# how to lu factorization?

**Answer # 1 #**

In numerical analysis and linear algebra, lower–upper (LU) decomposition or factorization factors a matrix as the product of a lower triangular matrix and an upper triangular matrix (see matrix decomposition). The product sometimes includes a permutation matrix as well. LU decomposition can be viewed as the matrix form of Gaussian elimination. Computers usually solve square systems of linear equations using LU decomposition, and it is also a key step when inverting a matrix or computing the determinant of a matrix. The LU decomposition was introduced by the Polish astronomer Tadeusz Banachiewicz in 1938.[1] To quote: "It appears that Gauss and Doolittle applied the method [of elimination] only to symmetric equations. More recent authors, for example, Aitken, Banachiewicz, Dwyer, and Crout … have emphasized the use of the method, or variations of it, in connection with non-symmetric problems … Banachiewicz … saw the point … that the basic problem is really one of matrix factorization, or “decomposition” as he called it."[2] It's also referred to as LR decomposition (factors into left and right triangular matrices).

Let A be a square matrix. An LU factorization refers to the factorization of A, with proper row and/or column orderings or permutations, into two factors – a lower triangular matrix L and an upper triangular matrix U:

In the lower triangular matrix all elements above the diagonal are zero, in the upper triangular matrix, all the elements below the diagonal are zero. For example, for a 3 × 3 matrix A, its LU decomposition looks like this:

Without a proper ordering or permutations in the matrix, the factorization may fail to materialize. For example, it is easy to verify (by expanding the matrix multiplication) that a 11 = ℓ 11 u 11 {\textstyle a_{11}=\ell _{11}u_{11}} . If a 11 = 0 {\textstyle a_{11}=0} , then at least one of ℓ 11 {\textstyle \ell _{11}} and u 11 {\textstyle u_{11}} has to be zero, which implies that either L or U is singular. This is impossible if A is nonsingular (invertible). This is a procedural problem. It can be removed by simply reordering the rows of A so that the first element of the permuted matrix is nonzero. The same problem in subsequent factorization steps can be removed the same way; see the basic procedure below.

It turns out that a proper permutation in rows (or columns) is sufficient for LU factorization. LU factorization with partial pivoting (LUP) refers often to LU factorization with row permutations only:

where L and U are again lower and upper triangular matrices, and P is a permutation matrix, which, when left-multiplied to A, reorders the rows of A. It turns out that all square matrices can be factorized in this form,[3] and the factorization is numerically stable in practice.[4] This makes LUP decomposition a useful technique in practice.

An LU factorization with full pivoting involves both row and column permutations:

where L, U and P are defined as before, and Q is a permutation matrix that reorders the columns of A.[5]

A Lower-diagonal-upper (LDU) decomposition is a decomposition of the form

where D is a diagonal matrix, and L and U are unitriangular matrices, meaning that all the entries on the diagonals of L and U are one.

Above we required that A be a square matrix, but these decompositions can all be generalized to rectangular matrices as well.[6] In that case, L and D are square matrices both of which have the same number of rows as A, and U has exactly the same dimensions as A. Upper triangular should be interpreted as having only zero entries below the main diagonal, which starts at the upper left corner. Similarly, the more precise term for U is that it is the "row echelon form" of the matrix A.

We factor the following 2-by-2 matrix:

One way to find the LU decomposition of this simple matrix would be to simply solve the linear equations by inspection. Expanding the matrix multiplication gives

This system of equations is underdetermined. In this case any two non-zero elements of L and U matrices are parameters of the solution and can be set arbitrarily to any non-zero value. Therefore, to find the unique LU decomposition, it is necessary to put some restriction on L and U matrices. For example, we can conveniently require the lower triangular matrix L to be a unit triangular matrix (i.e. set all the entries of its main diagonal to ones). Then the system of equations has the following solution:

Substituting these values into the LU decomposition above yields

Any square matrix A {\textstyle A} admits LUP and PLU factorizations.[3] If A {\textstyle A} is invertible, then it admits an LU (or LDU) factorization if and only if all its leading principal minors[7] are nonzero[8] (for example [ 0 1 1 0 ] {\displaystyle {\begin{bmatrix}0&1\\1&0\end{bmatrix}}} does not admit an LU or LDU factorization). If A {\textstyle A} is a singular matrix of rank k {\textstyle k} , then it admits an LU factorization if the first k {\textstyle k} leading principal minors are nonzero, although the converse is not true.[9]

If a square, invertible matrix has an LDU (factorization with all diagonal entries of L and U equal to 1), then the factorization is unique.[8] In that case, the LU factorization is also unique if we require that the diagonal of L {\textstyle L} (or U {\textstyle U} ) consists of ones.

