Main Content

lu

LU factorization

Description

example

[L,U] = lu(A) returns an upper triangular matrix U and a matrix L, such that A = L*U. Here, L is a product of the inverse of the permutation matrix and a lower triangular matrix.

example

[L,U,P] = lu(A) returns an upper triangular matrix U, a lower triangular matrix L, and a permutation matrix P, such that P*A = L*U. The syntax lu(A,'matrix') is identical.

example

[L,U,p] = lu(A,'vector') returns the permutation information as a vector p, such that A(p,:) = L*U.

example

[L,U,p,q] = lu(A,'vector') returns the permutation information as two row vectors p and q, such that A(p,q) = L*U.

example

[L,U,P,Q,R] = lu(A) returns an upper triangular matrix U, a lower triangular matrix L, permutation matrices P and Q, and a scaling matrix R, such that P*(R\A)*Q = L*U. The syntax lu(A,'matrix') is identical.

example

[L,U,p,q,R] = lu(A,'vector') returns the permutation information in two row vectors p and q, such that R(:,p)\A(:,q) = L*U.

example

lu(A) returns the matrix that contains the strictly lower triangular matrix L (the matrix without its unit diagonal) and the upper triangular matrix U as submatrices. Thus, lu(A) returns the matrix U + L - eye(size(A)), where L and U are defined as [L,U,P] = lu(A). The matrix A must be square.

Examples

collapse all

Compute the LU factorization of this matrix. Because the numbers are not symbolic objects, you get floating-point results.

M = [2 -3 -1; 1/2 1 -1; 0 1 -1];
[L, U] = lu(M)
L =
    1.0000         0         0
    0.2500    1.0000         0
         0    0.5714    1.0000

U =
    2.0000   -3.0000   -1.0000
         0    1.7500   -0.7500
         0         0   -0.5714

Now convert this matrix to a symbolic object, and compute the LU factorization.

M = sym(M);
[L, U] = lu(M)
L =
[   1,   0, 0]
[ 1/4,   1, 0]
[   0, 4/7, 1]
 
U =
[ 2,  -3,   -1]
[ 0, 7/4, -3/4]
[ 0,   0, -4/7]

Return the lower and upper triangular matrices and the permutation matrix by providing three output arguments.

syms a
[L, U, P] = lu(sym([0 0 a; a 2 3; 0 a 2]))
L =
[ 1, 0, 0]
[ 0, 1, 0]
[ 0, 0, 1]
U =
[ a, 2, 3]
[ 0, a, 2]
[ 0, 0, a]
P =
     0     1     0
     0     0     1
     1     0     0

Return the permutation information as a vector by using the 'vector' flag.

syms a
A = [0 0 a; a 2 3; 0 a 2];
[L, U, p] = lu(A, 'vector')
L =
[ 1, 0, 0]
[ 0, 1, 0]
[ 0, 0, 1]
U =
[ a, 2, 3]
[ 0, a, 2]
[ 0, 0, a]
p =
     2     3     1

Check that A(p,:) = L*U by using isAlways.

isAlways(A(p,:) == L*U)
ans =
  3×3 logical array
     1     1     1
     1     1     1
     1     1     1

Restore the permutation matrix P from the vector p.

P = zeros(3, 3);
for i = 1:3
    P(i, p(i)) = 1;
end
P
P =
     0     1     0
     0     0     1
     1     0     0

Return the permutation information as two vectors p and q.

syms a
A = [a, 2, 3*a; 2*a, 3, 4*a; 4*a, 5, 6*a];
[L, U, p, q] = lu(A, 'vector')
L =
[ 1, 0, 0]
[ 2, 1, 0]
[ 4, 3, 1]
U =
[ a,  2,  3*a]
[ 0, -1, -2*a]
[ 0,  0,    0]
p =
     1     2     3
q =
     1     2     3

Check that A(p, q) = L*U by using isAlways.

isAlways(A(p, q) == L*U)
ans =
  3×3 logical array
     1     1     1
     1     1     1
     1     1     1

Return the lower and upper triangular matrices, permutation matrices, and scaling matrix.

syms a
A = [0, a; 1/a, 0; 0, 1/5; 0,-1];
[L, U, P, Q, R] = lu(A)
L =
[ 1,       0, 0, 0]
[ 0,       1, 0, 0]
[ 0, 1/(5*a), 1, 0]
[ 0,    -1/a, 0, 1]
U =
[ 1/a, 0]
[   0, a]
[   0, 0]
[   0, 0]
P =
     0     1     0     0
     1     0     0     0
     0     0     1     0
     0     0     0     1
Q =
     1     0
     0     1
R =
[ 1, 0, 0, 0]
[ 0, 1, 0, 0]
[ 0, 0, 1, 0]
[ 0, 0, 0, 1]

Check that P*(R\A)*Q = L*U by using isAlways.

isAlways(P*(R\A)*Q == L*U)
ans =
  4×2 logical array
     1     1
     1     1
     1     1
     1     1

Return the permutation information as vectors p and q by using the 'vector' flag. Also, compute the scaling matrix R.

syms a
A = [0, a; 1/a, 0; 0, 1/5; 0,-1];
[L, U, p, q, R] = lu(A,'vector')
L =
[ 1,       0, 0, 0]
[ 0,       1, 0, 0]
[ 0, 1/(5*a), 1, 0]
[ 0,    -1/a, 0, 1]
U =
[ 1/a, 0]
[   0, a]
[   0, 0]
[   0, 0]
p =
     2     1     3     4
q =
     1     2
R =
[ 1, 0, 0, 0]
[ 0, 1, 0, 0]
[ 0, 0, 1, 0]
[ 0, 0, 0, 1]

Check that R(:,p)\A(:,q) = L*U by using isAlways.

isAlways(R(:,p)\A(:,q) == L*U)
ans =
  4×2 logical array
     1     1
     1     1
     1     1
     1     1

Return triangular matrices as submatrices by specifying one or no output arguments.

syms a
A = [0 0 a; a 2 3; 0 a 2];
lu(A)
ans =
[ a, 2, 3]
[ 0, a, 2]
[ 0, 0, a]

Verify that the resulting matrix is equal to U + L - eye(size(A)), where L and U are defined as [L,U,P] = lu(A).

[L,U,P] = lu(A);
U + L - eye(size(A))
ans =
[ a, 2, 3]
[ 0, a, 2]
[ 0, 0, a]

Input Arguments

collapse all

Input, specified as a numeric or symbolic matrix.

More About

collapse all

LU Factorization of a Matrix

LU factorization expresses an m-by-n matrix A as P* A = L *U. Here, L is an m-by-m lower triangular matrix, U is an m-by-n upper triangular matrix, and P is a permutation matrix.

Permutation Vector

The permutation vector p contains numbers corresponding to row exchanges in the matrix A. For an m-by-m matrix, p represents the following permutation matrix with indices i and j ranging from 1 to m.

Pij=δpi,j={1 if j=pi0 if jpi

Tips

  • Calling lu for numeric arguments that are not symbolic objects invokes the MATLAB® lu function.

  • The thresh option supported by the MATLAB lu function does not affect symbolic inputs.

  • If you use 'matrix' instead of 'vector', then lu returns permutation matrices, as it does by default.

  • L and U are nonsingular if and only if A is nonsingular. lu also can compute the LU factorization of a singular matrix A. In this case, L or U is a singular matrix.

  • Most algorithms for computing LU factorization are variants of Gaussian elimination.

Version History

Introduced in R2013a

See Also

| | | | | |