=f(w\otimes v)$ for all $v$, $w\in V$. Suppose moreover that $f_l$ is not injective, so that there exists a $v_0\in V\setminus0$ such that $f_l(v_0)=0$, that is, for all $w\in V$ we have $f(v_0\otimes w)=0$. In other words, for all $w\in V$ we have that $\langle v_0\rangle\subseteq\ker f_r(w)$.

Let $\bar V=V/\langle v_0\rangle$ and let $\pi:V\to\bar V$ be the quotient map.

This implies that for each $w\in V$ there is a map $g(w)\in\bar V^*$ such that $f_r(w)=g(w)\circ\pi$. One easily sees that $g:V\to\bar V^*$ is linear, so we have a factorization $f_r=\pi^t\circ g$ where $\pi^t:\bar V^*\to V^*$ is the transposed of $\pi$. Since the rank of $g$ is at most $\dim V-1$, the rank of $f_r$ is almost $\dim V-1<\dim V^*$. It follows that $f_r$ is not surjective.

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    Sorry. I added the clarification.2011-02-21