# the non-singular matrix of an inner product

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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There is an errata file that doesn't seem to have this, but it also doesn't seem to have been updated for a long time: math.uiuc.edu/~rotman/errata.pdf – Jonas Meyer Mar 22 '11 at 18:50
Actually, those errata are for the 2nd edition, but this is still in the 3rd edition. I can't find errata for the 3rd edition. – Jonas Meyer Mar 22 '11 at 20:51
@Jonas Meyer: Thanks, I completely forgot that errata lists are sometimes published online. Could you please do not replace logical connectives with English words, I find logical connectives easier to read? – beroal Mar 23 '11 at 8:53
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. – Jonas Meyer Mar 23 '11 at 14:22

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.
In the OP example, $(w,v)=0$ for all $w$ does imply $v=0$. – Did Mar 22 '11 at 18:47