Calculate limit of A^n for large n where A=[1/2 0 0; 1/2 1 3/4; 0 0 1/4]
i have found eigenvalues but i cant find the eigenvectors
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 $\displaystyle \left( \begin{array}{c}
a \\
b \\
c \end{array} \right)$.
$\displaystyle {\color{red}\lambda = 1}$: $\displaystyle \frac{a}{2} = a \Rightarrow a = 0$, $\displaystyle b = b$, $\displaystyle \frac{c}{4} = c \Rightarrow c = 0$.
Therefore the eigenvector has the form $\displaystyle \left( \begin{array}{c}
0 \\
b \\
0 \end{array} \right)$ and a normalised eigenvector is $\displaystyle \left( \begin{array}{c}
0 \\
1 \\
0 \end{array} \right)$.
$\displaystyle {\color{red}\lambda = \frac{1}{2}}$: $\displaystyle \frac{a}{2} = \frac{a}{2} \Rightarrow a = a$, $\displaystyle \frac{c}{4} = \frac{c}{2} \Rightarrow c = 0$, $\displaystyle \frac{a}{2} + b = \frac{b}{2} \Rightarrow b = -a$ (since c = 0).
Therefore the eigenvector has the form $\displaystyle \left( \begin{array}{c}
a \\
-a \\
0 \end{array} \right)$ and a normalised eigenvector is $\displaystyle \frac{1}{\sqrt{2}} \left( \begin{array}{c}
1 \\
-1\\
0 \end{array} \right)$.
You should try finding the eigenvector for $\displaystyle \lambda = \frac{1}{4}$. A normalised eigenvector is $\displaystyle \frac{1}{\sqrt{2}} \left( \begin{array}{c}
0 \\
-1\\
1 \end{array} \right)$.
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.