find the vectors such that this matrix limit exists and compute the limits Suppose that $A$ is a symmetric $n \times n$ matrix with distinct eigenvalues $\lambda_1,...,\lambda_l, ( l \le n )$. Find the sets $X = \{ x \in \mathbb R^n: \lim_{ k \to \infty } ( x^t A^{2k} x) ^{ \frac{1}{k} } exists \}$, and $ L = \{ \lim_{ k \to \infty } ( x^t A^{2k} x) ^{ \frac{1}{k} } : x \in X \}$.

Since $A$ is symmetric I know I must use $A$ is orthogonally diagonalize to solve the problem. But then I get stuck. Any help is appreciated.
 A: You can write $A=PDP^{-1}$ and get
$$((P^{-1}x)^T (D^2)^k (P^{-1}x))^{1/k}$$
write $y=P^{-1}x$ and $\beta_i=\lambda_i^2$. The above quadratic form becomes
$\sqrt[k]{\sum y_i^2 \beta_i^k}$
which is a weighted $k$-th norm of the set $\beta_i$. The $k$-norm tends to $\max$ function in the limit $k\to \infty$. The term in the sum with the greatest $\beta_i$ will dominate all other terms (terms with $y_i=0$ are skipped in finding the maximum). So you can write it
$$\lim_{k\to \infty}\sqrt[k]{\sum y_i^2 \beta_i^k}=$$
$$\lim_{k\to \infty}\sqrt[k]{y_i^2 \beta_{max}^k}=$$
$$=\lim_{k\to\infty}\sqrt[k]{y_i^2}\beta_{max}=\beta_{max}$$
$$\quad \beta_{max}=\max(\{0\}\cup \{\beta_i|y_i>0\})$$
It seems to me that the limit exists for all $y_i$ (and therefore for all $x_i$, because $P$ is an orthogonal bijection), and the set $L$ of possible limits is simply the set of squares of eigenvalues,
$$L=\{0\}\cup\{\beta_i=\lambda_i^2\}$$
because you can get any of the eigenvalues to be maximum by setting the $y_i$ of all higher eigenvalues to zero.
This solution seems straight-forward but suspicius. If anyone finds a flaw, I'd be glad of a correction.
