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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Replying to @RobJLow
Guessing without checking: if the det is zero at least one eigenvalue is zero; the other eigenvalue equals the trace.
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Replying to @johncarlosbaez @RobJLow
Strictly speaking, quadratic forms have neither determinants nor eigenvalues; only linear transformations can have determinants and eigenvalues. Quadratic forms have discriminants, which are not scalar quantities, and quadratic forms have roots.
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Replying to @MathPrinceps @RobJLow
Laurens is right. I think Robert Low secretly meant "linear transformation". That's how I read his post - I didn't really notice the phrase "quadratic form".
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Replying to @johncarlosbaez @RobJLow
The distinction between quadratic forms and linear transformations ultimately boils down to the distinction between a vector space and its dual. When, because some special structure is present, a vector space and its dual are naturally isomorphic, the distinction doesn't matter.
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What college courses prepare one to learn this?
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Linear algebra: but one which does stuff like abstract vector spaces, dual space, inner product. A lot of intros are really matrix algebra, which can conceal the subtleties, so it might well be the second linear algebra course that gets to this.
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That explains it. Maybe one day I'll go back and get (a) proper degree(s) in mathematics.
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The key point, I think, is that a vector space and its dual are always isomorphic when they're finite-dimensional (although they're by no means always naturally isomorphic, which is important.) It's only when they're infinite-dimensional that the distinction becomes inevitable.
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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.
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