L U Decomposition

LU decomposition of a matrix is the factorization of a given square matrix into two triangular matrices, one upper triangular matrix and one lower triangular matrix, such that the product of these two matrices gives the original matrix. It was introduced by Alan Turing in 1948, who also created the Turing machine.

LU decomposition method of factorizing a matrix as a product of two triangular matrices has various applications such as a solution of a system of equations, which itself is an integral part of many applications such as finding current in a circuit and solution of discrete dynamical system problems; finding the inverse of a matrix and finding the determinant of the matrix.

What is L U Decomposition?

A square matrix A can be decomposed into two square matrices L and U such that A = L U where U is an upper triangular matrix formed as a result of applying the Gauss Elimination Method on A, and L is a lower triangular matrix with diagonal elements being equal to 1. 

For A  = [Tex]\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} [/Tex].

We have L =[Tex] \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix} [/Tex] and U = [Tex]\begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix} [/Tex];

Such that A = L U i.e., [Tex]\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 0 \end{array}\right] \cdot \left[\begin{array}{ccc} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{array}\right][/Tex]

Here the value of l₂₁, u₁₁, etc. can be compared and found.

What is Gauss Elimination Method?

Gaussian Elimination, also known as Gauss-Jordan Elimination, is a method used in linear algebra to solve systems of linear equations and to find the inverse of a matrix. It’s named after the mathematician Carl Friedrich Gauss and also the mathematician Wilhelm Jordan, who made significant contributions to its development.

According to the Gauss elimination method:

1. Any zero row should be at the bottom of the matrix.
2. The first non-zero entry of each row should be on the right-hand side of the first non-zero entry of the preceding row. This method reduces the matrix to row echelon form.

LU Decomposition Method

To factories any square matrix into two triangular matrices i.e., one is a lower triangular matrix and the other is an upper triangular matrix, we can use the following steps.

• Given a set of linear equations, first convert them into matrix form A X = C where A is the coefficient matrix, X is the variable matrix and C is the matrix of numbers on the right-hand side of the equations.
• Now, reduce the coefficient matrix A, i.e., the matrix obtained from the coefficients of variables in all the given equations such that for ‘n’ variables we have an nXn matrix, to row echelon form using Gauss Elimination Method. The matrix so obtained is U.
• To find L, we have two methods. The first one is to assume the remaining elements as some artificial variables, make equations using A = L U and solve them to find those artificial variables. The other method is that the remaining elements are the multiplier coefficients because of which the respective positions became zero in the U matrix. (This method is a little tricky to understand by words but would get clear in the example below)
• Now, we have A (the nXn coefficient matrix), L (the nXn lower triangular matrix), U (the nXn upper triangular matrix), X (the nX1 matrix of variables) and C (the nX1 matrix of numbers on the right-hand side of the equations).
• The given system of equations is A X = C. We substitute A = L U. Thus, we have L U X = C. We put Z = U X, where Z is a matrix or artificial variables and solve for L Z = C first and then solve for U X = Z to find X or the values of the variables, which was required.

Example of LU Decomposition

Solve the following system of equations using the LU Decomposition method:

[Tex]\begin{equation*} x_1 + x_2 + x_3 = 1 \end{equation*} \begin{equation*} 4x_1 + 3x_2 – x_3 = 6 \end{equation*} \begin{equation*} 3x_1 + 5x_2 + 3x_3 = 4 \end{equation*} [/Tex]

Solution: Here, we have A =

[Tex]\begin{bmatrix} 1 & 1 & 1 \\ 4 & 3 & -1 \\ 3 & 5 & 3 \end{bmatrix} , X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} [/Tex]

and

[Tex]C = \begin{bmatrix} 1 \\ 6 \\ 4 \end{bmatrix} [/Tex]

such that A X = C. Now, we first consider

[Tex]\begin{bmatrix} 1 & 1 & 1 \\ 4 & 3 & -1 \\ 3 & 5 & 3 \end{bmatrix} [/Tex]

and convert it to row echelon form using Gauss Elimination Method. So, by doing

[Tex]\begin{equation} R_2 \to R_2 – 4R_1 \end{equation}  [/Tex][Tex]\begin{equation} R_3 \to R_3 – 3R_1 \end{equation} [/Tex]

we get

[Tex]\begin{bmatrix} 1 & 1 & 1 \\ 4 & 3 & -1 \\ 3 & 5 & 3 \end{bmatrix} \sim [/Tex]

