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There is the example 4.4.ii in “A First Course in Abstract Algebra” by Rotman.

If $A$ is a symmetric matrix with coefficients in $k$, define an inner product by $(v,w)=v^T\cdot A\cdot w$. The reader may prove that this is an inner product and that it is non-degenerate iff $A$ is non-singular.

Non-degenerate := $\forall v. (v,v)=0 \to v=0$. Rotman's definition of an inner product does not include “non-degenerate” and positive definiteness.

$v=\begin{bmatrix}0\\1\end{bmatrix} \land A=\begin{bmatrix}0&1\\1&0\end{bmatrix} \to (v,v)=0 \land v\neq 0$. $A$ is non-singular, self-inverse. Am I missing something?

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    Feel free to edit. My main reason for editing was to improve the presentation of $v$ and $A$. I will try to refrain from replacing logical connectives in your posts in the future.2011-03-23

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You're right.

What Rotman calls an inner product is sometimes called a symmetric bilinear form. Because he doesn't require nonnegativity, the given definition of nondegeneracy does not correspond to the usual definition, which is that $(w,v)=0$ for all $w$ implies $v=0$ (or something equivalent). It seems he had the latter definition in mind when making that claim, even though the given definition is only equivalent in the nonnegative case. (To see that they are equivalent in the nonnegative case you can apply the Cauchy-Schwarz inequality.) So no, you're not missing something, you were paying good attention.

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    @Didier: I see it wasn't clear. I checked Rotman's book, and the definition of nondegenerate given there is exactly as given by the OP. So the claim in question is incorrect. It could be fixed by adding that the matrix should either be positive definite or negative definite.2011-03-22