In general, any square matrix A n × n {\displaystyle A_{n\times n}} could have one of the following:

In Case 3, one can approximate an LU factorization by changing a diagonal entry a j j {\displaystyle a_{jj}} to a j j ± ε {\displaystyle a_{jj}\pm \varepsilon } to avoid a zero leading principal minor.[10]

If A is a symmetric (or Hermitian, if A is complex) positive-definite matrix, we can arrange matters so that U is the conjugate transpose of L. That is, we can write A as

This decomposition is called the Cholesky decomposition. The Cholesky decomposition always exists and is unique — provided the matrix is positive definite. Furthermore, computing the Cholesky decomposition is more efficient and numerically more stable than computing some other LU decompositions.

For a (not necessarily invertible) matrix over any field, the exact necessary and sufficient conditions under which it has an LU factorization are known. The conditions are expressed in terms of the ranks of certain submatrices. The Gaussian elimination algorithm for obtaining LU decomposition has also been extended to this most general case.[11]

When an LDU factorization exists and is unique, there is a closed (explicit) formula for the elements of L, D, and U in terms of ratios of determinants of certain submatrices of the original matrix A.[12] In particular, D 1 = A 1 , 1 {\textstyle D_{1}=A_{1,1}} , and for i = 2 , … , n {\textstyle i=2,\ldots ,n} , D i {\textstyle D_{i}} is the ratio of the i {\textstyle i} -th principal submatrix to the ( i − 1 ) {\textstyle (i-1)} -th principal submatrix. Computation of the determinants is computationally expensive, so this explicit formula is not used in practice.

The following algorithm is essentially a modified form of Gaussian elimination. Computing an LU decomposition using this algorithm requires 2 3 n 3 {\displaystyle {\tfrac {2}{3}}n^{3}} floating-point operations, ignoring lower-order terms. Partial pivoting adds only a quadratic term; this is not the case for full pivoting.[13]

Given an N × N matrix A = ( a i , j ) 1 ≤ i , j ≤ N {\displaystyle A=(a_{i,j})_{1\leq i,j\leq N}} , define A ( 0 ) {\displaystyle A^{(0)}} as the matrix A {\displaystyle A} in which the necessary rows have been swapped to meet the desired conditions (such as partial pivoting) for the 1st column. The parenthetical superscript (e.g., ( 0 ) {\displaystyle (0)} ) of the matrix A {\displaystyle A} is the version of the matrix. The matrix A ( n ) {\displaystyle A^{(n)}} is the A {\displaystyle A} matrix in which the elements below the main diagonal have already been eliminated to 0 through Gaussian elimination for the first n {\displaystyle n} columns, and the necessary rows have been swapped to meet the desired conditions for the ( n + 1 ) t h {\displaystyle (n+1)^{th}} column.

We perform the operation r o w i = r o w i − ( ℓ i , n ) ⋅ r o w n {\displaystyle row_{i}=row_{i}-(\ell _{i,n})\cdot row_{n}} for each row i {\displaystyle i} with elements (labelled as a i , n ( n − 1 ) {\displaystyle a_{i,n}^{(n-1)}} where i = n + 1 , … , N {\displaystyle i=n+1,\dotsc ,N} ) below the main diagonal in the n-th column of A ( n − 1 ) {\displaystyle A^{(n-1)}} . For this operation,

We perform these row operations to eliminate the elements a i , n ( n − 1 ) {\displaystyle a_{i,n}^{(n-1)}} to zero. Once we have subtracted these rows, we may swap rows to provide the desired conditions for the ( n + 1 ) t h {\displaystyle (n+1)^{th}} column. We may swap rows here to perform partial pivoting, or because the element a n + 1 , n + 1 {\displaystyle a_{n+1,n+1}} on the main diagonal is zero (and therefore cannot be used to implement Gaussian elimination).

