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This implies that at a local minimum the Hessian is positive-semidefinite, and at a local maximum the Hessian is negative-semidefinite. For positive-semidefinite and negative-semidefinite Hessians the test is inconclusive (a critical point where the Hessian is semidefinite but not definite may be a local extremum or a saddle point).
Set P = ∅. Set R = {1, ..., n}. Set x to an all-zero vector of dimension n. Set w = A T (y − Ax). Let w R denote the sub-vector with indexes from R; Main loop: while R ≠ ∅ and max(w R) > ε: Let j in R be the index of max(w R) in w. Add j to P. Remove j from R. Let A P be A restricted to the variables included in P. Let s be vector of ...
The Gram matrix is positive semidefinite, and every positive semidefinite matrix is the Gramian matrix for some set of vectors. The fact that the Gramian matrix is positive-semidefinite can be seen from the following simple derivation:
The conjugate gradient method can be applied to an arbitrary n-by-m matrix by applying it to normal equations A T A and right-hand side vector A T b, since A T A is a symmetric positive-semidefinite matrix for any A. The result is conjugate gradient on the normal equations (CGN or CGNR). A T Ax = A T b
If the quadratic form f yields only non-negative values (positive or zero), the symmetric matrix is called positive-semidefinite (or if only non-positive values, then negative-semidefinite); hence the matrix is indefinite precisely when it is neither positive-semidefinite nor negative-semidefinite. A symmetric matrix is positive-definite if and ...
In mathematics, positive semidefinite may refer to: Positive semidefinite function; Positive semidefinite matrix; Positive semidefinite quadratic form;
Let be the set of real numbers and be the set of complex numbers. A function f : R → C {\displaystyle f:\mathbb {R} \to \mathbb {C} } is called positive semi-definite if for all real numbers x 1 , …, x n the n × n matrix
A Hermitian diagonally dominant matrix with real non-negative diagonal entries is positive semidefinite. This follows from the eigenvalues being real, and Gershgorin's circle theorem. If the symmetry requirement is eliminated, such a matrix is not necessarily positive semidefinite. For example, consider