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Definition of scalar product: if $g:V \times V\to\Bbb{R}$ is nondegenerate symmetric bilinear form, $g$ is a scalar product on $V$ vector space.
Here is my question If $g$ scalar product is indefinite there is always a degenerate subspace of $V$.

How can I prove this? How do I have to start with?

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2 Answers 2

Since $g$ is indefinite, there exists a non-zero $x$ such that $g(x,x)=0$. Consider the subspace generated by $x$.

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Put $M = \{x\in V| g(x,x) = 0\}.$ Since $x\mapsto g(x,x)$ is a seminorm on $V$, $M$ is a subspace of $V$. You can use $g$ to define an inner product on $V/M$.

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I don't understand this. $M$ does not have to be a subspace, e.g., if we have a basis $\{e,f\}$ with $g(e,e) = 1, g(f,f) = -1, g(e,f)=g(f,e) = 0$ then $e+f$ and $e-f$ are in $M$ but their sum is not. –  Ted Jan 5 '13 at 23:34
    
Are you allowing degneracy or calling off positive-semidefinitness altogether? –  ncmathsadist Jan 6 '13 at 0:54
    
The OP assumes $g$ is indefinite. The scalar product I defined is indefinite and non-degenerate. –  Ted Jan 6 '13 at 1:14
    
My answer only applies if the form is nonnegative definite. Otherwise, I don't think the zero-set of $x\mapsto g(x,x)$ will be a subspace. –  ncmathsadist Jan 6 '13 at 1:17

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