We define the final permutation matrix P {\displaystyle P} as the identity matrix which has all the same rows swapped in the same order as the A {\displaystyle A} matrix.

Once we have performed the row operations for the first N − 1 {\displaystyle N-1} columns, we have obtained an upper triangular matrix A ( N − 1 ) {\displaystyle A^{(N-1)}} which is denoted by U {\displaystyle U} . Note, we can denote A ( N − 1 ) {\displaystyle A^{(N-1)}} as U {\displaystyle U} because the N-th column of A ( N − 1 ) {\displaystyle A^{(N-1)}} has no conditions for which rows need to be swapped. We can also calculate the lower triangular matrix denoted as L {\textstyle L} , such that P A = L U {\displaystyle PA=LU} , by directly inputting the values of values of ℓ i , n {\displaystyle \ell _{i,n}} via the formula below.

If we are given the matrix

If we did not swap rows at all during this process, we can perform the row operations simultaneously for each column n {\displaystyle n} by setting A ( n ) := L n A ( n − 1 ) , {\displaystyle A^{(n)}:=L_{n}A^{(n-1)},} where L n {\displaystyle L_{n}} is the N × N identity matrix with its n-th column replaced by the transposed vector ( 0 ⋯ 0 1 − ℓ n + 1 , n ⋯ − ℓ N , n ) T . {\displaystyle {\begin{pmatrix}0&\dotsm &0&1&-\ell _{n+1,n}&\dotsm &-\ell _{N,n}\end{pmatrix}}^{\textsf {T}}.} In other words, the lower triangular matrix

Performing all the row operations for the first N − 1 {\displaystyle N-1} columns using the A ( n ) := L n A ( n − 1 ) {\displaystyle A^{(n)}:=L_{n}A^{(n-1)}} formula is equivalent to finding the decomposition

Now let's compute the sequence of L 1 − 1 ⋯ L N − 1 − 1 {\displaystyle L_{1}^{-1}\dotsm L_{N-1}^{-1}} . We know that L i − 1 {\displaystyle L_{i}^{-1}} has the following formula.

If there are two lower triangular matrices with 1s in the main diagonal, and neither have a non-zero item below the main diagonal in the same column as the other, then we can include all non-zero items at their same location in the product of the two matrices. For example:

( 1 0 0 0 0 77 1 0 0 0 12 0 1 0 0 63 0 0 1 0 7 0 0 0 1 ) ( 1 0 0 0 0 0 1 0 0 0 0 22 1 0 0 0 33 0 1 0 0 44 0 0 1 ) = ( 1 0 0 0 0 77 1 0 0 0 12 22 1 0 0 63 33 0 1 0 7 44 0 0 1 ) {\displaystyle \left({\begin{array}{ccccc}1&0&0&0&0\\77&1&0&0&0\\12&0&1&0&0\\63&0&0&1&0\\7&0&0&0&1\end{array}}\right)\left({\begin{array}{ccccc}1&0&0&0&0\\0&1&0&0&0\\0&22&1&0&0\\0&33&0&1&0\\0&44&0&0&1\end{array}}\right)=\left({\begin{array}{ccccc}1&0&0&0&0\\77&1&0&0&0\\12&22&1&0&0\\63&33&0&1&0\\7&44&0&0&1\end{array}}\right)}

Finally, multiply L i − 1 {\displaystyle L_{i}^{-1}} together and generate the fused matrix denoted as L {\textstyle L} (as previously mentioned). Using the matrix L {\textstyle L} , we obtain A = L U . {\displaystyle A=LU.}

It is clear that in order for this algorithm to work, one needs to have a n , n ( n − 1 ) ≠ 0 {\displaystyle a_{n,n}^{(n-1)}\neq 0} at each step (see the definition of ℓ i , n {\displaystyle \ell _{i,n}} ). If this assumption fails at some point, one needs to interchange n-th row with another row below it before continuing. This is why an LU decomposition in general looks like P − 1 A = L U {\displaystyle P^{-1}A=LU} .

