We're going to find the largest eigenvalue and eigenvector for the
matrix I did in class.  Do this using the power method.
> A = [ 1,3;3,4];

Choose a random first guess:
> z = rand(2,1)
z =
    0.9501
    0.2311
Apply the matrix to z_0.  This gives us w_1.
> w = A*z
w =
    1.6435
    3.7749

Define z_1 by making w_1 of unit length:
> z = w/sqrt(sum(w.*w))
z =
    0.3992
    0.9169

Apply the matrix to z_1.  This gives us w_2.
> w = A*z
w =
    3.1498
    4.8650

find the approximate eigenvalue using the first component of w_2 and z_1 
> w(1)/z(1)
ans =
    7.8905

find the approximate eigenvalue using the second component of w_2 and z_1 
> w(2)/z(2)
ans =
    5.3061

They're not very close.  Let's keep going.
Define z_2 by making w_2 of unit length:
> z = w/sqrt(sum(w.*w))
z =
    0.5435
    0.8394

Apply the matrix to z_2.  This gives us w_3.
> w = A*z
w =
    3.0618
    4.9881

find the approximate eigenvalue using the first component of w_2 and z_1 
> w(1)/z(1)
ans =
    5.6337
find the approximate eigenvalue using the second component of w_2 and z_1 
> w(2)/z(2)
ans =
    5.9423
They're closer...

Define z_3 by making w_3 of unit length:
> z = w/sqrt(sum(w.*w))
z =
    0.5231
    0.8523

Apply the matrix to z_2.  This gives us w_4.
> w = A*z
w =
    3.0799
    4.9784

find the approximate eigenvalue using the first component of w_2 and z_1 
> w(1)/z(1)
ans =
    5.8875
find the approximate eigenvalue using the second component of w_2 and z_1 
> w(2)/z(2)
ans =
    5.8414
They're closer still...

Now I'll do 100 applications and will end up with w_105 and z_104 (or
something)
> for i=1:100, w = A*z; z = w/sqrt(sum(w.*w)); end
> w = A*z
w =
    3.0777
    4.9798
> format long e

find the approximate eigenvalue using the first component of w_2 and z_1 
> w(1)/z(1)
ans =
    5.854101966249684e+000
find the approximate eigenvalue using the second component of w_2 and z_1 
> w(2)/z(2)
ans =
    5.854101966249685e+000
What's the difference?
> w(1)/z(1)-w(2)/z(2)
ans =
   -8.881784197001252e-016
It's at round-off.  Compare it to the exact eigenvalue:
> (5+3*sqrt(5))/2
ans =
    5.854101966249685e+000

Now let's do another example.  
> A = rand(10,10); z = rand(10,1);
This time, I'll just print out the first 20 approximations of the
largest eigenvalue.  I'll use the first component of z and w to 
define those approximations...
> for i=1:20, w = A*z; w(1)/z(1), z = w/sqrt(sum(w.*w)); end
ans =
    6.051586826088908e+000
ans =
    5.435530397147399e+000
ans =
    5.311031263643748e+000
ans =
    5.342101733909655e+000
ans =
    5.333892658002270e+000
ans =
    5.333979763866569e+000
ans =
    5.333843519358664e+000
ans =
    5.333860028497632e+000
ans =
    5.333860588474614e+000
ans =
    5.333860903524015e+000
ans =
    5.333860860194060e+000
ans =
    5.333860851754611e+000
ans =
    5.333860850631273e+000
ans =
    5.333860850680542e+000
ans =
    5.333860850725885e+000
ans =
    5.333860850729825e+000
ans =
    5.333860850729944e+000
ans =
    5.333860850729725e+000
ans =
    5.333860850729706e+000
ans =
    5.333860850729705e+000

After 20 iterations, we've converged to the level of round-off...