Given symmetric matrix $A,$ find when $\lim_{k\to\infty} (x^tA^{2k}x)^{1/k}$ exists Suppose that $A$ is a symmetric $n \times n$ real matrix with distinct eigenvalues $\lambda_1, \dots, \lambda_l, \; (l \leq n)$.  Find the sets
\begin{equation}
X = \bigg\{ x \in \mathbb{R^n} \; : \; \lim_{k\to\infty} (x^t A^{2k}x)^{1/k} \, \text{ exists}\bigg\}
\end{equation}
and
\begin{equation}
L = \bigg\{ \lim_{k\to\infty} (x^t A^{2k}x)^{1/k} \; : \; x \in X \bigg\}
\end{equation}
where $x^t$ denotes the transpose of $x.$

I did some incorrect mathematics to determine that $X = \mathbb{R^n},$ although it may be right, my methods were wrong.  I wrote something like this, 
$A = MDM^{-1}$ where $D$ is diagonal and $M$ consists of $A$'s eigenvectors, let them be normalized.  Since $A$ is symmetric, $M^{-1} = M^t$. Then if we illegally say \begin{align} \lim_{k\to\infty} [(x^t M) D^{2k} (M^t x)]^{1/k} &= \lim_{k\to\infty} [D^{2k} (M^t x)^t (M^t x)]^{1/k}\\
&= \lim_{k\to\infty}  D^2 ||M^t x||^{1/k}\\
&= \lim_{k\to\infty}  D^2 ||x||^{1/k}\\
\end{align}
meaning that the limit exists for all $x \in \mathbb{R}$ because $||x||^{1/k}$ converges regardless of the value of $||x||.$  I realize that one can not just factor out a matrix to take a norm like that, my guess is that I should break this down into vector components at some point.  I had difficulty figuring out $L$ too, any suggestions?

You can start with the simple case of $n = 1$ and $A = (a)$. The limit 
$$ \lim_{k \to \infty} \left( x^t \cdot a^{2k} \cdot x \right)^{\frac{1}{k}} = \lim_{k \to \infty} x^{\frac{2}{k}} a^2 $$
exists for all $x \in \mathbb{R}$ and so we have $X = \mathbb{R}$ and $L = \{ 0, a^2 \}$.
Now consider the next case of $n = 2$. Let $v_1,v_2$ be an orthonormal basis of eigenvectors of $\mathbb{R}^2$ of $A$. Any vector $x \in \mathbb{R}^2$ can be written as $x = a_1 v_1 + a_2 v_2$ uniquely and then
$$ \left( x^t A^{2k} x \right)^{\frac{1}{k}} = \left( (a_1 v_1 + a_2 v_2)^t (a_1 \lambda_1^{2k} v_1 + a_2 \lambda_2^{2k} v_2) \right)^{\frac{1}{k}} = a_1^{\frac{2}{k}} \lambda_1^2 + a_2^{\frac{2}{k}} \lambda_2^2. $$ 
Again, the limit exists for all $a_1,a_2$ (and so for all $x \in \mathbb{R}^2$) and $L = \{ 0, \lambda_1^2, \lambda_2^2, \lambda_1^2 + \lambda_2^2 \}$.
This analysis generalizes easily for arbitrary $n$.
