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Asked by Mark
on 24 Nov 2011

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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Answer by Walter Roberson
on 24 Nov 2011

Accepted answer

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

Answer by 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

http://www.derekroconnor.net/NA/LE/LE-2006-6.pdf

**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)`.

Mark
on 28 Nov 2011

Thank you very much. Still, I'm not quite sure how to solve for y in Ly=Pb, y=inv(L)*P*b? If so, what about x=invU)*(inv(L)*P*b)? Last, b=inv(U)*(inv(L)*P*b) and then return b in place of x?

Derek O'Connor
on 28 Nov 2011

Once you have L, U, and P, then Ax = b is replaced by LUx = Pb. This equation is solved in 2 steps, where y = Ux and c = Pb:

1. Solve Ly = c for y using forward substitution, because L is lower triangular.

2. Solve Ux = y for x using back substitution, because U is upper triangular.

Note y = L^(-1)c and x = U^(-1)y are mathematical statements and are never used to numerically solve these equations.

I think you may need to do some reading and thinking about LU factorization(decomposition).

Take a look at my webpage http://www.derekroconnor.net/NA/na2col.html for a list of textbooks and Section 5 Notes on Numerical Linear Algebra, pages 12, 13, and 21-25.

Or, better still, read the relevant chapter in Cleve Moler's book which is available online at http://www.mathworks.com/moler/

Derek O'Connor
on 28 Nov 2011

Oh dear. It has just struck me that this may be a homework problem and I have given the game away.

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