Surname 2
Vector x is taken to be known m× m matrix. A is a symmetric matrix, i.e. (A
T
=A). B is known.
We can denote the solution of this system by using vector x
∗
. We can prove that two vectors e
and f are conjugates if and only if e
T
A f=0. If A is symmetric non–zero, the left-hand side
gives the inner product. Vectors are conjugate if they are orthogonal in relation to the inner
product. The conjugate is a symmetric relation: if e is conjugate to f, then also f is conjugate to
e.
Taking the conjugate vectors g and k, we can use them to approximate the value for x.
We can also use the conjugate gradient method as an iterative method. With the iterative
method, we can approximate the solution of systems though it takes a long time to carry out
the iterations.
Let’s assume the initial values for x
∗
by x
0
. The values can be assumed without any
loss, x = 0. If we start with x, then the solution from the successive iterations can be used to
get the approximate value of the system of equation. The solution comes from the fact that any
answer to the system of the equations is unique. For example in the following equation, the
conjugate constraint has an algorithm which bears a resemblance to orthonormalization. After
a long derivation, the formula gives the following expression.
From the above expression M, the algorithm provides a direct explanation of the
conjugate gradient method. The algorithm above states has to have storage of all previous
directions and vectors; in addition to many matrix-vector manipulations, thus, they are complex
to compute manually. However, one matrix-vector manipulation is required in each iteration.
The algorithm is used for solving AX =b where A is a real and positive-definite matrix-vector.