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!

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.
