Solve for x in (A^k)*x=b (sequentially, LU factorization)

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Mark
Mark el 24 de Nov. de 2011
Comentada: Sheraline Lawles el 22 de Feb. de 2021
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

Walter Roberson
Walter Roberson el 24 de Nov. de 2011
However, I would suggest that LU will not help much. See instead http://www.maths.lse.ac.uk/Personal/martin/fme4a.pdf
  1 comentario
Nicholas Lamm
Nicholas Lamm el 9 de Jul. de 2018
Editada: Rena Berman el 9 de Jul. de 2018
A) Linking to the documentation is about the least helpful thing you can do and B) youre not even right, LU decomposition is great for solving matrices and is even cheaper in certain situations.

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Más respuestas (1)

Derek O'Connor
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
Derek O'Connor
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
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 sadly, the above link to Derek O'Connor's webpage is no longer active.

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