Taking limits of norms of a matrix raised to the nth power: Given a matrix 
$$ A =
        \begin{bmatrix}
        0 & 3  \\
        -2 & 5  \\
             \end{bmatrix}
$$
and a vector $x = \begin{bmatrix}1&0\end{bmatrix}$, 
compute $\lim_{n\to\infty} \frac{\|A^{n+2}(x)\|}{\|A^{n}(x)\|}$.
So, I've computed the eigenvalues and eigenvectors of $A$, and noted that since $S^{-1}AS = D$, where $D$ is the diagonal matrix with the eigenvalues of $A$ on the diagonal and the columns of $S$ are the eigenvectors of $A$, then $$A = SDS^{-1},$$
and that now $A^n$ is just simply $$A^n = SD^nS^{-1}.$$
But, how do I actually compute the above limit? I tried simplifying the norm expression and arrived at $\|A^2\|$ by "cancelling" an $\|A^n (x)\|$ from the numerator and denominator, since the norm function is multiplicative.  Is this valid?  I am not getting an answer that matches the solution.
My work at the end is $\|A^2\| = \| SD^2S^{-1}\|$.
Edit: I am sure that my answer is wrong, since I didn't even get to use the vector (1,0) (from the cancelling of $A^n(x)$.)
 A: Direct calculation gives
$$\|A^n x\|^2 = x' A^{\prime n} A^n x = x' S^{\prime -1}D^nS'SD^nS^{-1} x.$$
If the eigenvalues are $\lambda_1,\lambda_2$, then $D^n = \begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix}$, and so, letting $T = S'S$,
$$
D^n T D^n = \begin{pmatrix} \lambda_1^{2n} T_{11} & \sqrt{\lambda_1 \lambda_2}^{2n} T_{12} \\ \sqrt{\lambda_1 \lambda_2}^{2n} T_{21} & \lambda_2^{2n} T_{22} \end{pmatrix}.
$$
It follows that $\|A^n x\|^2 = \alpha \lambda_1^{2n} + \beta \sqrt{\lambda_1 \lambda_2}^{2n} + \gamma \lambda_2^{2n}$ for some $\alpha,\beta,\gamma$.
In your case $\lambda_1 = 3$ and $\lambda_2 = 2$, and so
$$
\frac{\|A^{n+2} x\|}{\|A^n x\|} = \sqrt{\frac{\alpha 3^{2n+4} + \beta \sqrt{6}^{2n+4} + \gamma 2^{2n+4}}{\alpha 3^{2n} + \beta \sqrt{6}^{2n} + \gamma 2^{2n}}}.
$$
If $\alpha \neq 0$, then the limit of this expression is
$$
\lim_{n\to\infty} \frac{\|A^{n+2} x\|}{\|A^n x\|} =
\lim_{n\to\infty} \sqrt{\frac{\alpha 3^{2n+4}(\tfrac{\beta}{\alpha}\sqrt{2/3}^{2n+4} + \tfrac{\gamma}{\alpha} (2/3)^{2n+4})}{\alpha 3^{2n}(\tfrac{\beta}{\alpha}\sqrt{2/3}^{2n} + \tfrac{\gamma}{\alpha} (2/3)^{2n})}} = 9.
$$
If $\alpha = 0$ and $\beta \neq 0$ then the limit is $6$, and if $\alpha = \beta = 0$ but $\gamma \neq 0$ then the limit is $4$. For a random $x$, $\alpha \neq 0$, and so the answer is probably $9$; to be sure, you'll have to calculate $\alpha$ explicitly.
