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

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


A can be diagonalized if there exists an invertible matrix P and diagonal matrix D such that `A=PDP^-1`


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




Find eigenvalues of the matrix `A`

`|A-lamdaI|=0`

 `(8-lamda)`  `-6`  `2` 
 `-6`  `(7-lamda)`  `-4` 
 `2`  `-4`  `(3-lamda)` 
 = 0


`:.(8-lamda)((7-lamda) × (3-lamda) - (-4) × (-4))-(-6)((-6) × (3-lamda) - (-4) × 2)+2((-6) × (-4) - (7-lamda) × 2)=0`

`:.(8-lamda)((21-10lamda+lamda^2)-16)+6((-18+6lamda)-(-8))+2(24-(14-2lamda))=0`

`:.(8-lamda)(5-10lamda+lamda^2)+6(-10+6lamda)+2(10+2lamda)=0`

`:. (40-85lamda+18lamda^2-lamda^3)+(-60+36lamda)+(20+4lamda)=0`

`:.(-lamda^3+18lamda^2-45lamda)=0`

`:.-lamda(lamda-3)(lamda-15)=0`

`:.lamda=0 or(lamda-3)=0 or(lamda-15)=0 `

`:.` The eigenvalues of the matrix `A` are given by `lamda=0,3,15`,



1. Eigenvectors for `lamda=0`


1. Eigenvectors for `lamda=0`

`A-lamdaI = `
`8``-6``2`
`-6``7``-4`
`2``-4``3`
 - `0` 
`1``0``0`
`0``1``0`
`0``0``1`


 = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`


Now, reduce this matrix
`R_1 larr R_1-:8`

 = 
 `1` `1=8-:8`
`R_1 larr R_1-:8`
 `-3/4` `-3/4=-6-:8`
`R_1 larr R_1-:8`
 `1/4` `1/4=2-:8`
`R_1 larr R_1-:8`
`-6``7``-4`
`2``-4``3`


`R_2 larr R_2+6xx R_1`

 = 
`1``-3/4``1/4`
 `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1`
 `5/2` `5/2=7+6xx-3/4`
`R_2 larr R_2+6xx R_1`
 `-5/2` `-5/2=-4+6xx1/4`
`R_2 larr R_2+6xx R_1`
`2``-4``3`


`R_3 larr R_3-2xx R_1`

 = 
`1``-3/4``1/4`
`0``5/2``-5/2`
 `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1`
 `-5/2` `-5/2=-4-2xx-3/4`
`R_3 larr R_3-2xx R_1`
 `5/2` `5/2=3-2xx1/4`
`R_3 larr R_3-2xx R_1`


`R_2 larr R_2-:5/2`

 = 
`1``-3/4``1/4`
 `0` `0=0-:5/2`
`R_2 larr R_2-:5/2`
 `1` `1=5/2-:5/2`
`R_2 larr R_2-:5/2`
 `-1` `-1=-5/2-:5/2`
`R_2 larr R_2-:5/2`
`0``-5/2``5/2`


`R_1 larr R_1+(3/4)xx R_2`

 = 
 `1` `1=1+(3/4)xx0`
`R_1 larr R_1+(3/4)xx R_2`
 `0` `0=-3/4+(3/4)xx1`
`R_1 larr R_1+(3/4)xx R_2`
 `-1/2` `-1/2=1/4+(3/4)xx-1`
`R_1 larr R_1+(3/4)xx R_2`
`0``1``-1`
`0``-5/2``5/2`


`R_3 larr R_3+(5/2)xx R_2`

 = 
`1``0``-1/2`
`0``1``-1`
 `0` `0=0+(5/2)xx0`
`R_3 larr R_3+(5/2)xx R_2`
 `0` `0=-5/2+(5/2)xx1`
`R_3 larr R_3+(5/2)xx R_2`
 `0` `0=5/2+(5/2)xx-1`
`R_3 larr R_3+(5/2)xx R_2`


The system associated with the eigenvalue `lamda=0`

`(A-0I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-1/2`
`0``1``-1`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-(1/2)x_3=0,x_2-x_3=0`

