Find the limit a matrix raised to $n$ when $n$ goes to infinity

Let $A$ be a $3\times3$ matrix such that

$$A \left( \begin{array}{ccc} 1 \\ 2 \\ 1 \end{array} \right)=\left( \begin{array}{ccc} 1 \\ 2 \\ 1 \end{array} \right),~~~A \left( \begin{array}{ccc} 2 \\ 2 \\ 0 \end{array} \right)=\left( \begin{array}{ccc} 1 \\ 1 \\ 0 \end{array} \right),~~A \left( \begin{array}{ccc} 3 \\ 0 \\ 6 \end{array} \right)=\left( \begin{array}{ccc} -1 \\ 0 \\ 2 \end{array} \right)$$

Find $$\lim_{n\to\infty}A^n \left( \begin{array}{ccc} 6 \\ 7 \\ 0 \end{array} \right)$$

So, do I first find $A$ by letting A =$\left( \begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&i \end{array} \right)$ and using the given information to solve the corresponding linear equations and then solve the actual problem of finding the limit? Is there a more efficient way of doing this? Also, I am not quite sure how to find the limit so any hints would be greatly appreciated. Thanks!

• Are you sure about the $-$ sign ?
– user65203
Apr 18 '16 at 8:22
• @YvesDaoust : I hope he's not. Apr 18 '16 at 8:26

Hint: Write the vector $(6,7,0)^{T}$ as a linear combination of $(1,2,1)^{T}$, $(2,2,0)^{T}$, and $(3,0,6)^{T}$. Then use the formulas from the first line.

This will give an equation like $A(c_{1}v_{1}+c_{2}v_{2}+c_{3}v_{3}) = c_{1}Av_{1}+c_{2}Av_{2}+c_{3}Av_{3}$. Your formulas at the beginning will give a nice way to write $Av_{i}$ (for example, $Av_{1}=v_{1}$), and should give a clue on how to get an answer for general $n$. You can take the limit from here.

Use diagonalization. $A=PDP^{-1}$, where $P=\pmatrix{1&2&3\cr 2&2&0\cr 1&0&6\cr}$ and $D=\pmatrix{1&0&0\cr 0&1/2&0\cr 0&0&-1/3\cr}$. Then $$\lim_{n\to\infty} A^n \pmatrix{6\cr 7\cr 0\cr} = \lim_{n\to\infty} (PD^n P^{-1}) \cdot \pmatrix{6\cr 7\cr 0\cr} = P \lim_{n\to\infty} D^n \cdot P^{-1} \cdot \pmatrix{6\cr 7\cr 0\cr} = P \cdot \pmatrix{1&0&0\cr 0&0&0\cr 0&0&0\cr} \cdot P^{-1} \cdot \pmatrix{6\cr 7\cr 0\cr}.$$

• thanks! could you explain how you figured out the limit of the diagonal matrix? Apr 18 '16 at 8:34
• Since $D$ is diagonal, $\pmatrix{1&0&0\cr 0&1/2&0\cr 0&0&-1/3\cr}^n = \pmatrix{1^n&0&0\cr 0&(1/2)^n&0\cr 0&0&(-1/3)^n\cr}$. Now take the limit of each entry of $D^n$. Apr 18 '16 at 8:47

HINT...having found the matrix $A$, you will need to diagonalize it so you have $A=P^{-1}DP$, and then $$A^n=P^{-1}D^nP$$

• I said that already. Apr 18 '16 at 8:20
• @CarlHeckman: yes, I must have been typing out my response as you were posting yours. :) Apr 18 '16 at 8:23

Hint:

Notice that with the given vectors, assuming no $-$ sign,

$$Au=u,Av=2v,Aw=\frac13w,$$ so that

$$A^nu=u,A^nv=2^nv,A^nw=\frac1{3^n}w.$$

Then if you decompose the fourth vectors as a linear combination of $u,v,w$, you should easily see how it is transformed by $A^n$.

If we keep the $-$ sign,

$$Aw=au+bv+cw$$ for some coefficients $a,b,c$, and
$$A^2w=au+2bv+c(au+bv+cw)=a(1+c)u+b(2+c)v+c^2w,\\ A^3w=a(1+c)u+b2(2+c)v+c(a(1+c)u+b(2+c)v+c^2w)=\\ a(1+c)^2+b(2+c)^2v+c^3w.$$