Let $\rho(P)$ be the spectral radius of $P$. Show $\rho( \dfrac{P}{ \rho(P) + \epsilon } ) < 1 \text{ for all } \epsilon >0.$

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. – Chris Eagle Feb 9 '12 at 12:57

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.
If we denote the characteristic polynomial of $P$ by $q( \lambda ).$ Then clearly $c \lambda, c \neq 0$ is a root of $q( \lambda )$ if and only if $\lambda$ is a root of $q(\lambda).$ On the other hand, $\lambda \in \sigma(P)$ is an eigenvalue of $P$ implies $Px = \lambda x, x \neq 0.$ This implies $P cx = c\lambda x,$ so I was wondering $c \lambda$ is eigenvalue of $P$ corresponding to which eigenvector $x$ or $cx?$ – Zizo Feb 9 '12 at 14:06
Can it be you're missing a $c$ in there (first equation)? Otherwise I cannot believe $\lambda$ being an eigenvalue implies $c\lambda$ being an eigenvalue of the same matrix, too. With $\sigma$ you mean the spectrum, right? – Christian Rau Feb 9 '12 at 14:27