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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Otherwise, however, it does matter. In particular, I would point to the theory of principal component analysis in statistics: correlations constitute a quadratic form, not a linear transformation, so PCA implicitly references an isomorphism between a vector space and its dual.
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