Find Power Method for finding dominant eigenvalue ...
`[[1,6,1],[1,2,0],[0,0,3]]`
`x_0` = 1,1,1

Solution:

`A=`	1 6 1 1 2 0 0 0 3

`x_0=`

1 1 1

`1^(st)` iteration :

Multiply the matrix by the vector

`A x_0 =`

1 6 1 1 2 0 0 0 3

1 1 1

`=`

8 3 3

Normalize the resulting vector
To normalize, divide each element of vector by its largest absolute value, which is `8`

`x_1=`

`1/8`

8 3 3

`=`

1 0.375 0.375

`2^(nd)` iteration :

Repeat the multiplication

`A x_1 =`

1 6 1 1 2 0 0 0 3

1 0.375 0.375

`=`

3.625 1.75 1.125

Normalize again
The largest absolute value is `3.625`

`x_2=`

`1/3.625`

3.625 1.75 1.125

`=`

1 0.4828 0.3103

`3^(rd)` iteration :

Repeat the multiplication

`A x_2 =`

1 6 1 1 2 0 0 0 3

1 0.4828 0.3103

`=`

4.2069 1.9655 0.931

Normalize again
The largest absolute value is `4.2069`

`x_3=`

`1/4.2069`

4.2069 1.9655 0.931

`=`

1 0.4672 0.2213

`4^(th)` iteration :

Repeat the multiplication

`A x_3 =`

1 6 1 1 2 0 0 0 3

1 0.4672 0.2213

`=`

4.0246 1.9344 0.6639

Normalize again
The largest absolute value is `4.0246`

`x_4=`

`1/4.0246`

4.0246 1.9344 0.6639

`=`

1 0.4807 0.165

`5^(th)` iteration :

Repeat the multiplication

`A x_4 =`

1 6 1 1 2 0 0 0 3

1 0.4807 0.165

`=`

4.0489 1.9613 0.4949

Normalize again
The largest absolute value is `4.0489`

`x_5=`

`1/4.0489`

4.0489 1.9613 0.4949

`=`

1 0.4844 0.1222

`6^(th)` iteration :

Repeat the multiplication

`A x_5 =`

1 6 1 1 2 0 0 0 3

1 0.4844 0.1222

`=`

4.0287 1.9688 0.3667

Normalize again
The largest absolute value is `4.0287`

`x_6=`

`1/4.0287`

4.0287 1.9688 0.3667

`=`

1 0.4887 0.091

`7^(th)` iteration :

Repeat the multiplication

`A x_6 =`

1 6 1 1 2 0 0 0 3

1 0.4887 0.091

`=`

4.0232 1.9774 0.2731

Normalize again
The largest absolute value is `4.0232`

`x_7=`

`1/4.0232`

4.0232 1.9774 0.2731

`=`

1 0.4915 0.0679

`8^(th)` iteration :

Repeat the multiplication

`A x_7 =`

1 6 1 1 2 0 0 0 3

1 0.4915 0.0679

`=`

4.0169 1.983 0.2036

Normalize again
The largest absolute value is `4.0169`

`x_8=`

`1/4.0169`

4.0169 1.983 0.2036

`=`

1 0.4937 0.0507

`9^(th)` iteration :

Repeat the multiplication

`A x_8 =`

1 6 1 1 2 0 0 0 3

1 0.4937 0.0507

`=`

4.0127 1.9873 0.1521

Normalize again
The largest absolute value is `4.0127`

`x_9=`

`1/4.0127`

4.0127 1.9873 0.1521

`=`

1 0.4953 0.0379

`10^(th)` iteration :

Repeat the multiplication

`A x_9 =`

1 6 1 1 2 0 0 0 3

1 0.4953 0.0379

`=`

4.0095 1.9905 0.1137

Normalize again
The largest absolute value is `4.0095`

`x_10=`

`1/4.0095`

4.0095 1.9905 0.1137

`=`

1 0.4965 0.0284

`11^(th)` iteration :

Repeat the multiplication

`A x_10 =`

1 6 1 1 2 0 0 0 3

1 0.4965 0.0284

`=`

4.0071 1.9929 0.0851

Normalize again
The largest absolute value is `4.0071`

`x_11=`

`1/4.0071`

4.0071 1.9929 0.0851

`=`

1 0.4973 0.0212

`12^(th)` iteration :

Repeat the multiplication

`A x_11 =`

1 6 1 1 2 0 0 0 3

1 0.4973 0.0212

`=`

4.0053 1.9947 0.0637

Normalize again
The largest absolute value is `4.0053`

`x_12=`

`1/4.0053`

4.0053 1.9947 0.0637

`=`

1 0.498 0.0159

`13^(th)` iteration :

Repeat the multiplication

`A x_12 =`

1 6 1 1 2 0 0 0 3

1 0.498 0.0159

`=`

4.004 1.996 0.0477

Normalize again
The largest absolute value is `4.004`

`x_13=`

`1/4.004`

4.004 1.996 0.0477

`=`

1 0.4985 0.0119

`14^(th)` iteration :

