Chol Update Algoryhtm Explanation
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Hello,
Can anyone explain me how this chol update algorythm works?
function [L] = cholupdate(L, x)
n = length(x);
for k = 1:n
r = sqrt(L(k, k)^2 + x(k)^2);
c = r / L(k, k);
s = x(k) / L(k, k);
L(k, k) = r;
if k < n
L((k+1):n, k) = (L((k+1):n, k) + s * x((k+1):n)) / c;
x((k+1):n) = c * x((k+1):n) - s * L((k+1):n, k);
end
end
end
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Antworten (1)
SAI SRUJAN
am 25 Sep. 2024
Hi John,
The cholupdate function updates the Cholesky factorization of a matrix after a rank-1 update. This is particularly useful when you want to efficiently adjust the Cholesky factor without starting from scratch, which can save a lot of computational time.
Cholesky factorization is a method for decomposing a symmetric positive definite matrix (A) into the product of a lower triangular matrix (L) and its transpose (L^T), such that (A = LL^T). This is called the Cholesky factorization, and the matrix (L) is known as the Cholesky factor.
Please go through the following code sample with comments to understand more about the cholupdate function,
% L: The current lower triangular Cholesky factor of the matrix A.
% x: The vector used for the update.
function [L] = cholupdate(L, x)
n = length(x);
% The function iterates over each element of the vector x.
for k = 1:n
% r: The new diagonal element using the Pythagorean theorem.
% c and s: These are cosine and sine-like terms that help maintain the orthogonality.
r = sqrt(L(k, k)^2 + x(k)^2);
c = r / L(k, k);
s = x(k) / L(k, k);
L(k, k) = r;
% For off-diagonal elements, it updates the elements below the diagonal in the k-th column of L
% and adjusts the vector x.
if k < n
L((k+1):n, k) = (L((k+1):n, k) + s * x((k+1):n)) / c;
x((k+1):n) = c * x((k+1):n) - s * L((k+1):n, k);
end
end
end
% The updated matrix L, which is the Cholesky factor of the new matrix A + x*x'.
This method is much more efficient than recomputing the Cholesky factorization from scratch, especially for large matrices. It's widely used in applications where matrices are frequently updated.
I hope this helps!
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