Solve for x in (A^k)*x=b (sequentially, LU factorization)
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Mark
le 24 Nov 2011
Commenté : Sheraline Lawles
le 22 Fév 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?
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Walter Roberson
le 24 Nov 2011
http://www.mathworks.com/help/techdoc/ref/lu.html for LU factorization.
However, I would suggest that LU will not help much. See instead http://www.maths.lse.ac.uk/Personal/martin/fme4a.pdf
1 commentaire
Nicholas Lamm
le 9 Juil 2018
Modifié(e) : Rena Berman
le 9 Juil 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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Derek O'Connor
le 28 Nov 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 commentaires
Derek O'Connor
le 28 Nov 2011
Oh dear. It has just struck me that this may be a homework problem and I have given the game away.
Sheraline Lawles
le 22 Fév 2021
Just a note... sadly, the above link to Derek O'Connor's webpage is no longer active.
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