LU decomposition

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:

${\displaystyle PA=LU,}$

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,[2] and the factorization is numerically stable in practice.[3] This makes LUP decomposition a useful technique in practice.

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

${\displaystyle PAQ=LU,}$

where L, U and P are defined as before, and Q is a permutation matrix that reorders the columns of A.[4]

An LDU decomposition is a decomposition of the form

${\displaystyle A=LDU,}$

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. 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.

Any square matrix ${\textstyle A}$ admits LUP and PLU factorizations.[2] If ${\textstyle A}$ is invertible, then it admits an LU (or LDU) factorization if and only if all its leading principal minors are nonzero.[5] If ${\textstyle A}$ is a singular matrix of rank ${\textstyle k}$, then it admits an LU factorization if the first ${\textstyle k}$ leading principal minors are nonzero, although the converse is not true.[6]

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.[5] In that case, the LU factorization is also unique if we require that the diagonal of ${\textstyle L}$ (or ${\textstyle U}$) consists of ones.

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

${\displaystyle A=LL^{*}.\,}$

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.[7]

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.[8] In particular, ${\textstyle D_{1}=A_{1,1}}$, and for ${\textstyle i=2,\ldots ,n}$, ${\textstyle D_{i}}$ is the ratio of the ${\textstyle i}$-th principal submatrix to the ${\textstyle (i-1)}$-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 ${\displaystyle {\tfrac {2}{3}}n^{3}}$ floating-point operations, ignoring lower-order terms. Partial pivoting adds only a quadratic term; this is not the case for full pivoting.[9]

Given an N × N matrix ${\displaystyle A=(a_{i,j})_{1\leq i,j\leq N}}$, define ${\displaystyle A^{(0)}:=A}$.

We eliminate the matrix elements below the main diagonal in the n-th column of A^(n − 1) by adding to the i-th row of this matrix the n-th row multiplied by

${\displaystyle -l_{i,n}:=-{\frac {a_{i,n}^{(n-1)}}{a_{n,n}^{(n-1)}}},\quad i=n+1,\ldots ,N.}$

This can be done by multiplying A^(n − 1) to the left with the lower triangular matrix

${\displaystyle L_{n}={\begin{pmatrix}1&&&&&\\&\ddots &&&&\\&&1&&&\\&&-l_{n+1,n}&&&\\&&\vdots &&\ddots &\\&&-l_{N,n}&&&1\end{pmatrix}},}$

that is, the N x N identity matrix with its n-th column replaced by the vector ${\displaystyle {\begin{pmatrix}0&\cdots &0&1&-l_{n+1,n}&\cdots &-l_{N,n}\end{pmatrix}}^{T}.}$

We set

${\displaystyle A^{(n)}:=L_{n}A^{(n-1)}.}$

After N − 1 steps, we eliminated all the matrix elements below the main diagonal, so we obtain an upper triangular matrix A^(N − 1). We find the decomposition

${\displaystyle A=L_{1}^{-1}L_{1}A^{(0)}=L_{1}^{-1}A^{(1)}=L_{1}^{-1}L_{2}^{-1}L_{2}A^{(1)}=L_{1}^{-1}L_{2}^{-1}A^{(2)}=\dotsb =L_{1}^{-1}\ldots L_{N-1}^{-1}A^{(N-1)}.}$

Denote the upper triangular matrix A^(N − 1) by U, and ${\textstyle L=L_{1}^{-1}\ldots L_{N-1}^{-1}}$. Because the inverse of a lower triangular matrix L_n is again a lower triangular matrix, and the multiplication of two lower triangular matrices is again a lower triangular matrix, it follows that L is a lower triangular matrix. Moreover, it can be seen that

${\displaystyle L={\begin{pmatrix}1&&&&&\\l_{2,1}&\ddots &&&&\\&\ddots &1&&&\\\vdots &\dots &l_{n+1,n}&1&&\\&&\vdots &\ddots &\ddots &\\l_{N,1}&\dots &l_{N,n}&\dots &l_{N,N-1}&1\end{pmatrix}}.}$

We obtain ${\displaystyle A=LU.}$

It is clear that in order for this algorithm to work, one needs to have ${\displaystyle a_{n,n}^{(n-1)}\neq 0}$ at each step (see the definition of ${\displaystyle l_{i,n}}$). 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 ${\displaystyle P^{-1}A=LU}$.

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 ${\textstyle A^{T}=L_{0}U_{0}}$ is the LU-decomposition obtained through the algorithm presented in this section, then by taking ${\textstyle L=U_{0}^{T}}$ and ${\textstyle U=L_{0}^{T}}$, we have that ${\displaystyle A=LU}$ is a Crout decomposition.

Cormen et al.[10] describe a recursive algorithm for LUP decomposition.

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

${\displaystyle P_{1}A=\left({\begin{array}{c|ccc}a&&w^{T}&\\\hline &&&\\v&&A'&\\&&&\end{array}}\right)}$,

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

${\displaystyle P_{1}A=\left({\begin{array}{c|ccc}1&&0&\\\hline &&&\\cv&&I_{n-1}&\\&&&\end{array}}\right)\left({\begin{array}{c|c}a&w^{T}\\\hline &\\0&A'-cvw^{T}\\&\end{array}}\right)}$.

Now we can recursively find an LUP decomposition ${\textstyle P'(A'-cvw^{T})=L'U'}$. Let ${\textstyle v'=P'v}$. Therefore

${\displaystyle \left({\begin{array}{c|ccc}1&&0&\\\hline &&&\\0&&P'&\\&&&\end{array}}\right)P_{1}A=\left({\begin{array}{c|ccc}1&&0&\\\hline &&&\\cv'&&L'&\\&&&\end{array}}\right)\left({\begin{array}{c|ccc}a&&w^{T}&\\\hline &&&\\0&&U'&\\&&&\end{array}}\right)}$,

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 ${\textstyle A}$ and a desired low rank ${\textstyle k}$, the randomized LU returns permutation matrices ${\textstyle P,Q}$ and lower/upper trapezoidal matrices ${\textstyle L,U}$ of size ${\textstyle m\times k}$ and ${\textstyle k\times n}$ respectively, such that with high probability ${\textstyle \Vert PAQ-LU\Vert _{2}\leq C\sigma _{k+1}}$, where ${\textstyle C}$ is a constant that depends on the parameters of the algorithm and ${\textstyle \sigma _{k+1}}$ is the ${\textstyle (k+1)}$th singular value of the input matrix ${\textstyle A}$.[11]

If two matrices of order n can be multiplied in time M(n), where M(n) ≥ n^a for some n > 2, then an LU decomposition can be computed in time O(M(n)).[12] This means, for example, that an O(n^2.376) 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

${\displaystyle Ax=b,}$

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

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

1. First, we solve the equation ${\textstyle Ly=Pb}$ for y.
2. Second, we solve the equation ${\textstyle Ux=y}$ for x.

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 ${\textstyle {\frac {2}{3}}n^{3}}$ floating-point operations if the matrix ${\textstyle A}$ has size ${\textstyle n}$. This makes it twice as fast as algorithms based on QR decomposition, which costs about ${\textstyle {\frac {4}{3}}n^{3}}$ floating-point operations when Householder reflections are used. For this reason, LU decomposition is usually preferred.[13]

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):

${\displaystyle AX=LUX=B.}$

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.[14]

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

${\displaystyle \det(A)=\det(P^{-1})\det(L)\det(U)=(-1)^{S}\left(\prod _{i=1}^{n}l_{ii}\right)\left(\prod _{i=1}^{n}u_{ii}\right).}$

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, ${\textstyle \det(A)}$ 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.