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consistes of lecture slides fr linear alzebra advanced class
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Ran Cheng
Department of Electrical and Computer EngineeringDepartment of Physics and Astronomy University of California, Riverside
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)
3 steps: needs 3 elementary matrices
E =
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
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
Symmetry significantly reduces computational cost!
U has unit diagonal elements:
21
2231
32 ` 33n 1
n 2 n 3 · · ·
nn
1 u 12 u 13 · · · u 1 n 1 u 23 · · · u 2 n 1...^ ...
... ... 1
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:
31
32 1n 1
n 2 `n 3 · · · 1
u 11 u 12 u 13 · · · u 1 n u 22 u 23 · · · u 2 n u 33...^ ...
... ... unn
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
the 1st pivot is 0 It cannot be factorized into the LU from! Solution: row exchange
Permutation matrix P =
, so PA =
Then the original problem Ax = b becomes LUx = Pb.
and I =
3-dimensional permutation group:
P 12 =
(^123) → 231 and I =
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 ]