Note that the decomposition obtained through this procedure is a Doolittle decomposition: the main diagonal of L is composed solely of 1s. If one would proceed by removing elements above the main diagonal by adding multiples of the columns (instead of removing elements below the diagonal by adding multiples of the rows), we would obtain a Crout decomposition, where the main diagonal of U is of 1s.

Another (equivalent) way of producing a Crout decomposition of a given matrix A is to obtain a Doolittle decomposition of the transpose of A. Indeed, if A T = L 0 U 0 {\textstyle A^{\textsf {T}}=L_{0}U_{0}} is the LU-decomposition obtained through the algorithm presented in this section, then by taking L = U 0 T {\textstyle L=U_{0}^{\textsf {T}}} and U = L 0 T {\textstyle U=L_{0}^{\textsf {T}}} , we have that A = L U {\displaystyle A=LU} is a Crout decomposition.

Cormen et al.[14] describe a recursive algorithm for LUP decomposition.

Given a matrix A, let P1 be a permutation matrix such that

where a ≠ 0 {\textstyle a\neq 0} , if there is a nonzero entry in the first column of A; or take P1 as the identity matrix otherwise. Now let c = 1 / a {\textstyle c=1/a} , if a ≠ 0 {\textstyle a\neq 0} ; or c = 0 {\textstyle c=0} otherwise. We have

Now we can recursively find an LUP decomposition P ′ ( A ′ − c v w T ) = L ′ U ′ {\textstyle P'\left(A'-cvw^{\textsf {T}}\right)=L'U'} . Let v ′ = P ′ v {\textstyle v'=P'v} . Therefore

which is an LUP decomposition of A.

It is possible to find a low rank approximation to an LU decomposition using a randomized algorithm. Given an input matrix A {\textstyle A} and a desired low rank k {\textstyle k} , the randomized LU returns permutation matrices P , Q {\textstyle P,Q} and lower/upper trapezoidal matrices L , U {\textstyle L,U} of size m × k {\textstyle m\times k} and k × n {\textstyle k\times n} respectively, such that with high probability ‖ P A Q − L U ‖ 2 ≤ C σ k + 1 {\textstyle \left\|PAQ-LU\right\|_{2}\leq C\sigma _{k+1}} , where C {\textstyle C} is a constant that depends on the parameters of the algorithm and σ k + 1 {\textstyle \sigma _{k+1}} is the ( k + 1 ) {\textstyle (k+1)} -th singular value of the input matrix A {\textstyle A} .[15]

If two matrices of order n can be multiplied in time M(n), where M(n) ≥ na for some a > 2, then an LU decomposition can be computed in time O(M(n)).[16] This means, for example, that an O(n2.376) algorithm exists based on the Coppersmith–Winograd algorithm.

Special algorithms have been developed for factorizing large sparse matrices. These algorithms attempt to find sparse factors L and U. Ideally, the cost of computation is determined by the number of nonzero entries, rather than by the size of the matrix.

These algorithms use the freedom to exchange rows and columns to minimize fill-in (entries that change from an initial zero to a non-zero value during the execution of an algorithm).

General treatment of orderings that minimize fill-in can be addressed using graph theory.

Given a system of linear equations in matrix form

we want to solve the equation for x, given A and b. Suppose we have already obtained the LUP decomposition of A such that P A = L U {\textstyle PA=LU} , so L U x = P b {\textstyle LU\mathbf {x} =P\mathbf {b} } .

In this case the solution is done in two logical steps:

In both cases we are dealing with triangular matrices (L and U), which can be solved directly by forward and backward substitution without using the Gaussian elimination process (however we do need this process or equivalent to compute the LU decomposition itself).

The above procedure can be repeatedly applied to solve the equation multiple times for different b. In this case it is faster (and more convenient) to do an LU decomposition of the matrix A once and then solve the triangular matrices for the different b, rather than using Gaussian elimination each time. The matrices L and U could be thought to have "encoded" the Gaussian elimination process.

