What Is An LU Factorization? - Nick Higham

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For A\in\mathbb{R}^{n\times n} with \alpha = a_{11} \ne 0 we can write

\notag A = \begin{bmatrix} \alpha & b^T \\ c & D \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ c/\alpha & I_{n-1} \end{bmatrix} \begin{bmatrix} \alpha & b^T \\ 0 & D - cb^T/\alpha \end{bmatrix} = : L_1U_1. \qquad (2)

The (n-1)\times (n-1) matrix S = D - cb^T/\alpha is called the Schur complement of \alpha in A.

The first row and column of L_1 and U_1 have the correct forms for a unit lower triangular matrix and an upper triangular matrix, respectively. If we can find an LU factorization S = L_2U_2 then

\notag A = \begin{bmatrix} 1 & 0 \\ c/\alpha & L_2 \end{bmatrix} \begin{bmatrix} \alpha & b^T \\ 0 & U_2 \end{bmatrix}

is an LU factorization of A. Note that this is simply another way to express the kji algorithm above.

For several matrix structures it is immediate that \alpha \ne 0. If we can show that the Schur complement inherits the same structure then it follows by induction that we can compute the factorization for S, and so an LU factorization of A exists. Classes of matrix for which a_{11} \ne 0 and A being in the class implies the Schur complement S is also in the class include

  • symmetric positive definite matrices,
  • M-matrices,
  • matrices (block) diagonally dominant by rows or columns.

(The proofs of these properties are nontrivial.) Note that the matrix (1) is row diagonally dominant, as is its U factor, as must be the case since its rows are contained in Schur complements.

We say that A has upper bandwidth q if a_{ij} = 0 for j>i+q and lower bandwidth p if a_{ij} = 0 for i>j+p. Another use of (2) is to show that L and U inherit the bandwidths of A.

Theorem 2. Let A\in\mathbb{R}^{n\times n} have lower bandwidth p and upper bandwidth q. If A has an LU factorization then L has lower bandwidth p and U has upper bandwidth q.

Proof. In (2), the first column of L_1 and the first row of U_1 have the required structure and S has upper bandwidth q and lower bandwidth p, since c and b have only p and q nonzero components, respectively. The result follows by induction.

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