Solve for x in (A^k)*x=b (sequentially, LU factorization)
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Without computing A^k, solve for x in (A^k)*x=b.
A) Sequentially? (Pseudocode)
for n=1:k
x=A\b;
b=x;
end
Is the above process correct?
B) LU factorizaion?
How is this accompished?
Respuesta aceptada
Más respuestas (1)
Derek O'Connor
el 28 de Nov. de 2011
Contrary to what Walter says, LU Decomposition is a great help in this problem. See my solution notes to Lab Exercise 6 --- LU Decomposition and Matrix Powers
Additional Information
Here is the Golub-Van Loan Algorithm for solving (A^k)*x = b
[L,U,P] = lu(A);
for m = 1:k
y = L\(P*b);
x = U\y;
b = x;
end
Matlab's backslash operator "\" is clever enough to figure out that y = L\(P*b) is forward substitution, while x = U\y is back substitution, each of which requires O(n^2) work.
Total amount of work is: O(n^3) + k*O(n^2) = O(n^3 + k*n^2)
If k << n then this total is effectively O(n^3).
4 comentarios
Mark
el 28 de Nov. de 2011
Derek O'Connor
el 28 de Nov. de 2011
Once you have L, U, and P, then Ax = b is replaced by LUx = Pb. This equation is solved in 2 steps, where y = Ux and c = Pb:
1. Solve Ly = c for y using forward substitution, because L is lower triangular.
2. Solve Ux = y for x using back substitution, because U is upper triangular.
Note y = L^(-1)c and x = U^(-1)y are mathematical statements and are never used to numerically solve these equations.
I think you may need to do some reading and thinking about LU factorization(decomposition).
Take a look at my webpage http://www.derekroconnor.net/NA/na2col.html for a list of textbooks and Section 5 Notes on Numerical Linear Algebra, pages 12, 13, and 21-25.
Or, better still, read the relevant chapter in Cleve Moler's book which is available online at http://www.mathworks.com/moler/
Derek O'Connor
el 28 de Nov. de 2011
Oh dear. It has just struck me that this may be a homework problem and I have given the game away.
Sheraline Lawles
el 22 de Feb. de 2021
Just a note... sadly, the above link to Derek O'Connor's webpage is no longer active.
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