#Sylvester's criterion for 2x2 matrices

33 messages · Page 1 of 1 (latest)

flat lagoon
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I'm having a hard time with the proof for Sylvester's criterion, and am wondering if there's a more intuitive version for only 2x2 matrices. Why is it that for a 2x2 matrix M to be positive definite (i.e. all eigenvalues positive), a_11 > 0 and det(M) > 0?

onyx slateBOT
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grave summit
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for 2x2 this is very quick to verify by hand

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remember that by definition A is positive definite if x'Ax > 0 for all nonzero x

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as for "intuitive"

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whenever you multiply a vector with a positive semi definite matrix, the angle between the original vector and the result does not exceed pi/2

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so "inuitively", a positive semi definite transformation attempts to keep the vector within certain boundaries

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@flat lagoon

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tried to illustrate

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the blue region is allowed for the vector Ax

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but the red one is impossible

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this is precisely the condition x'Ax > 0 (or nonnegative for positive semi definite case)

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i think you can extrapolate then what positive definite means

flat lagoon
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i'm assuming you mean show that this > 0 given ad - bc > 0 and a > 0

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well assuming x1 and x2 aren't both 0

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got up to this step of completing the square and now i'm lost

flat lagoon
flat lagoon
grave summit
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you are forgetting something

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Sylvester is guaranteed to work on Hermitian matrices

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you have a,b,c,d all freely picked right now

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@flat lagoon

flat lagoon
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ooh

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yeah that makes sense now

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thanks a lot

grave summit
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that's why it works in analysis optimization problems, we assume the function is sufficiently smooth hence the hessian is hermitian

flat lagoon
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right, makes sense

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+close

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# flat lagoon +close
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# strong lark

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