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12. Cholesky Decomposition example ( Enter your problem )
  1. Example `[[6,15,55],[15,55,225],[55,225,979]]`
  2. Example `[[6,-2,2],[-2,3,-1],[2,-1,3]]`
  3. Example `[[25,15,-5],[15,18,0],[-5,0,11]]`
  4. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
Other related methods
  1. Transforming matrix to Row Echelon Form
  2. Transforming matrix to Reduced Row Echelon Form
  3. Rank of matrix
  4. Characteristic polynomial of matrix
  5. Eigenvalues
  6. Eigenvectors (Eigenspace)
  7. Triangular Matrix
  8. LU decomposition using Gauss Elimination method of matrix
  9. LU decomposition using Doolittle's method of matrix
  10. LU decomposition using Crout's method of matrix
  11. Diagonal Matrix
  12. Cholesky Decomposition
  13. QR Decomposition (Gram Schmidt Method)
  14. QR Decomposition (Householder Method)
  15. LQ Decomposition
  16. Pivots
  17. Singular Value Decomposition (SVD)
  18. Moore-Penrose Pseudoinverse
  19. Power Method for dominant eigenvalue
  20. determinants using Sarrus Rule
  21. determinants using properties of determinants
  22. Row Space
  23. Column Space
  24. Null Space

2. Example `[[6,-2,2],[-2,3,-1],[2,-1,3]]`
(Previous example)
4. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
(Next example)

3. Example `[[25,15,-5],[15,18,0],[-5,0,11]]`





Find Cholesky Decomposition ...
`[[25,15,-5],[15,18,0],[-5,0,11]]`


Solution:
Cholesky decomposition : `A=L*L^T`, Every symmetric positive definite matrix A can be decomposed into a product of a unique lower triangular matrix L and its transpose.


Here matrix is symmetric positive definite, so Cholesky decomposition is possible.

`A` = 
`25``15``-5`
`15``18``0`
`-5``0``11`


A matrix is positive definite if Determinants of all upper-left sub-matrices are positive.

Test method 2: Determinants of all upper-left sub-matrices are positive.
`A` = 
`25``15``-5`
`15``18``0`
`-5``0``11`


 `25` 
`=25`


 `25`  `15` 
 `15`  `18` 
`=225`


 `25`  `15`  `-5` 
 `15`  `18`  `0` 
 `-5`  `0`  `11` 
`=2025`


Determinants are `25,225,2025`

Here all determinants are positive, so matrix is positive semi-definite.




Formula
`l_(ki)=(a_(ki) - sum_{j=1}^{i-1} l_(ij) * l_(kj))/(l_(ii))`

`l_(kk)=sqrt(a_(kk)-sum_{j=1}^{k-1} l_(kj)^2)`

Here `A` = 
2515-5
15180
-5011


`l_(11)=sqrt(a_(11))=sqrt(25)=5`

`l_(21)=(a_(21))/l_(11)=(15)/(5)=3`

`l_(22)=sqrt(a_(22)-l_(21)^2)=sqrt(18-(3)^2)=sqrt(18-9)=3`

`l_(31)=(a_(31))/l_(11)=(-5)/(5)=-1`

`l_(32)=(a_(32)-l_(31) xx l_(21))/l_(22)=(0-(-1)xx(3))/(3)=(0-(-3))/(3)=1`

`l_(33)=sqrt(a_(33)-l_(31)^2-l_(32)^2)=sqrt(11-(-1)^2-(1)^2)=sqrt(11-2)=3`

So `L` = 
`l_(11)``0``0`
`l_(21)``l_(22)``0`
`l_(31)``l_(32)``l_(33)`
 = 
500
330
-113


`L xx L^T` = 
500
330
-113
 `xx` 
53-1
031
003
 = 
2515-5
15180
-5011


and `A` = 
2515-5
15180
-5011



This material is intended as a summary. Use your textbook for detail explanation.
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2. Example `[[6,-2,2],[-2,3,-1],[2,-1,3]]`
(Previous example)
4. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
(Next example)





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