TITLE: How to find the limit of this matrix function
QUESTION [6 upvotes]: Let $A$ be $n\times n$ real symmetric matrix that is positive definite. Let $x\in\mathbb{R^n}, \space x\ne 0$. Prove that the following limit
  $$
\lim_{m\to\infty}\dfrac{x^TA^{m+1}x}{x^TA^{m}x}
$$
  exists and find that limit.

My thought: since $A$ is symmetric, its eigenvalues are real. The solution should be related to eigenvalues of $A$. But I don't know how to do it and need help.

REPLY [2 votes]: By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e
$$
P^{T}AP=\pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n}
$$
where $\lambda_i>0$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$.
So we have
$$
P^{T}A^{m}P=\pmatrix{\lambda_1^m \\ & \ddots \\ && \lambda_n^m}
$$
Let $x=Py$. Then
\begin{align}
\dfrac{x^TA^{m+1}x}{x^TA^{m}x}&=\dfrac{y^TP^TA^{m+1}Py}{y^TP^TA^{m}Py}
\\
&=\dfrac{\sum\limits_{k=1}^n\lambda_k^{m+1}y_k^2}{\sum\limits_{k=1}^n\lambda_k^{m}y_k^2}
\\
&=\dfrac{\sum\limits_{k=1}^n\lambda_k(\lambda_k/\lambda_{max})^{m}y_k^2}{\sum\limits_{k=1}^n(\lambda_k/\lambda_{max})^my_k^2}
\end{align}
where $\lambda_{max}=\max\{\lambda_1,\cdots,\lambda_n\}$.
Suppose $\lambda_{k_i}=\lambda_{max}$. Then
$$
\lim_{m\to\infty}\dfrac{x^TA^{m+1}x}{x^TA^{m}x}=\dfrac{\sum\limits_{i=1}^l\lambda_{k_i}y_{k_i}^2}{\sum\limits_{i=1}^ly_{k_i}^2}=\lambda_{max}
$$