1. ## matrix exponents

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

2. Hi

You should try to compute $A^2,\,A^3,\,A^4$ to see what happens and then you'll be able to show by induction what is $A^n$ in function of $n$.

3. 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

4. Originally Posted by Jojo123
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

Let the eigenvector be $\left( \begin{array}{c}
a \\
b \\
c \end{array} \right)$
.

${\color{red}\lambda = 1}$: $\frac{a}{2} = a \Rightarrow a = 0$, $b = b$, $\frac{c}{4} = c \Rightarrow c = 0$.

Therefore the eigenvector has the form $\left( \begin{array}{c}
0 \\
b \\
0 \end{array} \right)$
and a normalised eigenvector is $\left( \begin{array}{c}
0 \\
1 \\
0 \end{array} \right)$
.

${\color{red}\lambda = \frac{1}{2}}$: $\frac{a}{2} = \frac{a}{2} \Rightarrow a = a$, $\frac{c}{4} = \frac{c}{2} \Rightarrow c = 0$, $\frac{a}{2} + b = \frac{b}{2} \Rightarrow b = -a$ (since c = 0).

Therefore the eigenvector has the form $\left( \begin{array}{c}
a \\
-a \\
0 \end{array} \right)$
and a normalised eigenvector is $\frac{1}{\sqrt{2}} \left( \begin{array}{c}
1 \\
-1\\
0 \end{array} \right)$
.

You should try finding the eigenvector for $\lambda = \frac{1}{4}$. A normalised eigenvector is $\frac{1}{\sqrt{2}} \left( \begin{array}{c}
0 \\
-1\\
1 \end{array} \right)$
.

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.

5. Originally Posted by Kiwi_Dave
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 Mr F adds: to two (maybe three) decimal places of accuracy.
..