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Given a bilinear form $f(a,b)$, we call it degenerate if for some $a$, $f(a,-)=0$. In the finite dimensional case, we have a matrix representation and can deduce from matrix theory that if $f$ is degenerate, there is some $b$ such that $f(-,b)=0$. But in the infinite dimensional case, it does not always apply.

Question is, does the equality hold if the vector space is infinite dimensional, or is there a counterexample?

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en.wikipedia.org/wiki/Bilinear_form#Maps_to_the_dual_space says you should have both conditions. –  Mark S. Jan 11 '13 at 15:07
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up vote 2 down vote accepted

Yes, $\exists a\ne0\ f(a,−)=0$ and $\exists b\ne0 \ f(−,b)=0$ are different conditions. For example, define a bilinear form $f$ on the Hilbert space $\ell_2$ of sequences $a=(a_1,a_2,\dots)$ by $$f(a,b)=\sum_{n=1}^\infty a_{n+1}b_n$$ Then $f(a,-)=0$ when $a=(1,0,0,\dots)$. However, $f(-,b)$ is a nonzero form for every $b\ne 0$.

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