Repeat the multiplication

`A x_13 =`

1 6 1 1 2 0 0 0 3

1 0.4985 0.0119

`=`

4.003 1.997 0.0357

Normalize again
The largest absolute value is `4.003`

`x_14=`

`1/4.003`

4.003 1.997 0.0357

`=`

1 0.4989 0.0089

`15^(th)` iteration :

Repeat the multiplication

`A x_14 =`

1 6 1 1 2 0 0 0 3

1 0.4989 0.0089

`=`

4.0022 1.9978 0.0268

Normalize again
The largest absolute value is `4.0022`

`x_15=`

`1/4.0022`

4.0022 1.9978 0.0268

`=`

1 0.4992 0.0067

`16^(th)` iteration :

Repeat the multiplication

`A x_15 =`

1 6 1 1 2 0 0 0 3

1 0.4992 0.0067

`=`

4.0017 1.9983 0.0201

Normalize again
The largest absolute value is `4.0017`

`x_16=`

`1/4.0017`

4.0017 1.9983 0.0201

`=`

1 0.4994 0.005

`17^(th)` iteration :

Repeat the multiplication

`A x_16 =`

1 6 1 1 2 0 0 0 3

1 0.4994 0.005

`=`

4.0013 1.9987 0.0151

Normalize again
The largest absolute value is `4.0013`

`x_17=`

`1/4.0013`

4.0013 1.9987 0.0151

`=`

1 0.4995 0.0038

`18^(th)` iteration :

Repeat the multiplication

`A x_17 =`

1 6 1 1 2 0 0 0 3

1 0.4995 0.0038

`=`

4.0009 1.9991 0.0113

Normalize again
The largest absolute value is `4.0009`

`x_18=`

`1/4.0009`

4.0009 1.9991 0.0113

`=`

1 0.4996 0.0028

`19^(th)` iteration :

Repeat the multiplication

`A x_18 =`

1 6 1 1 2 0 0 0 3

1 0.4996 0.0028

`=`

4.0007 1.9993 0.0085

Normalize again
The largest absolute value is `4.0007`

`x_19=`

`1/4.0007`

4.0007 1.9993 0.0085

`=`

1 0.4997 0.0021

`20^(th)` iteration :

Repeat the multiplication

`A x_19 =`

1 6 1 1 2 0 0 0 3

1 0.4997 0.0021

`=`

4.0005 1.9995 0.0063

Normalize again
The largest absolute value is `4.0005`

`x_20=`

`1/4.0005`

4.0005 1.9995 0.0063

`=`

1 0.4998 0.0016

`21^(st)` iteration :

Repeat the multiplication

`A x_20 =`

1 6 1 1 2 0 0 0 3

1 0.4998 0.0016

`=`

4.0004 1.9996 0.0048

Normalize again
The largest absolute value is `4.0004`

`x_21=`

`1/4.0004`

4.0004 1.9996 0.0048

`=`

1 0.4999 0.0012

`22^(nd)` iteration :

Repeat the multiplication

`A x_21 =`

1 6 1 1 2 0 0 0 3

1 0.4999 0.0012

`=`

4.0003 1.9997 0.0036

Normalize again
The largest absolute value is `4.0003`

`x_22=`

`1/4.0003`

4.0003 1.9997 0.0036

`=`

1 0.4999 0.0009

`23^(rd)` iteration :

Repeat the multiplication

`A x_22 =`

1 6 1 1 2 0 0 0 3

1 0.4999 0.0009

`=`

4.0002 1.9998 0.0027

Normalize again
The largest absolute value is `4.0002`

`x_23=`

`1/4.0002`

4.0002 1.9998 0.0027

`=`

1 0.4999 0.0007

`24^(th)` iteration :

Repeat the multiplication

`A x_23 =`

1 6 1 1 2 0 0 0 3

1 0.4999 0.0007

`=`

4.0002 1.9998 0.002

Normalize again
The largest absolute value is `4.0002`

`x_24=`

`1/4.0002`

4.0002 1.9998 0.002

`=`

1 0.4999 0.0005

`25^(th)` iteration :

Repeat the multiplication

`A x_24 =`

1 6 1 1 2 0 0 0 3

1 0.4999 0.0005

`=`

4.0001 1.9999 0.0015

Normalize again
The largest absolute value is `4.0001`

`x_25=`

`1/4.0001`

4.0001 1.9999 0.0015

`=`

1 0.5 0.0004

`26^(th)` iteration :

Repeat the multiplication

`A x_25 =`

1 6 1 1 2 0 0 0 3

1 0.5 0.0004

`=`

4.0001 1.9999 0.0011

Normalize again
The largest absolute value is `4.0001`

`x_26=`

`1/4.0001`

4.0001 1.9999 0.0011

`=`

1 0.5 0.0003

`:.` The dominant eigenvalue `lamda=4.0001~=4`

and the dominant eigenvector is :

`=`

1 0.5 0.0003

`~=`

1 0.5 0

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