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linalg(deprecated)

 LUdecomp
 LU decomposition of a matrix Calling Sequence LUdecomp(A) LUdecomp(A, arg2, arg3, ...) LUdecomp(A, P='p', L='l', U='u' , U1='u1', R='r', rank='ran', det='d') Parameters

 A - rectangular matrix arg.i - (optional) of the form name=val P='p' - (optional) the pivot factor L='l' - (optional) the unit lower triangular factor U='u' - (optional) the upper triangular factor U1='u1' - (optional) the modified U factor R='r' - (optional) the row-reduced factor rank='ran' - (optional) the rank of A det='d' - (optional) the determinant of U1 Description

 • Important: The linalg package has been deprecated. Use the superseding command LinearAlgebra[LUDecomposition], instead.
 - For information on migrating linalg code to the new packages, see examples/LinearAlgebraMigration.
 • The routine LUdecomp computes both the traditional LU decomposition and a modified PLU1R decomposition of the matrix A.  It returns U.
 • For matrices of floating-point entries, a partial (row) pivoting method is used.  For symbolic computation, pivoting is done only when a leading entry is zero.  The pivot (permutation) matrix is returned as P.
 • The basic decomposition generates a square unit lower triangular L factor and an upper triangular U factor with the same dimensions as A so that A = P*L*U.
 • The decomposition further factors U into U1*R where U1 is square upper triangular and R is the unique reduced row-echelon form of the matrix A.  In this case A = P*L*U1*R.
 • The det parameter is defined to be the determinant of the U1 factor. This is non-zero for all matrices.  When A has full rank, this will coincide with the determinant of A.
 • The rank parameter will be the rank of A.  This will be of questionable value if A is composed of floating-point values and is ill-conditioned.
 • U1 is assumed to be non-singular.  The matrix R, i.e. the row-echelon factor of A, will only be correct if this is the case.  If A has parameters in its entries, and for some values of the parameters det(U1) = 0, then the row-echelon form must be recomputed for these values of the parameters.
 • Note that this paper names the U1 factor simply U.
 • The command with(linalg,LUdecomp) allows the use of the abbreviated form of this command. Examples

Important: The linalg package has been deprecated. Use the superseding command LinearAlgebra[LUDecomposition], instead.

 > $\mathrm{with}\left(\mathrm{linalg}\right):$
 > $A≔\mathrm{matrix}\left(\left[\left[1,-2,3,1\right],\left[2,k,6,6\right],\left[-1,3,k-3,0\right]\right]\right)$
 ${A}{≔}\left[\begin{array}{cccc}{1}& {-2}& {3}& {1}\\ {2}& {k}& {6}& {6}\\ {-1}& {3}& {k}{-}{3}& {0}\end{array}\right]$ (1)
 > $x≔\mathrm{LUdecomp}\left(A,L='l',U='u',\mathrm{U1}='\mathrm{u1}',R='r',P='p',\mathrm{det}='d',\mathrm{rank}='\mathrm{ran}'\right)$
 ${x}{≔}\left[\begin{array}{cccc}{1}& {-2}& {3}& {1}\\ {0}& {k}{+}{4}& {0}& {4}\\ {0}& {0}& {k}& \frac{{k}}{{k}{+}{4}}\end{array}\right]$ (2)

show the results

 > $\mathrm{rank}\left(x\right)$
 ${3}$ (3)
 > $\mathrm{evalm}\left(x\right)$
 $\left[\begin{array}{cccc}{1}& {-2}& {3}& {1}\\ {0}& {k}{+}{4}& {0}& {4}\\ {0}& {0}& {k}& \frac{{k}}{{k}{+}{4}}\end{array}\right]$ (4)
 > $\mathrm{evalm}\left(l\right)$
 $\left[\begin{array}{ccc}{1}& {0}& {0}\\ {2}& {1}& {0}\\ {-1}& \frac{{1}}{{k}{+}{4}}& {1}\end{array}\right]$ (5)

verify the decomposition (p is the identity).

