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=n∑i=1αiˆpi.