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Given that $A$ is symmetric $nxn$ matrix.

Show that $\lim_{k \rightarrow \infty} (x^tA^{2k}x)^{1/k}$ exists for all $x \in R^n$ and possible limit values are the eigenvalues of A.

Since A is symmetric you can find an orthonormal basis. So A is similar to the diagonal matrix B. So you get $\lim_{k \rightarrow \infty} (\sum_{i=0}^n x_i^2 \lambda_i^{2k})^{1/k}$. So I can show that this last thing is bounded above and below, but it isn't monotone in k, so how should I show that it's convergent and find what it could possibly converge to?

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1 Answer 1

Hint: Of the $\lambda_i$ whose corresponding $x_i$ are nonzero, suppose $\lambda_l$ is the largest eigenvalue in absolute value. Then $\sum_{i=0}^n x_i^2 \lambda_i^{2k}$ is dominated by the $i=l$ term.

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I guess I'm being dense. This was how I got the upper bound which I mentioned in my post, but this sequence isn't monotone so we aren't getting convergence. –  Stuart Sep 15 '12 at 3:46
    
Assume for convenience $|\lambda_1|$ is the largest, with $|\lambda_j\| \le r |\lambda_1|$ and $|x_i| \le s |x_1|$ for $j > 1$ where $0 < r < 1$. Then $$x_1^2 \lambda_1^{2k} \le \sum_{j=1}^k x_j^2 \lambda_j^{2k} \le (1 + (n-1) s^2 r^{2k}) \lambda_1^{2k}$$ As $k \to \infty$, $r^{2k} \to 0$ ... –  Robert Israel Sep 16 '12 at 4:33

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