# LU decomposition

(Difference between revisions)
 Revision as of 18:34, 28 July 2007 (view source)Dimitar (Talk | contribs) (→Important Considerations)← Older edit Latest revision as of 09:22, 21 November 2011 (view source)Yonah (Talk | contribs) (small typo fixed: first we solve intermediary vectory y and then x based on this y) (One intermediate revision not shown) Line 16: Line 16: 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 - :$U\vec{x} = \vec{b}$ + :$U\vec{x} = \vec{y}$ 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 57: Line 57: Here a Fortran code fragment for LU decomposition written by  D. Partenov Here a Fortran code fragment for LU decomposition written by  D. Partenov - do j = 1, n + do j = 1, n do i = 1, j do i = 1, j if (i == 1) then if (i == 1) then a(i,j) = a(i, j) a(i,j) = a(i, j) else else - suma = 0.0 + suma = 0.0 do k = 1, i-1 do k = 1, i-1 - suma = suma + a(i,k)*a(k,j) + suma = suma + a(i,k)*a(k,j) end do end do a(i,j) = a(i,j) - suma a(i,j) = a(i,j) - suma Line 83: Line 83: end do end do end if end if - end do + end do - + y(1) = b(1) - y(1) = b(1) + do i = 2, n - + - do i = 2, n + suma = 0.0 suma = 0.0 do j = 1, i-1 do j = 1, i-1 Line 93: Line 91: end do end do y(i) = b(i) - suma y(i) = b(i) - suma - end do + end do - + i = n - i = n + j = n - j = n + x(i) = y(i)/a(i,j) - x(i) = y(i)/a(i,j) + do s = 1, n -1 - + - do s = 1, n -1 + i = n - s i = n - s suma = 0.0 suma = 0.0 Line 106: Line 102: end do end do x(i) = (y(i) - suma)/a(i,i) x(i) = (y(i) - suma)/a(i,i) - end do + end do == References == == References == *{{reference-book|author=Golub and Van Loan|year=1996|title=Matrix Computations|rest= 3rd edition, The Johns Hopkins University Press, Baltimore}} *{{reference-book|author=Golub and Van Loan|year=1996|title=Matrix Computations|rest= 3rd edition, The Johns Hopkins University Press, Baltimore}} *[http://en.wikipedia.org/wiki/LU_decomposition Wikipedia article ''LU decomposition] *[http://en.wikipedia.org/wiki/LU_decomposition Wikipedia article ''LU decomposition]

## Description

Consider the system of equations $A\vec{x}=\vec{b}$, where $A$ is an $n\times n$ nonsingular matrix. $A$ may be decomposed into an lower triangular part $L$ and an upper triangular part $U$ 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 right-hand side (more than one $\vec{b}$) is to be used.

The algorithm is relatively straightforward - first, we determine the upper and lower triangular parts:

$A = LU.$

Then,

$A \vec{x} = (LU) \vec{x} = L(U\vec{x})=L\vec{y},$

where $y=Ux$. Once we solve the system

$L\vec{y} = \vec{b},$

we will be able to find the solution to the original system by solving

$U\vec{x} = \vec{y}$

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

$y_i = {1 \over {l_{ii} }}\left( {b_i - \sum\limits_{j = 1}^{i-1} {l_{ij} y_j } } \right), i = 1,2,\ldots,n$

and the backward substitution process may be expressed as

$x_i = {1 \over {u_{ii} }}\left( {y_i - \sum\limits_{j = i + 1}^n {u_{ij} x_j } } \right), i = n,n-1,\ldots,1.$

LU decomposition essentially stores the operations of Gaussian elimination in "higher-level" form (see Golub and Van Loan), so repeated solutions using the same left-hand side are computed without repetition of operations that are independent of the right-hand side.

## Algorithm

Forward substitution

for k:=1 step until n do
for i:=1 step until k-1
$b_k=b_k-l_{ki}b_{i}$
end loop (i)
$b_{k}=b_{k}/l_{kk}$
end loop (k)

Backward substitution

for k:=n stepdown until 1 do
for i:=k+1 step until n
$b_k=b_k-u_{ki}b_{i}$
end loop (i)
$x_{k}=b_{k}/u_{kk}$
end loop (k)

## Important Considerations

As with Gaussian elimination, LU decomposition is probably best used for relatively small, relatively non-sparse systems of equations (with small and non-sparse 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 lefthand-side matrix and many right-hand 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, i-1
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, n-j
i = j + s
if (j == 1) then
a(i,j) = a(i,j)/a(j,j)
else
suma = 0.0
do k = 1, j-1
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, i-1
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