Lecture 15 - 15 Conjugate Gradients This method for...

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15 Conjugate Gradients This method for symmetric positive definite matrices is considered to be the “original” Krylov subspace method. It was proposed by Hestenes and Stiefel in 1952, and is motivated by the following theorem. Theorem 15.1 If A is symmetric positive definite, then solving A x = b is equivalent to minimizing the quadratic form ϕ ( x ) = 1 2 x T A x - x T b . Proof We will consider changes of ϕ along a certain ray x + α p with α R , and fixed direction vector p = 0 . First we show that ϕ ( x + α p ) > ϕ ( x ) if A symmetric positive definite. ϕ ( x + α p ) = 1 2 ( x + α p ) T A ( x + α p ) - ( x + α p ) T b = 1 2 x T A x + 1 2 x T A ( α p ) + 1 2 ( α p ) T A x + 1 2 ( α p ) T A ( α p ) - x T b - α p T b A T = A = 1 2 x T A x - x T b = ϕ ( x ) + α p T A x - α p T b + 1 2 α 2 p T A p = ϕ ( x ) + α p T ( A x - b ) + 1 2 α 2 p T A p > 0 . Thus, we see that ϕ (as a quadratic function in α with positive leading coefficient) will have to have a minimum along the ray x + α p . We now decide what the value of α at this minimum is. A necessary condition (and also sufficient since the coefficient of α 2 is positive) is d ϕ ( x + α p ) = 0 . To this end we compute d ϕ ( x + α p ) = p T ( A x - b ) + α p T A p , which has its root at ˆ α = p T ( b - A x ) p T A p . The corresponding minimum value is ϕ ( x + ˆ α p ) = ϕ ( x ) - p T ( b - A x ) 2 2 p T A p 0 . The last equation shows that ϕ ( x + ˆ α p ) < ϕ ( x ) if and only if p T ( b - A x ) = 0 , 116
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i.e., p is not orthogonal to the residual r = b - A x . To see the equivalence with the solution of the linear system A x = b we need to consider two possibilities: 1. x is such that A x = b . Then ϕ ( x + ˆ α p ) = ϕ ( x ) and ϕ ( x ) is the minimum value. 2. x is such that A x = b . Then ϕ ( x + ˆ α p ) < ϕ ( x ), i.e., there exists a direction p such that p T ( b - A v ) = 0 and ϕ ( x ) is not the minimum. The preceding proof actually suggests a rough iterative algorithm: Take x 0 = 0 , r 0 = b , p 0 = r 0 for n = 1 , 2 , 3 , . . . Compute a step length α n = ( p T n - 1 r n - 1 ) / ( p T n - 1 A p n - 1 ) Update the approximate solution x n = x n - 1 + α n p n - 1 Update the residual r n = r n - 1 - α n A p n - 1 Find a new search direction p n end Note that at this point we have not specified how to pick the search directions p n .
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