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Conjugate gradient method

Assume now that we have a symmetric positive-definite matrix ˆA of size n×n. At each iteration i+1 we obtain the conjugate direction of a vector

ˆxi+1=ˆxi+αiˆpi.

We assume that ˆpi is a sequence of n mutually conjugate directions. Then the ˆpi form a basis of Rn and we can expand the solution $ \hat{A}\hat{x} = \hat{b}$ in this basis, namely

ˆx=ni=1αiˆpi.