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Let $P$ be a square matrix and $ \rho(P)$ the spectral radius of $P$. How to show \begin{align} \rho\left( \dfrac{P}{ \rho(P) + \epsilon } \right) < 1 \text{ for all } \epsilon >0. \end{align}

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    Start by thinking about the relationship between the spectrum of $P$ and the spectrum of $\lambda P$, for $\lambda$ a scalar.2012-02-09

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Let us elaborate on the comment by Chris: by the definition of an eigenvalue you can show that $ \lambda \in\sigma(P) \text{ if and only if } c\lambda\in \sigma(cP) $ where $c$ is any non-zero constant. Next, $\rho(P) = \max\{|\lambda|:\lambda\in\sigma(P)\}$ so $\rho(cP) = |c|\rho(P)$.

Finally, put $ c = \frac{1}{\rho(P)+\epsilon} $ to get $ \rho(cP) = \frac{\rho(P)}{\rho(P)+\epsilon} $ and you only need to upper-bound the latter ratio.

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    @Sulaiman: you're welcome2012-02-09