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!
 A: 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.
A: 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
$$\lim_{n\to\infty} A^n \pmatrix{6\cr 7\cr 0\cr} = \lim_{n\to\infty} (PD^n P^{-1}) \cdot \pmatrix{6\cr 7\cr 0\cr} = P \lim_{n\to\infty} D^n \cdot P^{-1} \cdot \pmatrix{6\cr 7\cr 0\cr} = P \cdot \pmatrix{1&0&0\cr 0&0&0\cr 0&0&0\cr} \cdot P^{-1} \cdot \pmatrix{6\cr 7\cr 0\cr}.$$
A: HINT...having found the matrix $A$, you will need to diagonalize it so you have $A=P^{-1}DP$, and then $$A^n=P^{-1}D^nP$$
A: 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,
The third equation will read
$$Aw=au+bv+cw$$ for some coefficients $a,b,c$, and
$$A^2w=au+2bv+c(au+bv+cw)=a(1+c)u+b(2+c)v+c^2w,\\
A^3w=a(1+c)u+b2(2+c)v+c(a(1+c)u+b(2+c)v+c^2w)=\\
a(1+c)^2+b(2+c)^2v+c^3w.$$
You see the pattern.
