Hi
You should try to compute to see what happens and then you'll be able to show by induction what is in function of .
I expect that the eigenvalues that you got were 1, 0.5 and 0.25?
The eigenvectors are found by solving
[0.5-y 0 0;0.5 1-y 0.75;0 0 0.25-y]x=0 (where y is an eigenvalue)
or in matlab by typing
[V,D]=eig(A)
To give the three eigenvectors:
[0 1 0], [0.707 -0.707 0], [0 -0.707 0.707]
Now A = V*D*inv(V) and it is easy to calculate A^n
Where's your trouble here?
Let the eigenvector be .
: , , .
Therefore the eigenvector has the form and a normalised eigenvector is .
: , , (since c = 0).
Therefore the eigenvector has the form and a normalised eigenvector is .
You should try finding the eigenvector for . A normalised eigenvector is .
So now you can easily diagonalise your matrix (well, you still have to get the inverse of a 3 x 3 matrix but that's just a mechanical process) and then easily raise to any power you want.