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A block LU factorisation turns a linear solve into a Schur-complement solve

Example

For $A = (\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 2 \end{array})$

with the $1 + 2$ block split, the Schur complement is

$S = (\begin{array}{cc} \frac{5}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{3}{2} \end{array}),$

and

$A = (\begin{array}{ccc} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{2} & 0 & 1 \end{array}) (\begin{array}{ccc} 2 & 1 & 1 \\ 0 & \frac{5}{2} & \frac{1}{2} \\ 0 & \frac{1}{2} & \frac{3}{2} \end{array}) .$

For the right-hand side $b = (4, 5, 4)^{T}$ , the solve $A x = b$ reduces to the Schur-complement solve and yields $x = (1, 1, 1)^{T}$ .

Facts & Assumptions

Given: The displayed matrix $A$ , its block split, the right-hand side $b$ , and the candidate factorisation.

[L1]

An invertible leading block yields the block LU factorisation through its Schur complement (An invertible leading block yields block LU through its Schur complement).

[L2]

Triangular systems are solved by forward and backward substitution (Forward and backward substitution are correct, unique, and quadratic in scalar operations).

Verification

technique · direct

1.1

The leading block is $A_{11} = [2]$ , so $S = (\begin{array}{cc} 3 & 1 \\ 1 & 2 \end{array}) - \frac{1}{2} (\binom{1}{1}) (1 1)$ , which is the displayed Schur complement. The block-LU formula of [L1] gives the displayed factorisation.

givenL1algebra

2.1

Solve $L y = b$ : $y_{1} = 4$ , $y_{2} = 5 - \frac{1}{2} \cdot 4 = 3$ , $y_{3} = 4 - \frac{1}{2} \cdot 4 = 2$ . Then solve $U x = y$ from the bottom: $\frac{3}{2} x_{3} = 2 - \frac{1}{2} x_{2}$ , $\frac{5}{2} x_{2} + \frac{1}{2} x_{3} = 3$ , and $2 x_{1} + x_{2} + x_{3} = 4$ . This gives $x_{3} = 1$ , $x_{2} = 1$ , and $x_{1} = 1$ .

step 1.1L2algebra

3.1

Steps 1.1-2.1 verify both the factorisation and the Schur-complement solve.

step 1.1step 2.1∎

Depends on

Used by

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Sources

David Bindel, CS 4220: Numerical Analysis, Blocked LU and Cholesky (standard reference, not scraped)