Splitting a Matrix with Choleski Method: A Simple Example

[niko2000]

Hi,
Could anybody describe a procedure for splitting a matrix using a choleski method?
Could anyone show this on a simple example:
Let's say we have a matrix A:
[2 1 0 0]
[1 2 1 0]
[0 1 2 1]
[0 0 1 2]
and matrix b:
[1]
[0]
[0]
[1]
How to get x:
Ax = b
using abovementioned method?

 

[Wong]

ha...I just learned choleski factorisation last semester in the numerical analysis course!

To apply choleski factorisation, the matrix ought to be postive definite. This is true for the matrix you wrote. To explain the procedure one needs to know how to "partition" a matrix. I hope that you know it already.

Cholesky factorisation says that for a positive definite matrix P, it is possible to factorise it in the form,

P = LL^{T}

where L is lower triangular and "...^{T} denotes transpose. To develop a recursive procedure for the factorisation, we partition the matrix in the form,

P = \left(\begin{array}{cc}a & b^{T}\\b & C\end{array}\right)
L = \left(\begin{array}{cc} d &0\\e & F\end{array}\right)

where a, d are scalars, b, e are column vectors. Then,

P = \left(\begin{array}{cc}a & b^{T}\\b & C\end{array}\right) = LL^{T} = \left(\begin{array}{cc}d^{2} & de^{T}\\de & ee^{T}+FF^{T}\end{array}\right)

So d=sqrt(a), e=b/d, and F*F^T = C*C^T - e*e^T. Remember that what we want is d, e and F. Now we get d and e. To get F, just notice that F*F^T is positive definite and so we may apply the previous step again until we get all entries of F.

To illustrate, use your example \left(\begin{array}{cccc}2&1&0&0\\1&2&1&0\\0&1&2&1\\0&0&1&2\end{array}\right). Then a = 2, b^T=[1 0 0], C = \left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right). Thus d = sqrt(2), e^T = 1/sqrt(2)*[1 0 0] and similarly find F*F^T. Now treat F*F^T as the positive definite matrix and continue the previous steps until you get the factorisation.

To solve A*x=b, notice that once we get the factorisation A=L*L^T, the system becomes L*L^T*x=b, where L is lower triangular. We first solve L*y=b for y by "forward substitution" and then solve L^T*x=y by "backward substitution".

To illustrate, suppose we have the following system,

2x = 4
x + y = 6
x + y +z = 10

Then one may first solve the first equation to get x=2 and then substitute it into the second equation to get y=4. Finally put x=2, y=4 into the third equation to get z=4.

Hope that helps

 

[M98Ranger]

Sure, I would be happy to explain the procedure for splitting a matrix using the Choleski method and provide a simple example. The Choleski method is a numerical algorithm used to decompose a symmetric, positive definite matrix into a lower triangular matrix and its transpose. This decomposition is useful for solving systems of linear equations, such as the one shown in your example, where we have a matrix A and vector b and we want to find the vector x that satisfies the equation Ax = b.

To begin, we first need to check that our matrix A is symmetric and positive definite. In your example, A is symmetric, so we just need to check if it is positive definite. This means that all of the eigenvalues of A must be positive. In your case, we can easily see that the eigenvalues are 3, 3, 3, and 3, so A is indeed positive definite.

Once we have verified that A is positive definite, we can start the Choleski decomposition. The first step is to find the lower triangular matrix L such that A = LL^T. This can be done by using the following formula:

L = [√a11 0 0 0]
[ a21/√a11 √(a22 - a21^2/a11) 0 0]
[ a31/√a11 (a32 - a31a21/a11)/√(a22 - a21^2/a11) √(a33 - a31^2/a11) 0]
[ a41/√a11 (a42 - a41a21/a11)/√(a22 - a21^2/a11) (a43 - a41a31/a11 - a42a32/a22)/√(a33 - a31^2/a11) √(a44 - a41^2/a11 - a42^2/a22 - a43^2/a33)]

In this formula, aij represents the element in the ith row and jth column of A. Using this formula, we can calculate L for our example matrix A:

L = [√2 0 0 0]
[ 1/√2 √(3/2) 0 0]
[ 0 1/√3 √(2/3) 0]
[ 0