Find Cholesky Decomposition ...
`[[6,-2,2],[-2,3,-1],[2,-1,3]]`

Solution:

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

`A`

=

`6` `-2` `2` `-2` `3` `-1` `2` `-1` `3`

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`

=

`6` `-2` `2` `-2` `3` `-1` `2` `-1` `3`

`6`

`=6`

`6`   `-2`   `-2`   `3`

`=14`

`6`   `-2`   `2`   `-2`   `3`   `-1`   `2`   `-1`   `3`

`=32`

Determinants are `6,14,32`

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

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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`

=

6 -2 2 -2 3 -1 2 -1 3

`l_(11)=sqrt(a_(11))=sqrt(6)=2.4495`

`l_(21)=(a_(21))/l_(11)=(-2)/(2.4495)=-0.8165`

`l_(22)=sqrt(a_(22)-l_(21)^2)=sqrt(3-(-0.8165)^2)=sqrt(3-0.6667)=1.5275`

`l_(31)=(a_(31))/l_(11)=(2)/(2.4495)=0.8165`

`l_(32)=(a_(32)-l_(31) xx l_(21))/l_(22)=(-1-(0.8165)xx(-0.8165))/(1.5275)=(-1-(-0.6667))/(1.5275)=-0.2182`

`l_(33)=sqrt(a_(33)-l_(31)^2-l_(32)^2)=sqrt(3-(0.8165)^2-(-0.2182)^2)=sqrt(3-0.7143)=1.5119`

So `L`

=

`l_(11)` `0` `0` `l_(21)` `l_(22)` `0` `l_(31)` `l_(32)` `l_(33)`

=

2.4495 0 0 -0.8165 1.5275 0 0.8165 -0.2182 1.5119

`L xx L^T`

=

2.4495 0 0 -0.8165 1.5275 0 0.8165 -0.2182 1.5119

`xx`

2.4495 -0.8165 0.8165 0 1.5275 -0.2182 0 0 1.5119

=

6 -2 2 -2 3 -1 2 -1 3

and `A`

=

6 -2 2 -2 3 -1 2 -1 3

Solution is possible.

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