# Math Help - pwer method

1. How can I solve a problem
find by power method larger egien value of the matrix
[ 4 1 -1 ]
2 3 -1
-2 1 5

Help me as soon as possible4

2. Originally Posted by emn
How can I solve a problem
find by power method larger egien value of the matrix
[ 4 1 -1 ]
2 3 -1
-2 1 5

Help me as soon as possible4
Let $\bold{A}$ be a matrix with an eigen value of absolute value greater than that of its other eigen values, then the ittereation:

$\bold{b}_{k+1}=\frac{\bold{Ab}_{k}}{|\bold{Ab}_k|} \ \ \ \ \ ...(1)$

converges to the eigen vector corresponding to the eigen vector of largest absolute value.

If $\bold{b}$ is this limit the largest eigen value can be found from any component of the equation:

$\bold{Ab}=\lambda \bold{b} \ \ \ \ \ \ \ ...(2)$

So you start with a random vector $\bold{b}_0$ and apply the itteration in $(1)$ to it untill a satisfactory lefel of convergence is observed, then use any non-zero component of $\bold{b}$ and equation $(2)$ to find the eigen value.

Now this requires a lot of computation and I expect you are supposed to use some sort of tool to perform the itteration.

RonL

3. Originally Posted by CaptainBlack
Let $\bold{A}$ be a matrix with an eigen value of absolute value greater than that of its other eigen values, then the ittereation:

$\bold{b}_{k+1}=\frac{\bold{Ab}_{k}}{|\bold{Ab}_k|} \ \ \ \ \ ...(1)$

converges to the eigen vector corresponding to the eigen vector of largest absolute value.

If $\bold{b}$ is this limit the largest eigen value can be found from any component of the equation:

$\bold{Ab}=\lambda \bold{b} \ \ \ \ \ \ \ ...(2)$

So you start with a random vector $\bold{b}_0$ and apply the itteration in $(1)$ to it untill a satisfactory lefel of convergence is observed, then use any non-zero component of $\bold{b}$ and equation $(2)$ to find the eigen value.

Now this requires a lot of computation and I expect you are supposed to use some sort of tool to perform the itteration.

RonL
This can be mechanised as in the following:

Code:
>A=[4,1,-1;2,3,-1;-2,1,5]
4             1            -1
2             3            -1
-2             1             5
>
>function itterate(A)
$bm=random(3,1);$  repeat
    bp=A.bm;
$bp=bp/sqrt(sum((bp^2)'));$    l=A.bp/bp;
$lambda=l(1);$    if totalmax(abs(bm-bp))<1e-6
$break$    endif
$bm=bp;$  end
$return {lambda,bp}$endfunction
>
>{lambda,b}=itterate(A);lambda," ",b
6

0.577351
0.577351
-0.577349
>
RonL