The cost of solving a system of linear equations is approximately 2 3 n 3 {\textstyle {\frac {2}{3}}n^{3}} floating-point operations if the matrix A {\textstyle A} has size n {\textstyle n} . This makes it twice as fast as algorithms based on QR decomposition, which costs about 4 3 n 3 {\textstyle {\frac {4}{3}}n^{3}} floating-point operations when Householder reflections are used. For this reason, LU decomposition is usually preferred.[17]

When solving systems of equations, b is usually treated as a vector with a length equal to the height of matrix A. In matrix inversion however, instead of vector b, we have matrix B, where B is an n-by-p matrix, so that we are trying to find a matrix X (also a n-by-p matrix):

We can use the same algorithm presented earlier to solve for each column of matrix X. Now suppose that B is the identity matrix of size n. It would follow that the result X must be the inverse of A.[18]

Given the LUP decomposition A = P − 1 L U {\textstyle A=P^{-1}LU} of a square matrix A, the determinant of A can be computed straightforwardly as

The second equation follows from the fact that the determinant of a triangular matrix is simply the product of its diagonal entries, and that the determinant of a permutation matrix is equal to (−1)S where S is the number of row exchanges in the decomposition.

In the case of LU decomposition with full pivoting, det ( A ) {\textstyle \det(A)} also equals the right-hand side of the above equation, if we let S be the total number of row and column exchanges.

The same method readily applies to LU decomposition by setting P equal to the identity matrix.

References

Computer code

Online resources

**Answer # 2 #**

Consider the system of equations in three variables:

a11x1 + a12x2 + a13x3 = b1

a21x1 + a22x2 + a23x3 = b2

a31x1 + a32x2 + a33x3 = b3

These can be written in the form of AX = B as:

Here,

Now follow the steps given below to solve the above system of linear equations by LU Decomposition method.

Step 1: Generate a matrix A = LU such that L is the lower triangular matrix with principal diagonal elements being equal to 1 and U is the upper triangular matrix.

That means,

Step 2: Now, we can write AX = B as:

LUX = B….(1)

Step 3: Let us assume UX = Y….(2)

Step 4: From equations (1) and (2), we have;

LY = B

On solving this equation, we get y1, y2, y3.

Step 5: Substituting Y in equation (2), we get UX = Y

By solving equation, we get X, x1, x2, x3.

The above process is also called the Method of Triangularisation.

Let’s understand how to solve the system of linear equations in three variables by LU Decomposition method with the help of an solved example given below.

Example:

Solve the system of equations x1 + x2 + x3 = 1, 3x1 + x2 – 3x3 = 5 and x1 – 2x2 – 5x3 = 10 by LU decomposition method.

Solution:

Given system of equations are:

x1 + x2 + x3 = 1

3x1 + x2 – 3x3 = 5

x1 – 2x2 – 5x3 = 10

These equations are written in the form of AX = B as:

Step 1: Let us write the above matrix as LU = A.

That means,

By expanding the left side matrices, we get;

Thus, by equating the corresponding elements, we get;

u11 = 1, u12 = 1, u13 = 1

l21u11 = 3,

l21u12 + u22 = 1,

u21u13 + u23 = -3

l31u11 = 1,

l31u12 + l32u22 = -2,

l31u13 + l32u23 + u33 = -5

Solving these equations, we get;

u22 = -2, u23 = -6, u33 = 3

l21 = 3, l31 = 1, l32 = 3/2

Step 2: LUX = B

Step 3: Let UX = Y

Step 4: From the previous two steps, we have LY = B

Thus,

So,

y1 = 1

3y1 + y2 = 5

y1 + (3/2)y2 + y3 = 10

Solving these equations, we get;

y1 = 1, y2 = 2, y3 = 6

Step 5: Now, consider UX = Y. So,

By expanding this equation, we get;

x1 + x2 + x3 = 1

-2x2 – 6x3 = 2

3x3 = 6

Solving these equations, we can get;

x3 = 2, x2 = -7 and x1 = 6

Therefore, the solution of the given system of equations is (6, -7, 2).

To learn similar methods of solving the system of linear equations, visit byjus.com today!

