Linear algebra nugget of the day. You can classify a 2d quadratic form without finding eigenvalues. Det tells you if the eigenvalues are the same sign and if det is positive, trace tells you which sign. Exercise: what if det is 0?
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True. But classifying linear maps from a vector space, up to isomorphism, is completely different from classifying bilinear forms, up to isomorphism, even in the finite-dimensional case. So: let duals be duals!
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Exactly right, of course -- and this is the very point that led me to make my original contribution to this exchange. As Sylvester first showed, non-degenerate quadratic forms up to isomorphism are classified (up to sign) by their signature -- far simpler than linear maps.
End of conversation
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