#Sylvester's criterion for 2x2 matrices
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for 2x2 this is very quick to verify by hand
remember that by definition A is positive definite if x'Ax > 0 for all nonzero x
as for "intuitive"
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
so "inuitively", a positive semi definite transformation attempts to keep the vector within certain boundaries
@flat lagoon
tried to illustrate
the blue region is allowed for the vector Ax
but the red one is impossible
this is precisely the condition x'Ax > 0 (or nonnegative for positive semi definite case)
i think you can extrapolate then what positive definite means
thanks, but ngl i'm having trouble verifying it by hand too
i'm assuming you mean show that this > 0 given ad - bc > 0 and a > 0
well assuming x1 and x2 aren't both 0
got up to this step of completing the square and now i'm lost
also not sure how to connect this to sylvester's condition
I managed to show it's true using the trace and determinant for bc > 0, but not sure about the case when bc < 0
you are forgetting something
Sylvester is guaranteed to work on Hermitian matrices
you have a,b,c,d all freely picked right now
@flat lagoon
that's why it works in analysis optimization problems, we assume the function is sufficiently smooth hence the hessian is hermitian
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