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

Solution:

`A=`	1 2 0 -2 1 2 1 3 1

`x_0=`

1 1 1

`1^(st)` iteration :

Multiply the matrix by the vector

`A x_0 =`

1 2 0 -2 1 2 1 3 1

1 1 1

`=`

3 1 5

Normalize the resulting vector
To normalize, divide each element of vector by its largest absolute value, which is `5`

`x_1=`

`1/5`

3 1 5

`=`

0.6 0.2 1

`2^(nd)` iteration :

Repeat the multiplication

`A x_1 =`

1 2 0 -2 1 2 1 3 1

0.6 0.2 1

`=`

1 1 2.2

Normalize again
The largest absolute value is `2.2`

`x_2=`

`1/2.2`

1 1 2.2

`=`

0.4545 0.4545 1

`3^(rd)` iteration :

Repeat the multiplication

`A x_2 =`

1 2 0 -2 1 2 1 3 1

0.4545 0.4545 1

`=`

1.3636 1.5455 2.8182

Normalize again
The largest absolute value is `2.8182`

`x_3=`

`1/2.8182`

1.3636 1.5455 2.8182

`=`

0.4839 0.5484 1

`4^(th)` iteration :

Repeat the multiplication

`A x_3 =`

1 2 0 -2 1 2 1 3 1

0.4839 0.5484 1

`=`

1.5806 1.5806 3.129

Normalize again
The largest absolute value is `3.129`

`x_4=`

`1/3.129`

1.5806 1.5806 3.129

`=`

0.5052 0.5052 1

`5^(th)` iteration :

Repeat the multiplication

`A x_4 =`

1 2 0 -2 1 2 1 3 1

0.5052 0.5052 1

`=`

1.5155 1.4948 3.0206

Normalize again
The largest absolute value is `3.0206`

`x_5=`

`1/3.0206`

1.5155 1.4948 3.0206

`=`

0.5017 0.4949 1

`6^(th)` iteration :

Repeat the multiplication

`A x_5 =`

1 2 0 -2 1 2 1 3 1

0.5017 0.4949 1

`=`

1.4915 1.4915 2.9863

Normalize again
The largest absolute value is `2.9863`

`x_6=`

`1/2.9863`

1.4915 1.4915 2.9863

`=`

0.4994 0.4994 1

`7^(th)` iteration :

Repeat the multiplication

`A x_6 =`

1 2 0 -2 1 2 1 3 1

0.4994 0.4994 1

`=`

1.4983 1.5006 2.9977

Normalize again
The largest absolute value is `2.9977`

`x_7=`

`1/2.9977`

1.4983 1.5006 2.9977

`=`

0.4998 0.5006 1

`8^(th)` iteration :

Repeat the multiplication

`A x_7 =`

1 2 0 -2 1 2 1 3 1

0.4998 0.5006 1

`=`

1.501 1.501 3.0015

Normalize again
The largest absolute value is `3.0015`

`x_8=`

`1/3.0015`

1.501 1.501 3.0015

`=`

0.5001 0.5001 1

`9^(th)` iteration :

Repeat the multiplication

`A x_8 =`

1 2 0 -2 1 2 1 3 1

0.5001 0.5001 1

`=`

1.5002 1.4999 3.0003

Normalize again
The largest absolute value is `3.0003`

`x_9=`

`1/3.0003`

1.5002 1.4999 3.0003

`=`

0.5 0.4999 1

`10^(th)` iteration :

Repeat the multiplication

`A x_9 =`

1 2 0 -2 1 2 1 3 1

0.5 0.4999 1

`=`

1.4999 1.4999 2.9998

Normalize again
The largest absolute value is `2.9998`

`x_10=`

`1/2.9998`

1.4999 1.4999 2.9998

`=`

0.5 0.5 1

`:.` The dominant eigenvalue `lamda=2.9998~=3`

and the dominant eigenvector is :

`=`

0.5 0.5 1

---

---