LU Decomposition

LU decomposition is another approach to Gauss-Jordan elimination. It is generalized as A = LU.


Description

If a matrix A can be decomposed into lower and upper triangular matrices L and U, the system can generally be solved by:

  1. Let y = Ux.

  2. Reformulate the original system as Ly = b. This system's first equation simply states the value of y1. The second equation gives y2 in terms of y1, which can be can be solved by substitution. And so on.

  3. Solve for y.

  4. Reformulate the original problem as Ux = y. As before, this system's last equation simply states the value of xn. And so on.

  5. Solve for x.

Furthermore, note that the inverse matrix A-1 can be calculated as U-1L-1.

julia> using LinearAlgebra

julia> A = [1 2 1; 3 8 1; 0 4 1]
3×3 Matrix{Int64}:
 1  2  1
 3  8  1
 0  4  1

julia> lu(A, NoPivot())
LU{Float64, Matrix{Float64}, Vector{Int64}}
L factor:
3×3 Matrix{Float64}:
 1.0  0.0  0.0
 3.0  1.0  0.0
 0.0  2.0  1.0
U factor:
3×3 Matrix{Float64}:
 1.0  2.0   1.0
 0.0  2.0  -2.0
 0.0  0.0   5.0

Note that NoPivot() must be specified because Julia wants to do a more efficient factorization.


Computation of Elimination Matrices

Gauss-Jordan elimination begins with identifying the pivot and transforming this:

┌        ┐
│ [1] 2 1│
│  3  8 1│
│  0  4 1│
└        ┘

...into this:

┌         ┐
│ [1] 2  1│
│  0  2 -2│
│  0  4  1│
└         ┘

This transformation was ultimately a linear combination of rows: subtracting three of row 1 from row 2. This can be reformulated with matrices.

julia> A = [1 2 1; 3 8 1; 0 4 1]
3×3 Matrix{Int64}:
 1  2  1
 3  8  1
 0  4  1

julia> E2_1 = [1 0 0; -3 1 0; 0 0 1]
3×3 Matrix{Int64}:
  1  0  0
 -3  1  0
  0  0  1

julia> E2_1 * A
3×3 Matrix{Int64}:
 1  2   1
 0  2  -2
 0  4   1

This elimination matrix is called E2 1 because it eliminated cell (2,1), the elimination cell. An elimination matrix is always the identity matrix with the negated multiplier in the elimination cell.

The Gauss-Jordan approach continues with subtracting two of row 2 from row 3. Formulated as matrices instead:

julia> E3_2 = [1 0 0; 0 1 0; 0 -2 1]
3×3 Matrix{Int64}:
 1   0  0
 0   1  0
 0  -2  1

julia> E3_2 * E2_1 * A
3×3 Matrix{Int64}:
 1  2   1
 0  2  -2
 0  0   5


Decomposition as LU

In this specific example, elimination can be written out as (E3 2E2 1)A = U.

A preferable form is A = LU, where L takes on the role of all elimination matrices. L may be a lower triangular matrix. In this specific example, L = E2 1-1E3 2-1.

See below how E3 2E2 1 is messier than E2 1-1E3 2-1 despite needing to compute inverses:

julia> E3_2 * E2_1
3×3 Matrix{Int64}:
  1   0  0
 -3   1  0
  6  -2  1

julia> convert(Matrix{Int64}, inv(E2_1) * inv(E3_2))
3×3 Matrix{Int64}:
 1  0  0
 3  1  0
 0  2  1

Furthermore, in this expression, elimination matrices are iteratively appended instead of prepended.


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LinearAlgebra/LUDecomposition (last edited 2026-01-07 01:54:41 by DominicRicottone)