Solve for x in (A^k)*x=b (sequentially, LU factorization)
13 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Mark
am 24 Nov. 2011
Kommentiert: Sheraline Lawles
am 22 Feb. 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?
0 Kommentare
Akzeptierte Antwort
Walter Roberson
am 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 Kommentar
Nicholas Lamm
am 9 Jul. 2018
Bearbeitet: Rena Berman
am 9 Jul. 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.
Weitere Antworten (1)
Derek O'Connor
am 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 Kommentare
Derek O'Connor
am 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
am 22 Feb. 2021
Just a note... sadly, the above link to Derek O'Connor's webpage is no longer active.
Siehe auch
Kategorien
Mehr zu Linear Algebra finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!