`=>x_1=(1/2)x_3,x_2=x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=0` is

`v=`
`(1/2)x_3`
`x_3`
`x_3`


Let `x_3=1`

`v_1=`
`1/2`
`1`
`1`
`v_1=`
`1/2`
`1`
`1`


2. Eigenvectors for `lamda=3`




2. Eigenvectors for `lamda=3`

`A-lamdaI = `
`8``-6``2`
`-6``7``-4`
`2``-4``3`
 - `3` 
`1``0``0`
`0``1``0`
`0``0``1`


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


 = 
`5``-6``2`
`-6``4``-4`
`2``-4``0`


Now, reduce this matrix
interchanging rows `R_1 harr R_2`

 = 
`-6``4``-4`
`5``-6``2`
`2``-4``0`


`R_1 larr R_1-:-6`

 = 
 `1` `1=-6-:-6`
`R_1 larr R_1-:-6`
 `-2/3` `-2/3=4-:-6`
`R_1 larr R_1-:-6`
 `2/3` `2/3=-4-:-6`
`R_1 larr R_1-:-6`
`5``-6``2`
`2``-4``0`


`R_2 larr R_2-5xx R_1`

 = 
`1``-2/3``2/3`
 `0` `0=5-5xx1`
`R_2 larr R_2-5xx R_1`
 `-8/3` `-8/3=-6-5xx-2/3`
`R_2 larr R_2-5xx R_1`
 `-4/3` `-4/3=2-5xx2/3`
`R_2 larr R_2-5xx R_1`
`2``-4``0`


`R_3 larr R_3-2xx R_1`

 = 
`1``-2/3``2/3`
`0``-8/3``-4/3`
 `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1`
 `-8/3` `-8/3=-4-2xx-2/3`
`R_3 larr R_3-2xx R_1`
 `-4/3` `-4/3=0-2xx2/3`
`R_3 larr R_3-2xx R_1`


`R_2 larr R_2-:-8/3`

 = 
`1``-2/3``2/3`
 `0` `0=0-:-8/3`
`R_2 larr R_2-:-8/3`
 `1` `1=-8/3-:-8/3`
`R_2 larr R_2-:-8/3`
 `1/2` `1/2=-4/3-:-8/3`
`R_2 larr R_2-:-8/3`
`0``-8/3``-4/3`


`R_1 larr R_1+(2/3)xx R_2`

 = 
 `1` `1=1+(2/3)xx0`
`R_1 larr R_1+(2/3)xx R_2`
 `0` `0=-2/3+(2/3)xx1`
`R_1 larr R_1+(2/3)xx R_2`
 `1` `1=2/3+(2/3)xx1/2`
`R_1 larr R_1+(2/3)xx R_2`
`0``1``1/2`
`0``-8/3``-4/3`


`R_3 larr R_3+(8/3)xx R_2`

 = 
`1``0``1`
`0``1``1/2`
 `0` `0=0+(8/3)xx0`
`R_3 larr R_3+(8/3)xx R_2`
 `0` `0=-8/3+(8/3)xx1`
`R_3 larr R_3+(8/3)xx R_2`
 `0` `0=-4/3+(8/3)xx1/2`
`R_3 larr R_3+(8/3)xx R_2`


The system associated with the eigenvalue `lamda=3`

`(A-3I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``1`
`0``1``1/2`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1+x_3=0,x_2+(1/2)x_3=0`

`=>x_1=-x_3,x_2=-(1/2)x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=3` is

`v=`
`-x_3`
`-(1/2)x_3`
`x_3`


Let `x_3=1`

`v_2=`
`-1`
`-1/2`
`1`
`v_2=`
`-1`
`-1/2`
`1`


3. Eigenvectors for `lamda=15`




3. Eigenvectors for `lamda=15`

`A-lamdaI = `
`8``-6``2`
`-6``7``-4`
`2``-4``3`
 - `15` 
`1``0``0`
`0``1``0`
`0``0``1`


 = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`
 - 
`15``0``0`
`0``15``0`
`0``0``15`