 > $\mathrm{map}\left(\mathrm{normal},\mathrm{evalm}\left(l\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&*\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}u\right)\right)$
 $\left[\begin{array}{cccc}{1}& {-2}& {3}& {1}\\ {2}& {k}& {6}& {6}\\ {-1}& {3}& {k}{-}{3}& {0}\end{array}\right]$ (6)
 > $\mathrm{evalm}\left(r\right)$
 $\left[\begin{array}{cccc}{1}& {0.}& {0.}& \frac{{9}{+}{k}}{{k}{+}{4}}\\ {0}& {1}& {0.}& \frac{{4}}{{k}{+}{4}}\\ {0}& {0}& {1}& \frac{{1}}{{k}{+}{4}}\end{array}\right]$ (7)
 > $\mathrm{evalm}\left(\mathrm{u1}\right)$
 $\left[\begin{array}{ccc}{1}& {-2}& {3}\\ {0}& {k}{+}{4}& {0}\\ {0}& {0}& {k}\end{array}\right]$ (8)
 > $\mathrm{det}\left(\mathrm{u1}\right)$
 $\left({k}{+}{4}\right){}{k}$ (9)

det(u1) = k*(4+k) so u1 is singular when k=-4 or k=0, we investigate these separately

 > $x≔\mathrm{LUdecomp}\left(\mathrm{subs}\left(k=0,\mathrm{op}\left(A\right)\right),R='r',\mathrm{det}='d',\mathrm{rank}='\mathrm{ran}'\right)$
 ${x}{≔}\left[\begin{array}{cccc}{1}& {-2}& {3}& {1}\\ {0}& {4}& {0}& {4}\\ {0}& {0}& {0}& {0}\end{array}\right]$ (10)
 > $\mathrm{evalm}\left(r\right)$
 $\left[\begin{array}{cccc}{1}& {0.}& {3}& {3}\\ {0}& {1}& {0}& {1}\\ {0}& {0}& {0}& {0}\end{array}\right]$ (11)
 > $x≔\mathrm{LUdecomp}\left(\mathrm{subs}\left(k=-4,\mathrm{op}\left(A\right)\right),R='r',\mathrm{det}='d',\mathrm{rank}='\mathrm{ran}'\right)$
 ${x}{≔}\left[\begin{array}{cccc}{1}& {-2}& {3}& {1}\\ {0}& {1}& {-4}& {1}\\ {0}& {0}& {0}& {4}\end{array}\right]$ (12)
 > $\mathrm{evalm}\left(r\right)$
 $\left[\begin{array}{cccc}{1}& {0.}& {-5}& {0.}\\ {0}& {1}& {-4}& {0.}\\ {0}& {0}& {0}& {1}\end{array}\right]$ (13)
 > $d$
 ${4}$ (14)
 > $\mathrm{ran}$
 ${3}$ (15)

and floating point gaussian elimination

 > $A≔\mathrm{matrix}\left(\mathrm{evalf}\left(\left[\left[1,2,3\right],\left[2,4,5\right],\left[3,6,7\right]\right]\right)\right)$
 ${A}{≔}\left[\begin{array}{ccc}{1.}& {2.}& {3.}\\ {2.}& {4.}& {5.}\\ {3.}& {6.}& {7.}\end{array}\right]$ (16)
 > $U≔\mathrm{LUdecomp}\left(A,P='p',L='l'\right)$
 ${U}{≔}\left[\begin{array}{ccc}{3.}& {6.}& {7.}\\ {0.}& {0.}& {0.666666666666667}\\ {0.}& {0.}& {0.}\end{array}\right]$ (17)
 > $\mathrm{evalm}\left(p\right)$
 $\left[\begin{array}{ccc}{0.}& {1.}& {0.}\\ {0.}& {0.}& {1.}\\ {1.}& {0.}& {0.}\end{array}\right]$ (18) References

 Corless, Robert M.; Jeffrey, David; and Nerenberg, M. A. H. "The Row Echelon Decomposition of a Matrix." University of Western Ontario Tech Report AM-91-01, Department of Applied Mathematics, 1991.