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

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Mark on 24 Nov 2011
Commented: Sheraline Lawles on 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?

Walter Roberson on 24 Nov 2011
However, I would suggest that LU will not help much. See instead http://www.maths.lse.ac.uk/Personal/martin/fme4a.pdf
Nicholas Lamm on 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.

Derek O'Connor on 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
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).
Sheraline Lawles on 22 Feb 2021
Just a note... sadly, the above link to Derek O'Connor's webpage is no longer active.