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If b is an indefinite symmetric bilinear form is it nondegenerate?
And conversely if b is nondegenerate is it positive/negative definite or indefinite?
How can i start to prove this? Note:Edited and changed the question

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A bilinear form is continuous, so nondegenerate $\leftrightarrow$ positive or negative definite. EDIT:I assumed finite dimensions. –  tst Jan 5 '13 at 14:09
    
But I don't use continuity here.I have a b bilinear form on V (that is finite dimensional real vector space ). –  Serkan Hafız Yaray Jan 5 '13 at 14:21
    
I don't think that a bilinear form in finite dimensions can be discontinuous. –  tst Jan 5 '13 at 14:27
    
The point is that a bilinear form in FD is actually a matrix, let it be $B$. Let $b$ be the largest entry of the matrix and $B_m$ the matrix with all entries equal to $b$. Then obviously $||B||\le ||B_m||$ but we already know that $||B_m||$ is finite, so $||B||$ is finite also, so $B$ is bounded thus continuous. –  tst Jan 5 '13 at 14:31
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The examples of the zero matrix (indefinite but degenerate), the identity matrix (nondegenerate and positive definite) and the diagonal matrix $(1,-1)$ (nondegenerate and indefinite) resolve both questions in the negative. –  whuber Jan 5 '13 at 18:21

1 Answer 1

$b$ is indefinite iff its matrix representation $B$ has both positive and negative eigenvalues; $b$ is nondegenerate iff $B$ has no zero eigenvalues. So,

  • an indefinite $b$ can be degenerate (e.g. $B=\operatorname{diag}(1,0,-1)$);
  • a nondegenerate $b$ can be positive definite, negative definite or indefinite (examples: consider $B=1,\ B=-1$ and $B=\operatorname{diag}(1,-1)$.)
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