Gaussian Elimination and LU Factorization: Linear Algebra Techniques in EE 230, Lecture notes of Linear Algebra

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EE 230 Lecture 2

Ran Cheng

Department of Electrical and Computer EngineeringDepartment of Physics and Astronomy University of California, Riverside

Linear equations

2 x + y + z = 5 ¿ 4 x − 6 y = − 2 ¡ − 2 x + 7 y + 2 z = 9 ¬ ¡− 2 ׿: − 8 y − 2 z = − 12 √ ¬+¿: 8 y + 3z = 14 ƒ √+ƒ: z = 2 Back substitution: z = 2 into √ → y = 1, then both into ¿ → x = 1 In augmented matrix:  

Pivots cannot be zero Can break down even if A is non singular

In general, elementary row operations include: Adding m×(row-j) to (row-i) Eij (m) → I with the (i, j) location replaced by m Eij (−m) = Eij (m)−^1 and |Eij (m)| = 1 Exchanging row-i with row-j Tij → I with 1 on (i, i) & (j, j) moved to (i, j) & (j, i) Tij = T (^) ij− 1 and |Tij | = − 1 Multiplying row-i by m Ti (m) → I with the (i, i) component replaced by m Ti

m

= T (^) i− 1 (m)

Previous example

A =

 ≡ U

3 steps: needs 3 elementary matrices

E =

 F =

 G =

Note that [E , F ] = 0 but [E , G ] 6 = 0. GFEA = U ⇒ A = (GFE )−^1 U = E −^1 F −^1 G −^1 U Define L ≡ (GFE )−^1 =

 (^) , then A = LU

LU (Triangular) factorization

Example

Features of LU factorization: LU factorization is not unique A = LU = LDD−^1 U = (LD)(D−^1 U) ≡ ˜L U˜ where D is an arbitrary diagonal and non-singular matrix. Either L or U has unit diagonal elements, the factorization is unique.

Generally, if L has unit diagonal elements, then U is decomposed as

If A = AT^ , then U = LT^ , so A = LDLT^ or A = LLT^ with L = L

D

Symmetry significantly reduces computational cost!

U has unit diagonal elements:

L =

` 11

21 22

31 32 ` 33

n 1n 2 n 3 · · ·nn

^ U^ =

1 u 12 u 13 · · · u 1 n 1 u 23 · · · u 2 n 1...^ ...

... ... 1

Crout Algorithm

i ≥ j : `ij = aij − ∑^ j−^1 k=

`ik ukj i = 1 · · · n, & j = 1 · · · i

i < j : uij = `^1 ii

aij − ∑^ i k=

`ik ukj

i = 1 · · · n − 1 , & j = i + 1, · · · n

L has unit diagonal elements:

L =

` 21 1

31 32 1

n 1n 2 `n 3 · · · 1

^ U^ =

u 11 u 12 u 13 · · · u 1 n u 22 u 23 · · · u 2 n u 33...^ ...

... ... unn

Doolittle Algorithm

i ≤ j : uij = aij −

∑^ i−^1 k=

`ik ukj i = 1 · · · n, & j = i · · · n

i > j : `ij = (^) u^1 jj

aij −

∑^ j−^1 k=

`ik ukj

i = 2 · · · n, & j = 1, · · · i − 1

Breakdown of LU factorization

A =

[ 0 2

]

the 1st pivot is 0 It cannot be factorized into the LU from! Solution: row exchange

Permutation matrix P =

[ 0 1

]

, so PA =

[ 1 3

]

= LU

Then the original problem Ax = b becomes LUx = Pb.

2-dimensional permutation group

P =

[ 0 1

]

and I =

[ 1 0

]

3-dimensional permutation group:

P 12 =

 P 23 =

 P 13 =

P 12 P 23 = P 23 P 13 = P 13 P 12 =

P 23 P 12 = · · · =

 (^123) → 231 and I =

Example

A =

 → P 12 P 23 A =

How many ways to rotate tires for a 18-wheel truck?

Gauss-Jordan method of calculating A−^1 :

[ A | I ] =

Since A = LU ⇒ U = L−^1 A L−^1 acting on [ A | I ] → [^ U | L−^1 ]  

U−^1 acting on [^ U | L−^1 ]^ → [^ I | U−^1 L−^1 ]^ = [^ I | A−^1 ]