Find Power Method for finding dominant eigenvalue ...
`[[3,2],[1,4]]`
`x_0` = -1,1

Solution:

`A=`

3 2 1 4

`x_0=`

-1 1

`1^(st)` iteration :

Multiply the matrix by the vector

`A x_0 =`

3 2 1 4

-1 1

`=`

-1 3

Normalize the resulting vector
To normalize, divide each element of vector by its largest absolute value, which is `3`

`x_1=`

`1/3`

-1 3

`=`

-0.3333 1

`2^(nd)` iteration :

Repeat the multiplication

`A x_1 =`

3 2 1 4

-0.3333 1

`=`

1 3.6667

Normalize again
The largest absolute value is `3.6667`

`x_2=`

`1/3.6667`

1 3.6667

`=`

0.2727 1

`3^(rd)` iteration :

Repeat the multiplication

`A x_2 =`

3 2 1 4

0.2727 1

`=`

2.8182 4.2727

Normalize again
The largest absolute value is `4.2727`

`x_3=`

`1/4.2727`

2.8182 4.2727

`=`

0.6596 1

`4^(th)` iteration :

Repeat the multiplication

`A x_3 =`

3 2 1 4

0.6596 1

`=`

3.9787 4.6596

Normalize again
The largest absolute value is `4.6596`

`x_4=`

`1/4.6596`

3.9787 4.6596

`=`

0.8539 1

`5^(th)` iteration :

Repeat the multiplication

`A x_4 =`

3 2 1 4

0.8539 1

`=`

4.5616 4.8539

Normalize again
The largest absolute value is `4.8539`

`x_5=`

`1/4.8539`

4.5616 4.8539

`=`

0.9398 1

`6^(th)` iteration :

Repeat the multiplication

`A x_5 =`

3 2 1 4

0.9398 1

`=`

4.8194 4.9398

Normalize again
The largest absolute value is `4.9398`

`x_6=`

`1/4.9398`

4.8194 4.9398

`=`

0.9756 1

`7^(th)` iteration :

Repeat the multiplication

`A x_6 =`

3 2 1 4

0.9756 1

`=`

4.9269 4.9756

Normalize again
The largest absolute value is `4.9756`

`x_7=`

`1/4.9756`

4.9269 4.9756

`=`

0.9902 1

`8^(th)` iteration :

Repeat the multiplication

`A x_7 =`

3 2 1 4

0.9902 1

`=`

4.9706 4.9902

Normalize again
The largest absolute value is `4.9902`

`x_8=`

`1/4.9902`

4.9706 4.9902

`=`

0.9961 1

`9^(th)` iteration :

Repeat the multiplication

`A x_8 =`

3 2 1 4

0.9961 1

`=`

4.9882 4.9961

Normalize again
The largest absolute value is `4.9961`

`x_9=`

`1/4.9961`

4.9882 4.9961

`=`

0.9984 1

`10^(th)` iteration :

Repeat the multiplication

`A x_9 =`

3 2 1 4

0.9984 1

`=`

4.9953 4.9984

Normalize again
The largest absolute value is `4.9984`

`x_10=`

`1/4.9984`

4.9953 4.9984

`=`

0.9994 1

`11^(th)` iteration :

Repeat the multiplication

`A x_10 =`

3 2 1 4

0.9994 1

`=`

4.9981 4.9994

Normalize again
The largest absolute value is `4.9994`

`x_11=`

`1/4.9994`

4.9981 4.9994

`=`

0.9997 1

`12^(th)` iteration :

Repeat the multiplication

`A x_11 =`

3 2 1 4

0.9997 1

`=`

4.9992 4.9997

Normalize again
The largest absolute value is `4.9997`

`x_12=`

`1/4.9997`

4.9992 4.9997

`=`

0.9999 1

`13^(th)` iteration :

Repeat the multiplication

`A x_12 =`

3 2 1 4

0.9999 1

`=`

4.9997 4.9999

Normalize again
The largest absolute value is `4.9999`

`x_13=`

`1/4.9999`

4.9997 4.9999

`=`

1 1

`:.` The dominant eigenvalue `lamda=4.9999~=5`

and the dominant eigenvector is :

`=`

1 1

`~=`

1 1

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