Let $A$ be an $n \times n$ real symmetric matrix with eigenvalues $\lambda_1,\ldots,\lambda_n.$ Find the set of $v \in \mathbb{R}^n$ such that $$\lim_{k\to\infty}\left<A^{2k}v,v\right>^{1/k}$$ exists and find the set of possible limits.
Since $A$ is symmetric there exists an orthonormal basis $v_1,\ldots,v_n$ of eigenvectors of $A$. Let $\lambda_j$ be the eigenvector corresponding to $v_j$. If $v \in \mathbb{R}^n$, then $$v = c_1v_1 + \cdots c_nv_n, c_j \in \mathbb{R}.$$ Since $<v_i, v_j> = 0$ for $i \neq j$ and $= 1$ if $i = j$, and $A^{2k}v_j = \lambda_j^{2k}v_j$, by the linearity of the inner product the limit becomes $$\lim_{k\to\infty} \left(c_1^2\lambda_1^{2k} + \cdots c_n^2\lambda_n^{2k}\right)^{1/k}.$$ This is where I'm stuck. I don't know how to evaluate the limit. I tried taking logs and using L'Hopital's rule, but that only gave another limit I couldn't compute. Since $$c_1^2\lambda_1^{2k} + \cdots + c_n^2\lambda_n^{2k} \leq (c_1^{2/k}\lambda_1^2 + \cdots + c_n^{2/k}\lambda_n^2)^k,$$ I know that if the limit exists it's bounded above by $\lambda_1^2 + \cdots \lambda_n^2$. Any help proceeding would be great.
I think you have a typo.
$\lambda_j$ is an eigenvalue not an eigenvector.
Assume that all the coefficients $c_i$ are non-zero.
Let $M=\max_{1 \leq i \leq n}\{c_i^2\}$ and $\lambda_{i_0}^{2}=\max_{1 \leq i \leq n}\{\lambda_i^{2}\}$
Now $$\sqrt[k]{(c_1^2\lambda_1^{2k} + \cdots + c_n^2\lambda_n^{2k})}\leq \sqrt[n]{M(\max_{1 \leq i \leq n}\{ \lambda_i^2\})^k}=\sqrt[k]{M \lambda_{i_0}^{2k}} \to \lambda_{i_0}^2$$
$$\sqrt[k]{(c_1^2\lambda_1^{2k} + \cdots + c_n^2\lambda_n^{2k})}\geq \sqrt[k]{c_{i_0}^2 \lambda_{i_0}^{2k}} \to \lambda_{i_0}^{2}$$
Thus from squeeze theorem the limit is $\lambda_{i_0}^{2}$
Now if there are zero coefficients $c_i$ then let $A=\{j \in\{1,2...n\}|c_j \neq 0\}$
So we have to compute the limit $$\lim_{k \to \infty} \sqrt[k]{\sum_{j \in A}c_j^{2} \lambda_j^{2k}}$$
which can be computed by follwing the previous argument.