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Find cholesky decomposition [[8,-6,2],[-6,7,-4],[2,-4,3]]

Solution:
Your problem `->` cholesky decomposition [[8,-6,2],[-6,7,-4],[2,-4,3]]


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 definate, so Cholesky decomposition is possible.

A matrix is positive definite if it’s symmetric and all its pivots are positive.

`A` = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`


Test method 1: Existence of all Positive Pivots.
First apply Gaussian Elimination method to find Pivots
`A` = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`


`R_2 larr R_2+3/4xx R_1`

 = 
`8``-6``2`
 `0` `0=-6+3/4xx8`
`R_2 larr R_2+3/4xx R_1`
 `5/2` `5/2=7+3/4xx-6`
`R_2 larr R_2+3/4xx R_1`
 `-5/2` `-5/2=-4+3/4xx2`
`R_2 larr R_2+3/4xx R_1`
`2``-4``3`


`R_3 larr R_3-1/4xx R_1`

 = 
`8``-6``2`
`0``5/2``-5/2`
 `0` `0=2-1/4xx8`
`R_3 larr R_3-1/4xx R_1`
 `-5/2` `-5/2=-4-1/4xx-6`
`R_3 larr R_3-1/4xx R_1`
 `5/2` `5/2=3-1/4xx2`
`R_3 larr R_3-1/4xx R_1`


`R_3 larr R_3+ R_2`

 = 
`8``-6``2`
`0``5/2``-5/2`
 `0` `0=0+0`
`R_3 larr R_3+ R_2`
 `0` `0=-5/2+5/2`
`R_3 larr R_3+ R_2`
 `0` `0=5/2+-5/2`
`R_3 larr R_3+ R_2`


Pivots are the first non-zero element in each row of this eliminated matrix.

`:.` Pivots are `8,5/2`

Here all pivots are positive, so matrix is positive definate.


Test method 2: Determinants of all upper-left sub-matrices are positive.
`A` = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`


 `8` 
`=8`


 `8`  `-6` 
 `-6`  `7` 
`=20`


 `8`  `-6`  `2` 
 `-6`  `7`  `-4` 
 `2`  `-4`  `3` 
`=0`


Dets are `8,20,0`

Here all determinants are positive, so matrix is positive definate.



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` = 
8-62
-67-4
2-43


`l_(11)=sqrt(a_(11))=sqrt(8)=2.8284`

`l_(21)=(a_(21))/l_(11)=(-6)/(2.8284)=-2.1213`

`l_(22)=sqrt(a_(22)-l_(21)^2)=sqrt(7-(-2.1213)^2)=sqrt(7-4.5)=1.5811`

`l_(31)=(a_(31))/l_(11)=(2)/(2.8284)=0.7071`

`l_(32)=(a_(32)-l_(31) xx l_(21))/l_(22)=(-4-(0.7071)xx(-2.1213))/(1.5811)=(-4-(-1.5))/(1.5811)=-1.5811`

`l_(33)=sqrt(a_(33)-l_(31)^2-l_(32)^2)=sqrt(3-(0.7071)^2-(-1.5811)^2)=sqrt(3-3)=0`

So `L` = 
`l_(11)``0``0`
`l_(21)``l_(22)``0`
`l_(31)``l_(32)``l_(33)`
 = 
2.828400
-2.12131.58110
0.7071-1.58110


`L xx L^T` = 
2.828400
-2.12131.58110
0.7071-1.58110
 `xx` 
2.8284-2.12130.7071
01.5811-1.5811
000
 = 
8-62
-67-4
2-43


and `A` = 
8-62
-67-4
2-43







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