Show that the spectral radius of the matrix A is less than or equal any natural norm, i.e:

$$\rho(A) \leq ||A||=\max_{||x||=1}{||Ax||}$$

where $\rho(A)=\max\{|\lambda|:\lambda \text{ is a eigenvalue of $A$}\}$.

I'm a little confused with this, because I'm not sure about the definition of natural norm of $A$. If $\rho(A)=\lambda^*$ and $u$ is such that $||u||=1$, then: \begin{align*} ||A||&\geq ||Au||\\ &=||\lambda^* u||\\ &=|\lambda^*| \end{align*} Is this ok?


  • 1
    $\begingroup$ Just assume $\rho(A) = |\lambda^\ast|$ instead of $\rho(A) = \lambda^\ast $ and your reasoning becomes correct. $\endgroup$ – TZakrevskiy Sep 15 '13 at 19:53
  • $\begingroup$ (1) you are mixing $u$ and $x$ when you say " $x$ is such that $||u||=1$". (2) why $||Au||=||\lambda u||$? $\endgroup$ – Maesumi Sep 15 '13 at 20:22
  • $\begingroup$ (1) Yes, thanks, I edited my question. (2) Do you mean $\lambda^*$? because $\lambda^*$ is an eigenvalue of $A$. $\endgroup$ – Hiperion Sep 15 '13 at 22:45
  • $\begingroup$ If $A$ is not symmetric, then you may not be able to find a unit eigenvector. $\endgroup$ – user539442 Apr 18 '18 at 21:59

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