LU decomposition

Let be $n\times n$ matrix . factorization is a procedure for decomposing into a product of a lower triangular matrix (diagonal elements of L are unity) and an upper triangular matrix such as with

$L=\begin{pmatrix} 1\\ l_{21} & 1\\ \vdots & \vdots & \ddots\\ l_{n1} & l_{n2} & \cdots & 1 \end{pmatrix}$

and

$U=\begin{pmatrix} u_{11} & u_{12} & \cdots & u_{1n}\\ & u_{22} & \cdots & u_{2n}\\ & & \ddots & u_{n-1,n}\\ & & & u_{nn} \end{pmatrix}$

Solution of linear system

For the resolution of linear system : , the system becomes

$LUx = b \Leftrightarrow \left\{\begin{array}{cc} Ly = b& (1),\\ Ux = y &(2). \end{array}\right.$

We solve the system (1) to find the vector

, then the system (2) to find the vector

. The resolution is facilitated by the triangular shape of the matrices.

$Ly = b \Leftrightarrow \left\{\begin{array}{ll} y_1 = b_1/l_{11}& \\ y_i = \displaystyle\frac{1}{l_{ii}}(b_i-\sum_{j=1}^{i-1}l_{ij}y_j) &\forall i=2,3,\ldots,n. \end{array}\right.$

$Ux = y \Leftrightarrow \left\{\begin{array}{ll} x_n = y_n/u_{nn}& \\ x_i = \displaystyle\frac{1}{u_{ii}}(y_i-\sum_{j=i+1}^{n}u_{ij}x_j) &\forall i=n-1,n-2,\ldots,1. \end{array}\right.$

Theorems

if an LU factorization exists, then it is unique.
An invertible matrix admits an LU factorization if and only if all its principal minors are non-zero (principal minor of order is the determiant of the matrix $(A)_{1\leq i,j\leq k}$ ).
If is only invertible, then can be written where is a permutation matrix.

LU Decomposition algorithm

We suppose that admits an LU factorization, the LU Decomposition algorithm is:

$k=1,\ldots,n-1\left\{ \begin{array}{l|ll} l_{ik}=0&i=1,\ldots,k-1 & \\ l_{kk}=1& & \\ l_{ik}=\displaystyle\frac{a_{ik}^{(k)}}{a_{kk}^{(k)}}&i=k+1,\ldots,n & \\ a_{ij}^{(k+1)}=a_{ij}^{(k)}&i=1,\ldots,k & j=1,\ldots,n \\ a_{ij}^{(k+1)}=0 &i=k+1,\ldots,n & j=1,\ldots,k \\ a_{ij}^{(k+1)}=a_{ij}^{(k)}-l_{ik}a_{kj}^{(k)} & i=k+1,\ldots,n &j=k+1,\ldots,n \end{array}\right.$

$\begin{array}{lll} l_{in}=0&i=1,\ldots,n-1&l_{nn}=1\\ U=(a^{(n)}_{ij})_{1\leq i,j\leq n}&L=(l_{ij})_{1\leq i,j\leq n} & \end{array}$

Calculating Matrix Determinant

The LU decomposition also makes it possible to calculate the determinant of , which is equal to the product of the diagonal elements of the matrix if admits an LU factorization since

$det(A) = det(L) \times det(U)=1\times det(U)=det(U)$

Solution of linear system

Theorems

LU Decomposition algorithm

Calculating Matrix Determinant

Sections

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