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Solving Linear Eqns 8/26/2003 page 1 © Dennis L. Bricker Dept of Mechanical & Industrial Engineering University of Iowa

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Page 1: Solving Linear Eqns - University of Iowa

Solving Linear Eqns 8/26/2003 page 1

© Dennis L. BrickerDept of Mechanical & Industrial EngineeringUniversity of Iowa

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Solving Linear Eqns 8/26/2003 page 2

Contents

• Preliminarieso Elementary Matrix Operationso Elementary Matriceso Echelon Matriceso Rank of Matrices

• Elimination Methodso Gauss Eliminationo Gauss-Jordan Elimination

• Factorization Methodso Product Form of Inverse (PFI)o LU Factorizationo LDLT Factorizationo Cholesky LLT Factorization

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CHOLESKY FACTORIZATION

Suppose that A is a symmetric & positive definite matrix.

Then the Cholesky factorization of A is

ˆ T̂A L L=where L̂ is a lower triangular matrix.

Computation:Suppose that we have the factorization

TA L D L=Then if 0i

iD ≥ , we can define a new diagonal matrix D̂ whereˆ i ii iD D≡

Then ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆTT T TA L D L L D D L LD LD L L= = = = where ˆ ˆL LD=

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Example:

We wish to find the Cholesky factorization of the matrix

2 0 10 1 11 1 2

A =

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Cholesky factorization…

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The lower triangular matrix L is found by performing (on the

identity matrix) the inverse of the row operations used to reduce

the A matrix:

3 3 1

3 3 2

1 0 012 0 1 0

1 1 12

R R RL

R R R

← + ⇒ = ← +

We now have the LU factorization of matrix A:

1 0 0 2 0 10 1 0 0 1 11 11 1 0 02 2

A LU

= =

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Define the diagonal matrix D:

1

1 0 02 0 0 20 1 0 0 1 0

1 0 0 20 0 2

D D−

= ⇒ =

Note that

1

1 0 0 2 0 12ˆ 0 1 0 0 1 1

0 0 2 10 0 211 0 2

0 1 10 0 1

U D U−

= =

=

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And so,

11 01 0 0 2 0 0 20 1 0 0 1 0 0 1 11 1 0 0 11 1 0 02 2

TA LDL

= =

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Define the diagonal matrix D̂ where ˆ i ii iD D≡ :

2 0 0ˆ 0 1 0

10 02

D

=

Then compute1 0 0 2 0 0 2 0 0

ˆ ˆ 0 1 0 0 1 0 0 1 01 1 1 11 1 0 0 12 2 2 2

L LD

= = =

So the Cholesky factorization is

12 02 0 0 2ˆ ˆ 0 1 0 0 1 1

1 1 11 0 02 2 2

TA LL

= =