Download - Specials Methods
SPECIAL METHODS
THOMAS METHOD
This method emerges as a simplification of an LU factorization of a tridiagonal matrix.
We know that a positive definite matrix A has a unique symmetric square root F such that :
Now if we do not insist on symmetry, there is a very large set of (non symmetric) matrices G such that :
and which may also be regarded as "square roots" of A. The positive definite matrix A is then said to be factored into the "square" of its square root.
One of these factorizations is of particular interest, both from theoretical and practical standpoints: Cholesky Decomposition, which is expressed as follows.
THOMAS METHOD
• Let A be a positive definite matrix.
Then there exists a unique lower triangular matrix L with positive diagonal elements such that : A = LL’
Based on the matrix product shown above gives the following expressions:
THOMAS METHOD
00
,
1
,11,,
1,1
1,11,
111
n
nnnnnnn
nnn
nn
nnn
cya
Where
ULbU
cU
U
aL
bU
As far as making the sweep from k = 2 to n leads to the following:
THOMAS METHOD
kkkkkkk
kkk
kk
kkk
ULbU
cU
U
aL
,11,,
1,1
1,11,
If Lux and Ux = r = d then Ld = r, therefore:
THOMAS METHOD
11,
11
2
kkkkk dLrd
nUntilkfrom
rd
Finally we solve Ux = d from a backward place:
nn
nn U
dx
Where
,
,
kk
n
kjjkjk
k U
xUd
x
untilnkfor
,
1
,11
THOMAS METHOD
Solve the following system using the method of Thomas:
Vectors are identified a, b, c and r as follows:
EXAMPLE
4
9
17
2
2
4
32
321
21
xx
xxx
xx
4
9
17
021
121
401
3
2
1
333
222
111
r
r
r
cab
cab
cab
9
7)1(*
9
21
1
9
2
9
2
3
9)4(*)2(1
4
21
2
2
1
2332333
223
22
332
1221222
112
11
221
111
ULbU
cU
U
aL
kFor
ULbU
cU
U
aL
kFor
bU
We obtain the following equalities:
EXAMPLE
Now once known L and U, Ld = r is solved by a progressive replacement:
4
9
17
192
12
1
3
2
1
d
d
d
EXAMPLE
914)25)(92(4
3
25)17)(2(9
2
17
23233
12122
11
dLrd
KPara
dLrd
KPara
rd
Finally Ux = d is solved by replacing regressive:
EXAMPLE
914
25
17
97
19
41
3
2
1
x
x
x
5
1
)3*4(17)(
12
39
)2*1(25)(
21
297
914
11
21211
22
32322
33
33
U
xUdx
nkPara
U
xUdx
nkPara
U
dx
So the solution vector is:
EXAMPLE
5
3
2
x
If A is only positive semi-definite, the diagonal elements of L can only be said to be non negative.
The Cholesky factorization can be symbolically represented by :
CHOLESKY METHODS
nnnnnnnn
nnnnnnnn
nnnnnnnn
nnn
nnn
nn
nnnn
nnnnnn
nnn
nnn
nnnnnnnn
nnnnnn
nnnn
aaaaa
aaaaa
aaaaa
aaaaa
aaaaa
L
LL
LLL
LLLL
LLLLL
LLLLL
LLLL
LLL
LL
L
,1,2,2,1,
,11,12,12,11,1
,21,22,22,21,2
,21,22,22221
,11,12,11211
,
1,1,1
2,2,12,2
2,2,12,222
1,1,11,22111
,1,2,2,1,
1,12,12,11,1
2,22,21,2
2221
11
L LT
A =LLTA
The Cholesky factorization is the prefered numerical method for calculating :
The inverse, and the determinant of a positive definite matrix (in particular of a covariance matrix), as well as for the simulation of a random multivariate normal variable.
CHOLESKY METHODS
From the product of the n-th row of L by the n-th column of LT we have:
1
1
2,
1
1
2,
2
21,
22,
22,
21,
2
221,
22,
22,
21,
n
jjnnnnn
n
jjnnnnn
nnnnnnnnnn
nnnnnnnnnn
LaL
LaL
LLLLaL
aLLLLL
CHOLESKY METHODS
1
1
2,
k
jjkkkkk LaL
Making the sweep from k = 1 to n we have:
CHOLESKY METHODS
2
1,1,1,1,
1,1
2,12,2,12,1,11,1,1,
1,1,11,2,12,2,12,1,11,
n
jjnjnnnnn
nn
nnnnnnnnnnnn
nnnnnnnnnnnnnn
LLaL
L
LLLLLLaL
aLLLLLLLL
On the other hand if we multiply the n-th row of L by the column (n-1) LT we have:
CHOLESKY METHODS
By scanning for k = 1 to n we have:
CHOLESKY METHODS
11
1
1,,,,
kiwhere
LLaLi
jjijkikik
Apply Cholesky for symmetric matrix decomposed as follows:
For k = 1:
211
121
112
A
21111 aL
EXAMPLE
For k = 2:
For k = 3:
EXAMPLE
23)21(2
12
1
22212222
11
2121
LaL
iwhenL
aL
32)61()21(2
26123
)21(*)21(1
12
1
22232
2313333
22
21313232
11
3131
LLaL
iwhenL
LLaL
iwhenL
aL
326121
02321
002
L
EXAMPLE
Finally, as a result of decomposition was found that:
BIBLIOGRAPHY
http://www.google.com.co/imgres?imgurl=http://userweb.cs.utexas.edu/~plapack/icpp98/img38.gif&imgrefurl=http://www.cs.utexas.edu/~plapack/icpp98/node2.html&usg=__lcX8ioMpUm91zLexBn6t8JE72_Q=&h=823&w=751&sz=20&hl=es&start=3&um=1&itbs=1&tbnid=cWGkKDGz2Os9EM:&tbnh=144&tbnw=131&prev=/images%3Fq%3Dcholesky%26um%3D1%26hl%3Des%26ndsp%3D20%26tbs%3Disch:1