Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer
    
In the OP example, $(w,v)=0$ for all $w$ does imply $v=0$. –  Did Mar 22 '11 at 18:47
    
@Didier: Yes, and with the usual definition of nondegenerate (not the one Rotman gives) the claim would be true. –  Jonas Meyer Mar 22 '11 at 18:52
    
Failing to have Rotman's book at hand, I shall follow you on this... Thanks. –  Did Mar 22 '11 at 18:57
    
@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. –  Jonas Meyer Mar 22 '11 at 19:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.