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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. –  Chris Eagle Feb 9 '12 at 12:57

1 Answer 1

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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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
@Sulaiman Eigen*space* is a vector space :) –  Ilya Feb 9 '12 at 14:17
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
@II y a. Shukran which means thank you in my language :) –  Zizo Feb 9 '12 at 14:28
@ChristianRau yes –  Zizo Feb 9 '12 at 14:29

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