matlib (version 0.9.8)

LU: LU Decomposition

Description

LU computes the LU decomposition of a matrix, \(A\), such that \(P A = L U\), where \(L\) is a lower triangle matrix, \(U\) is an upper triangle, and \(P\) is a permutation matrix.

Usage

LU(A, b, tol = sqrt(.Machine$double.eps), verbose = FALSE, ...)

Value

A list of matrix components of the solution, P, L and U. If b

is supplied, the vectors \(d\) and x are also returned.

Arguments

A

coefficient matrix

b

right-hand side vector. When supplied the returned object will also contain the solved \(d\) and x elements

tol

tolerance for checking for 0 pivot

verbose

logical; if TRUE, print intermediate steps

...

additional arguments passed to showEqn

Author

Phil Chalmers

Details

The LU decomposition is used to solve the equation \(A x = b\) by calculating \(L(Ux - d) = 0\), where \(Ld = b\). If row exchanges are necessary for \(A\) then the permutation matrix \(P\) will be required to exchange the rows in \(A\); otherwise, \(P\) will be an identity matrix and the LU equation will be simplified to \(A = L U\).

Examples

  A <- matrix(c(2, 1, -1,
               -3, -1, 2,
               -2,  1, 2), 3, 3, byrow=TRUE)
  b <- c(8, -11, -3)
  (ret <- LU(A)) # P is an identity; no row swapping
  with(ret, L %*% U) # check that A = L * U
  LU(A, b)
  
  LU(A, b, verbose=TRUE)
  LU(A, b, verbose=TRUE, fractions=TRUE)

  # permutations required in this example
  A <- matrix(c(1,  1, -1,
                2,  2,  4,
                1, -1,  1), 3, 3, byrow=TRUE)
  b <- c(1, 2, 9)
  (ret <- LU(A, b))
  with(ret, P %*% A)
  with(ret, L %*% U)