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I have a set of square matrices $A_i \in \mathbb{R}^{n \times n}$ for $i=1,\ldots,N$, such that $[A_i]_{jk} \ge 0$ for all $i$ and coordinates $j,k$.

If the largest eigenvalue of each $A_i$ is smaller than 1, is it going to be true also for $$\frac{1}{N} \sum_{i=1}^N A_i?$$

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I think the answer is true, from triangle inequality. The spectral norm of each $A_i$ is smaller than 1, and therefore, the spectral norm of the sum is going less than $N$, dividing by $1/N$ yields the necessary result. Should have put more thought into it, thanks. –  kloop Oct 21 '12 at 9:04

1 Answer 1

The answer doesn't follow from triangle inequality as you stated above, because in your problem you have only stated that all eigenvalues are smaller than $1$; you have not imposed any condition on how negative they can be. However, your condition can still be stated as $\langle Ax, x \rangle \le \|x\|^2$, so by the linearity of the inner product that same condition will hold for the "mean" matrix.

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