Kiran Kedlaya's A < B PDF

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Note that the Hessian is symmetric because of the symmetry of mixed partials, so this statement makes sense. Proof. The function f is convex if and only if its restriction to each line is convex, and the second derivative along a line through x in the direction of y is (up to a scale factor) just Hy · y evaluated at x. So f is convex if and only if Hy · y > 0 for all nonzero y, that is, if H is positive definite. The bad news about this criterion is that determining whether a matrix is positive definite is not a priori an easy task: one cannot check M x · x ≥ 0 for every vector, so it seems one must compute all of the eigenvalues of M , which can be quite a headache.

Xn , we call the directional derivative along the i-th standard basis vector the partial derivative of f with respect to i and denote it by ∂f . In other words, the partial derivative is the derivative of f as a function of xi along, ∂xi regarding the other variables as constants. TOTAL DERIVATIVE Caveat! Since the derivative is not a “function” in our restricted sense (it has takes values in a vector space, not R) we cannot take a “second derivative”—yet. ASSUMING the derivative exists, it can be computed by taking partial derivatives along a basis.

The good news is that there is a very nice criterion for positive definiteness of a symmetric matrix, due to Sylvester, that saves a lot of work. Theorem 28 (Sylvester’s criterion). An n × n symmetric matrix of real numbers is positive definite if and only if the determinant of the upper left k × k submatrix is positive for k = 1, . . , n. Proof. By the M x · x definition, the upper left k × k submatrix of a positive definite matrix is positive definite, and by the eigenvalue definition, a positive definite matrix has positive determinant.

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A < B by Kiran Kedlaya

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