There is a simple canonical form of a symmetric and antisymmetric bilibear forms. Is there a canonical form for a general bilinear form?
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1$\begingroup$ To improve your question, why not include in it the canonical forms you have in mind for the first sentence. Also, are you restricting the characteristic in any way (e.g., do you allow bilinear forms in characteristic 2)? $\endgroup$– KConradCommented Dec 27, 2012 at 5:40
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1$\begingroup$ This question should have a canonical answer in the literature... Hopefully someone knows! $\endgroup$– Mariano Suárez-ÁlvarezCommented Dec 27, 2012 at 5:54
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1$\begingroup$ I think this may be interesting for you: mathoverflow.net/questions/105870/… $\endgroup$– QfwfqCommented Dec 27, 2012 at 13:58
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2 Answers
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If the characteristic of the ground field is not 2, then a bilinear form B on a vector space V can be written uniquely in the form B = A + S, where A is an antisymmetric bilinear form and S is a symmetric bilinear form. Then you will generally have invariants of the pair to deal with. For example, there will be the ranks of A and S and, for S, the type of the quadratic form. In a generic case, for example, when S is nondegenerate, one can then turn A into a linear map L from V into itself via the rule A(x,y) = S(x,Ly) and then the characteristic polynomial of L as a linear map will provide invariants of the original B. Any canonical form will have to be rich enough to capture such invariants.
answered Dec 27, 2012 at 5:51
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People in matrix analysis would call this a "canonical form under congruence". Take a look at http://arxiv.org/abs/0709.2473; the solution is stated there.
answered Dec 27, 2012 at 10:45
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3$\begingroup$ More precisely, the solution when the ground field is $\mathbb{C}$ is stated there. $\endgroup$ Commented Dec 27, 2012 at 14:04