**Answer # 3 #**

- Step 1: Generate a matrix A = LU such that L is the lower triangular matrix with principal diagonal elements being equal to 1 and U is the upper triangular matrix.
- Step 2: Now, we can write AX = B as:
- Step 3: Let us assume UX = Y….( .
- Step 4: From equations (1) and (2), we have;

**Answer # 4 #**

This 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.Basically, the LU decomposition method comes in handy whenever it is possible to model the problem to be solved into matrix form. Conversion to the matrix form and solving with triangular matrices makes it easy to do calculations in the process of finding the solution.

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 = , we have L = and U = ; such that A = L U.

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

Gauss Elimination MethodAccording to the Gauss Elimination method:

Steps for LU Decomposition:

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

Solution: Here, we have

A = and such that A X = C.

Now, we first consider and convert it to row echelon form using Gauss Elimination Method.

So, by doing

(1)

(2)

we get

Now, by doing

(3)

we get

(Remember to always keep ‘ – ‘ sign in between, replace ‘ + ‘ sign by two ‘ – ‘ signs)

Hence, we get L = and U =

(notice that in L matrix, is from (1), is from (2) and is from (3))

Now, we assume Z and solve L Z = C.

So, we have

Solving, we get , and .

Now, we solve U X = Z

Therefore, we get ,

Thus, the solution to the given system of linear equations is , , and hence the matrix X =

Exercise:In the LU decomposition of the matrix

, 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) 7For Solution, see https://www.geeksforgeeks.org/gate-gate-cs-2015-set-1-question-28/

This article is compiled by Nishant Arora. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

**Answer # 5 #**

Consider the system of equations in three variables:

a11x1 + a12x2 + a13x3 = b1

a21x1 + a22x2 + a23x3 = b2

a31x1 + a32x2 + a33x3 = b3

These can be written in the form of AX = B as:

Here,

Now follow the steps given below to solve the above system of linear equations by LU Decomposition method.

Step 1: Generate a matrix A = LU such that L is the lower triangular matrix with principal diagonal elements being equal to 1 and U is the upper triangular matrix.

That means,

Step 2: Now, we can write AX = B as:

LUX = B….(1)

Step 3: Let us assume UX = Y….(2)

Step 4: From equations (1) and (2), we have;

LY = B

On solving this equation, we get y1, y2, y3.

Step 5: Substituting Y in equation (2), we get UX = Y

By solving equation, we get X, x1, x2, x3.

The above process is also called the Method of Triangularisation.

Let’s understand how to solve the system of linear equations in three variables by LU Decomposition method with the help of an solved example given below.

Example:

Solve the system of equations x1 + x2 + x3 = 1, 3x1 + x2 – 3x3 = 5 and x1 – 2x2 – 5x3 = 10 by LU decomposition method.

Solution:

Given system of equations are:

x1 + x2 + x3 = 1

3x1 + x2 – 3x3 = 5

x1 – 2x2 – 5x3 = 10

These equations are written in the form of AX = B as:

Step 1: Let us write the above matrix as LU = A.

That means,

By expanding the left side matrices, we get;

Thus, by equating the corresponding elements, we get;

u11 = 1, u12 = 1, u13 = 1

l21u11 = 3,

l21u12 + u22 = 1,

u21u13 + u23 = -3

l31u11 = 1,

l31u12 + l32u22 = -2,

l31u13 + l32u23 + u33 = -5

Solving these equations, we get;

u22 = -2, u23 = -6, u33 = 3

l21 = 3, l31 = 1, l32 = 3/2

Step 2: LUX = B

Step 3: Let UX = Y

Step 4: From the previous two steps, we have LY = B

Thus,

So,

y1 = 1

3y1 + y2 = 5

y1 + (3/2)y2 + y3 = 10

Solving these equations, we get;

y1 = 1, y2 = 2, y3 = 6

Step 5: Now, consider UX = Y. So,

By expanding this equation, we get;

x1 + x2 + x3 = 1

-2x2 – 6x3 = 2

3x3 = 6

Solving these equations, we can get;

x3 = 2, x2 = -7 and x1 = 6

Therefore, the solution of the given system of equations is (6, -7, 2).

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