  # LU Factorization

In Matlab there are several built-in functions provided for matrix factorization (also called decomposition).

 The name of the built-in function for a Lower-Upper decomposition is 'lu'. To get the LU factorization of a square matrix A, type the command '[L, U] = lu(A)'. Matlab returns a lower triangular matrix L and an upper triangular matrix  U such that L*U = A.

Suppose that we need to perform a factorization on

A= [ 1  2 -3
-3 -4 13
2 1 -5]

We can verify that if matrix

L = [ 1   0     0
-3   1     0
2  -1.5  1]

and matrix

U = [1   2  -3
0   2   4
0   0   7]

Then, factors of A are L and U, that is: A = L*U

The decomposition of the matrix A is an illustration of an important and well known theorem. If A is a nonsingular matrix that can be transformed into an upper diagonal form U by the application or row addition operations, then there exists a lower triangular matrix L such that A = LU.

Row addition operations can be represented by a product of elementary matrices. If n such operations are required, the matrix U is related to the matrix A in the following way:

U = En En-1 ... E2 E1 A

The lower triangular matrix L is found from

L = E1-1 E2-1 ... En-1

L will have ones on the diagonal. The off-diagonal elements are zeros above the diagonal, while the elements below the diagonal are the multipliers required to perform Gaussian elimination on the matrix A. The element lij
is equal to the multiplier used to eliminate the (i, j) position.

Example:

In Matlab, let's find the LU decomposition of the matrix

A = [-2 1 -3; 6 -1 8; 8 3 -7]

Write this instruction in the command window or within a script:

[L, U] = lu(A)

L =
-0.2500   -0.5385    1.0000
0.7500     1.0000           0
1.0000            0           0

U =
8.0000    3.0000   -7.0000
0     -3.2500   13.2500
0         0         2.3846

We can test the answer, by typing

L*U

And, finnally, the Matlab answer is:

ans =
-2.0000    1.0000   -3.0000
6.0000   -1.0000    8.0000
8.0000    3.0000   -7.0000

>>

Showing that A = L*U, indeed.

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