The problem of solving a set of linear equations with a symmetric positive definite matrix is equivalent to the problem of minimizing a quadratic function. Consider the problem of finding x ∈ R n satisfying Ax = b, where A ∈ R n×n , b ∈ R n and A is symmetric positive definite. The solution to this problem is also a solution of the optimization problem (P): min x∈R n f (x) = 1 2 x T Ax − b T x. (1.1) Consider the point ¯ x such that ∇ f (¯ x) = g(¯ x) = A ¯ x − b = 0. (1.2) We can show that (1.2) are the necessary optimality conditions for problem (1.1). Lemma 1.1. Suppose that A is a symmetric positive definite matrix. If ¯ x solves the problem (1.1) then (1.2) holds. Proof. Assume that g(¯ x) = ¯ r = 0 and evaluate f at the point ¯ x − α ¯ r, where α is some positive number: f (¯ x − α ¯ r) = 1 2 (¯ x − α ¯ r) T A (¯ x − α ¯ r) − b T (¯ x − α ¯ r) = 1 2 ¯ x T A ¯ x − α ¯ x T A¯ r + 1 2 α 2 ¯ r T A¯ r − b T ¯ x + αb T ¯ r
CITATION STYLE
Conjugate Direction Methods for Quadratic Problems. (2008). In Conjugate Gradient Algorithms in Nonconvex Optimization (pp. 1–62). Springer Berlin Heidelberg. https://doi.org/10.1007/978-3-540-85634-4_1
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