 = 
`-7``-6``2`
`-6``-8``-4`
`2``-4``-12`


Now, reduce this matrix
`R_1 larr R_1-:-7`

 = 
 `1` `1=-7-:-7`
`R_1 larr R_1-:-7`
 `6/7` `6/7=-6-:-7`
`R_1 larr R_1-:-7`
 `-2/7` `-2/7=2-:-7`
`R_1 larr R_1-:-7`
`-6``-8``-4`
`2``-4``-12`


`R_2 larr R_2+6xx R_1`

 = 
`1``6/7``-2/7`
 `0` `0=-6+6xx1`
`R_2 larr R_2+6xx R_1`
 `-20/7` `-20/7=-8+6xx6/7`
`R_2 larr R_2+6xx R_1`
 `-40/7` `-40/7=-4+6xx-2/7`
`R_2 larr R_2+6xx R_1`
`2``-4``-12`


`R_3 larr R_3-2xx R_1`

 = 
`1``6/7``-2/7`
`0``-20/7``-40/7`
 `0` `0=2-2xx1`
`R_3 larr R_3-2xx R_1`
 `-40/7` `-40/7=-4-2xx6/7`
`R_3 larr R_3-2xx R_1`
 `-80/7` `-80/7=-12-2xx-2/7`
`R_3 larr R_3-2xx R_1`


interchanging rows `R_2 harr R_3`

 = 
`1``6/7``-2/7`
`0``-40/7``-80/7`
`0``-20/7``-40/7`


`R_2 larr R_2-:-40/7`

 = 
`1``6/7``-2/7`
 `0` `0=0-:-40/7`
`R_2 larr R_2-:-40/7`
 `1` `1=-40/7-:-40/7`
`R_2 larr R_2-:-40/7`
 `2` `2=-80/7-:-40/7`
`R_2 larr R_2-:-40/7`
`0``-20/7``-40/7`


`R_1 larr R_1-(6/7)xx R_2`

 = 
 `1` `1=1-(6/7)xx0`
`R_1 larr R_1-(6/7)xx R_2`
 `0` `0=6/7-(6/7)xx1`
`R_1 larr R_1-(6/7)xx R_2`
 `-2` `-2=-2/7-(6/7)xx2`
`R_1 larr R_1-(6/7)xx R_2`
`0``1``2`
`0``-20/7``-40/7`


`R_3 larr R_3+(20/7)xx R_2`

 = 
`1``0``-2`
`0``1``2`
 `0` `0=0+(20/7)xx0`
`R_3 larr R_3+(20/7)xx R_2`
 `0` `0=-20/7+(20/7)xx1`
`R_3 larr R_3+(20/7)xx R_2`
 `0` `0=-40/7+(20/7)xx2`
`R_3 larr R_3+(20/7)xx R_2`


The system associated with the eigenvalue `lamda=15`

`(A-15I)`
`x_1`
`x_2`
`x_3`
 = 
`1``0``-2`
`0``1``2`
`0``0``0`
 
`x_1`
`x_2`
`x_3`
 = 
`0`
`0`
`0`


`=>x_1-2x_3=0,x_2+2x_3=0`

`=>x_1=2x_3,x_2=-2x_3`

`:.` eigenvectors corresponding to the eigenvalue `lamda=15` is

`v=`
`2x_3`
`-2x_3`
`x_3`


Let `x_3=1`

`v_3=`
`2`
`-2`
`1`
`v_3=`
`2`
`-2`
`1`




2. The eigenvectors compose the columns of matrix P
`:.P` = 
`1/2``-1``2`
`1``-1/2``-2`
`1``1``1`




1. The diagonal matrix D is composed of the eigenvalues
`:.D` = 
`0``0``0`
`0``3``0`
`0``0``15`


3. Now find `P^-1`

`|P|` = 
 `1/2`  `-1`  `2` 
 `1`  `-1/2`  `-2` 
 `1`  `1`  `1` 


 =
 `1/2` × 
 `-1/2`  `-2` 
 `1`  `1` 
 `+1` × 
 `1`  `-2` 
 `1`  `1` 
 `+2` × 
 `1`  `-1/2` 
 `1`  `1` 


`=1/2 xx (-1/2 × 1 - (-2) × 1) +1 xx (1 × 1 - (-2) × 1) +2 xx (1 × 1 - (-1/2) × 1)`

