LU decomposition
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m (→Algorithm) 
(small typo fixed: first we solve intermediary vectory y and then x based on this y) 

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we will be able to find the solution to the original system by solving  we will be able to find the solution to the original system by solving  
  :<math>U\vec{x} = \vec{  +  :<math>U\vec{x} = \vec{y}</math> 
The first solution is a foward substitution, while the second solution is a backward substitution. Both can be done efficiently once the factorization is available. The forward substitution may be expressed as  The first solution is a foward substitution, while the second solution is a backward substitution. Both can be done efficiently once the factorization is available. The forward substitution may be expressed as  
Line 29:  Line 29:  
x_i = {1 \over {u_{ii} }}\left( {y_i  \sum\limits_{j = i + 1}^n {u_{ij} x_j } } \right),  x_i = {1 \over {u_{ii} }}\left( {y_i  \sum\limits_{j = i + 1}^n {u_{ij} x_j } } \right),  
i = n,n1,\ldots,1.</math>  i = n,n1,\ldots,1.</math>  
+  
+  
+  LU decomposition essentially stores the operations of Gaussian elimination in "higherlevel" form (see [[#ReferencesGolub and Van Loan]]), so repeated solutions using the same lefthand side are computed without repetition of operations that are independent of the righthand side.  
== Algorithm ==  == Algorithm ==  
Line 51:  Line 54:  
If one has a single lefthandside matrix and many righthand side vectors, then LU decomposition would be a good solution procedure to consider. At the very least, it should be faster than solving each system separately with Gaussian elimination.  If one has a single lefthandside matrix and many righthand side vectors, then LU decomposition would be a good solution procedure to consider. At the very least, it should be faster than solving each system separately with Gaussian elimination.  
+  
+  Here a Fortran code fragment for LU decomposition written by D. Partenov  
+  
+  do j = 1, n  
+  do i = 1, j  
+  if (i == 1) then  
+  a(i,j) = a(i, j)  
+  else  
+  suma = 0.0  
+  do k = 1, i1  
+  suma = suma + a(i,k)*a(k,j)  
+  end do  
+  a(i,j) = a(i,j)  suma  
+  end if  
+  end do  
+  if ( j < n) then  
+  do s = 1, nj  
+  i = j + s  
+  if (j == 1) then  
+  a(i,j) = a(i,j)/a(j,j)  
+  else  
+  suma = 0.0  
+  do k = 1, j1  
+  suma = suma + a(i,k)*a(k,j)  
+  end do  
+  a(i,j) = (a(i,j)  suma)/a(j,j)  
+  end if  
+  end do  
+  end if  
+  end do  
+  y(1) = b(1)  
+  do i = 2, n  
+  suma = 0.0  
+  do j = 1, i1  
+  suma = suma + a(i,j)*y(j)  
+  end do  
+  y(i) = b(i)  suma  
+  end do  
+  i = n  
+  j = n  
+  x(i) = y(i)/a(i,j)  
+  do s = 1, n 1  
+  i = n  s  
+  suma = 0.0  
+  do j = i+1, n  
+  suma = suma + a(i,j)*x(j)  
+  end do  
+  x(i) = (y(i)  suma)/a(i,i)  
+  end do  
+  
+  == References ==  
+  *{{referencebookauthor=Golub and Van Loanyear=1996title=Matrix Computationsrest= 3rd edition, The Johns Hopkins University Press, Baltimore}}  
+  *[http://en.wikipedia.org/wiki/LU_decomposition Wikipedia article ''LU decomposition] 
Latest revision as of 09:22, 21 November 2011
Contents 
Description
Consider the system of equations , where is an nonsingular matrix. may be decomposed into an lower triangular part and an upper triangular part that will lead us to a direct procedure for the solution of the original system. This decomposition procedure is quite useful when more than one righthand side (more than one ) is to be used.
The algorithm is relatively straightforward  first, we determine the upper and lower triangular parts:
Then,
where . Once we solve the system
we will be able to find the solution to the original system by solving
The first solution is a foward substitution, while the second solution is a backward substitution. Both can be done efficiently once the factorization is available. The forward substitution may be expressed as
and the backward substitution process may be expressed as
LU decomposition essentially stores the operations of Gaussian elimination in "higherlevel" form (see Golub and Van Loan), so repeated solutions using the same lefthand side are computed without repetition of operations that are independent of the righthand side.
Algorithm
Add factorization here
Forward substitution
 for k:=1 step until n do
 for i:=1 step until k1

 end loop (i)

 for i:=1 step until k1
 end loop (k)
Backward substitution
 for k:=n stepdown until 1 do
 for i:=k+1 step until n

 end loop (i)

 for i:=k+1 step until n
 end loop (k)
Important Considerations
As with Gaussian elimination, LU decomposition is probably best used for relatively small, relatively nonsparse systems of equations (with small and nonsparse open to some interpretation). For larger and/or sparse problems, it would probably be best to either use an iterative method or use a direct solver package (e.g. DSCPACK) as opposed to writing one of your own.
If one has a single lefthandside matrix and many righthand side vectors, then LU decomposition would be a good solution procedure to consider. At the very least, it should be faster than solving each system separately with Gaussian elimination.
Here a Fortran code fragment for LU decomposition written by D. Partenov
do j = 1, n do i = 1, j if (i == 1) then a(i,j) = a(i, j) else suma = 0.0 do k = 1, i1 suma = suma + a(i,k)*a(k,j) end do a(i,j) = a(i,j)  suma end if end do if ( j < n) then do s = 1, nj i = j + s if (j == 1) then a(i,j) = a(i,j)/a(j,j) else suma = 0.0 do k = 1, j1 suma = suma + a(i,k)*a(k,j) end do a(i,j) = (a(i,j)  suma)/a(j,j) end if end do end if end do y(1) = b(1) do i = 2, n suma = 0.0 do j = 1, i1 suma = suma + a(i,j)*y(j) end do y(i) = b(i)  suma end do i = n j = n x(i) = y(i)/a(i,j) do s = 1, n 1 i = n  s suma = 0.0 do j = i+1, n suma = suma + a(i,j)*x(j) end do x(i) = (y(i)  suma)/a(i,i) end do
References
 Golub and Van Loan (1996), Matrix Computations, 3rd edition, The Johns Hopkins University Press, Baltimore.
 Wikipedia article LU decomposition