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?
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采纳的回答
Walter Roberson
2011-11-24
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 个评论
Nicholas Lamm
2018-7-9
编辑:Rena Berman
2018-7-9
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
2011-11-28
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 个评论
Derek O'Connor
2011-11-28
Oh dear. It has just struck me that this may be a homework problem and I have given the game away.
Sheraline Lawles
2021-2-22
Just a note... sadly, the above link to Derek O'Connor's webpage is no longer active.
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