[Tex]\begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 2 & 0 \end{bmatrix} [/Tex]

Now, by doing

[Tex]\begin{equation} R_3 \to R_3 – (-2)R_2 \end{equation} [/Tex]

We get

[Tex]\sim \begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 0 & -10 \end{bmatrix} [/Tex]

(Remember to always keep ‘ – ‘ sign in between, replace ‘ + ‘ sign by two ‘ – ‘ signs) Hence, we get L =

[Tex]\begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 3 & -2 & 1 \end{bmatrix} [/Tex]

and U =

[Tex]\begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 0 & -10 \end{bmatrix} [/Tex]

(notice that in L matrix,

[Tex]l_{21} = 4 [/Tex]

is from (1),

[Tex]l_{31} = 3 [/Tex]

is from (2) and

[Tex]l_{32} = -2 [/Tex]

is from (3)) Now, we assume Z

[Tex]= \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} [/Tex]

and solve L Z = C.

[Tex]\begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 3 & -2 & 1 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} [/Tex]

[Tex]= \begin{bmatrix} 1 \\ 6 \\ 4 \end{bmatrix} [/Tex]

So, we have

[Tex]z_1 = 1 , [/Tex]

[Tex]4z_1 + z_2 = 6 , [/Tex]

[Tex]3z_1 – 2z_2 + z_3 = 4 . [/Tex]

Solving, we get

[Tex]z_1 = 1 [/Tex]

,

[Tex]z_2 = 2 [/Tex]

and

[Tex]z_3 = 5 [/Tex]

. Now, we solve U X = Z

[Tex]\begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 0 & -10 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} [/Tex]

[Tex]= \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix} [/Tex]

Therefore, we get

[Tex]x_1 + x_2 + x_3 = 1 , [/Tex]

[Tex]-x_2 – 5x_3 = 2 [/Tex]

,

[Tex]-10x_3 = 5 . [/Tex]

Thus, the solution to the given system of linear equations is

[Tex]x_1 = 1 [/Tex]

,

[Tex]x_2 = 0.5 [/Tex]

,

[Tex]x_3 = -0.5 [/Tex]

and hence the matrix X =

[Tex]\begin{bmatrix} 1 \\ 0.5 \\ -0.5 \end{bmatrix} [/Tex]

Exercise on LU Decomposition

In the LU decomposition of the matrix

| 2 2 |
| 4 9 |

, if the diagonal elements of U are both 1, then the lower diagonal entry l22 of L is (GATE CS 2015) (A) 4 (B) 5 (C) 6 (D) 7

For Solution, See GATE | GATE-CS-2015 (Set 1) | Question 65.

FAQs on LU Decomposition

What is the LU decomposition method?

LU decomposition, short for Lower-Upper decomposition, is a matrix factorization technique used to break down a square matrix into the product of a lower triangular matrix (L) and an upper triangular matrix (U). It’s commonly employed to simplify solving systems of linear equations and calculating determinants.

Why is LU decomposition unique?

LU decomposition is unique because it provides a way to factorize a square matrix A into lower and upper triangular matrices (L and U) uniquely, allowing efficient solving of linear systems and determinant calculation.

How is LU decomposition calculated?

LU decomposition is calculated using Gaussian elimination, where you transform a square matrix A into lower (L) and upper (U) triangular matrices by performing row operations while keeping track of the changes in separate matrices. This process is iterative and continues until A is fully decomposed. The method with all the steps for LU decomposition is given in the article.

When LU decomposition is not possible?

LU decomposition may not be possible when the matrix A is singular (non-invertible) or when it requires pivoting for stability, but the pivot element becomes zero, causing division by zero during the decomposition process.

Are there any alternatives to LU decomposition?

Yes, alternatives to LU decomposition include Cholesky decomposition for symmetric positive definite matrices, QR decomposition for general matrices, and eigenvalue-based methods like spectral decomposition and singular value decomposition (SVD) for various matrix operations and applications.

Can LU decomposition be applied to non-square matrices?

LU decomposition is typically applied to square matrices. For rectangular matrices, QR decomposition is more commonly used. However, variations like LUP decomposition can handle rectangular matrices as well, where P is a permutation matrix.