show that $ \lim_{n\rightarrow\infty} v_{n}$exists.

Question

For $$A=\left(\begin{array}{cc}2/3&1/3\\ 1/3&2/3\end{array}\right).$$ and $$v=\left(\begin{array}{cc}a\\ b\end{array}\right)$$

we definie a sequence of vectors $v_{1}=v$,$ v_{n+1}=Av_{n}$. show that $ \lim_{n\rightarrow\infty} v_{n}$exists and is equal to $$v=\left(\begin{array}{cc}{(a+b)}/2\\ (a+b)/2\end{array}\right)$$

this is the orginal question

im completely struck as i don't know from where i have to start ,,as i was taking $v_{n+1} = Av_{n}$ ,taking $v_{n+1} = v_n = l$ now i got $l= Al$ as i got $A= I $ it gives me contradictionss,,,,

Pliz help me

Thanks in Advance

Answer 1

Approach 1: Try to Diagonalize $A$. For instance, suppose that you were trying to solve the problem not with $A$, but with $$ B = \begin{bmatrix} 1/2 & 0 \\ 0 & 1 \end{bmatrix}.$$ In what way would it be easier?

Approach 2: If you have learned about Markov matrices, apply the results for these kinds of matrices.

Note: Your approach of trying to solve $A v = v$, should not give you a contradiction. It should give you that $v$ is a multiple of the vector $ \begin{bmatrix} 1 \\ 1 \end{bmatrix},$ but that does not tell you that it equals $ \begin{bmatrix} \frac{a+b}{2} \\ \frac{a+b}{2} \end{bmatrix}.$

Answer 2

$$ A=(1/3)BDB^{-1}$$ where $$B= \left(\begin{array}{cc}1&1\\ 1&-1\end{array}\right)$$ the matrix of eigenvectors,and

$$D=\left(\begin{array}{cc}3&0\\ 0&1\end{array}\right)$$

is the diagonal matrix of eigenvalues.

Thus $$ A^n=(1/3)^nBD^nB^{-1}$$

Note that we can find $A^nv$ using the above information.

Answer 3

Alt. hint:   let $v_n=(a_n, b_n)^T$ then the recurrence can be written as:

$$ 3a_{n+1}=2a_n+b_{n} \\ 3b_{n+1}=a_n+2b_{n} $$

Substituting $b_n=3a_{n+1}-2a_n$ from the first equation into the second one gives:

$$9a_{n+2}-6a_{n+1}=a_n+6a_{n+1}-4a_n \;\;\iff\;\; 3a_{n+2}=4a_{n+1}-a_n \quad \begin{cases}a_0 = a \\ a_1 = (2a+b)/3\end{cases} $$

The latter is a regular linear homogeneous recurrence with constant coefficents which can be solved with the usual methods, or for a shortcut telescope $a_{n+2}-a_{n+1}=\frac{1}{3}\left(a_{n+1}-a_n\right)$ twice.