`=1/2 xx (-1/2 +2) +1 xx (1 +2) +2 xx (1 +1/2)`

`=1/2 xx (3/2) +1 xx (3) +2 xx (3/2)`

`= 3/4 +3 +3`

`=27/4`


`Adj(P)` = 
Adj
`1/2``-1``2`
`1``-1/2``-2`
`1``1``1`


 = 
 + 
 `-1/2`  `-2` 
 `1`  `1` 
 - 
 `1`  `-2` 
 `1`  `1` 
 + 
 `1`  `-1/2` 
 `1`  `1` 
 - 
 `-1`  `2` 
 `1`  `1` 
 + 
 `1/2`  `2` 
 `1`  `1` 
 - 
 `1/2`  `-1` 
 `1`  `1` 
 + 
 `-1`  `2` 
 `-1/2`  `-2` 
 - 
 `1/2`  `2` 
 `1`  `-2` 
 + 
 `1/2`  `-1` 
 `1`  `-1/2` 
T


 = 
`+((-1/2) × 1 - (-2) × 1)``-(1 × 1 - (-2) × 1)``+(1 × 1 - (-1/2) × 1)`
`-((-1) × 1 - 2 × 1)``+((1/2) × 1 - 2 × 1)``-((1/2) × 1 - (-1) × 1)`
`+((-1) × (-2) - 2 × (-1/2))``-((1/2) × (-2) - 2 × 1)``+((1/2) × (-1/2) - (-1) × 1)`
T


 = 
`+(-1/2 +2)``-(1 +2)``+(1 +1/2)`
`-(-1 -2)``+(1/2 -2)``-(1/2 +1)`
`+(2 +1)``-(-1 -2)``+(-1/4 +1)`
T


 = 
`3/2``-3``3/2`
`3``-3/2``-3/2`
`3``3``3/4`
T


 = 
`3/2``3``3`
`-3``-3/2``3`
`3/2``-3/2``3/4`


`"Now, "P^(-1)=1/|P| × Adj(P)`

 = `1/(27/4)` ×
`3/2``3``3`
`-3``-3/2``3`
`3/2``-3/2``3/4`


 = 
`2/9``4/9``4/9`
`-4/9``-2/9``4/9`
`2/9``-2/9``1/9`


`:.P^-1` = 
`2/9``4/9``4/9`
`-4/9``-2/9``4/9`
`2/9``-2/9``1/9`




4. Now verify that `A=PDP^(-1)`

`P×D`=
`1/2``-1``2`
`1``-1/2``-2`
`1``1``1`
×
`0``0``0`
`0``3``0`
`0``0``15`


=
`1/2×0-1×0+2×0``1/2×0-1×3+2×0``1/2×0-1×0+2×15`
`1×0-1/2×0-2×0``1×0-1/2×3-2×0``1×0-1/2×0-2×15`
`1×0+1×0+1×0``1×0+1×3+1×0``1×0+1×0+1×15`


=
`0+0+0``0-3+0``0+0+30`
`0+0+0``0-3/2+0``0+0-30`
`0+0+0``0+3+0``0+0+15`


=
`0``-3``30`
`0``-3/2``-30`
`0``3``15`


`(P × D)×(P^-1)`=
`0``-3``30`
`0``-3/2``-30`
`0``3``15`
×
`2/9``4/9``4/9`
`-4/9``-2/9``4/9`
`2/9``-2/9``1/9`


=
`0×2/9-3×-4/9+30×2/9``0×4/9-3×-2/9+30×-2/9``0×4/9-3×4/9+30×1/9`
`0×2/9-3/2×-4/9-30×2/9``0×4/9-3/2×-2/9-30×-2/9``0×4/9-3/2×4/9-30×1/9`
`0×2/9+3×-4/9+15×2/9``0×4/9+3×-2/9+15×-2/9``0×4/9+3×4/9+15×1/9`


=
`0+4/3+20/3``0+2/3-20/3``0-4/3+10/3`
`0+2/3-20/3``0+1/3+20/3``0-2/3-10/3`
`0-4/3+10/3``0-2/3-10/3``0+4/3+5/3`


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


`:.P*D*P^-1` = 
`8``-6``2`
`-6``7``-4`
`2``-4``3`




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







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