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When I read page 24 in Iterative Methods for Sparse Linear Systems, Second Edition, I can not understand the following statement: (My major is not math)

Let $A \in \mathbb{C}^{n \times n}$ have eigenvalues $\lambda_{1},\dots,\lambda_{n}$, $\rho(A)$ its spectral radius and $\gamma(A)$ its numerical radius. I want to prove the relationships $$ \rho(A) \le \gamma(A) \le \| A \|_2 \qquad \text{and} \qquad \frac{\| A \|_2}{2} \le \gamma(A) \le \| A \|_2. $$

ViktorStein
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Yulong Ao
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1 Answers1

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Take an eigenvalue $\lambda$ where the spectral radius is attained, and the corresponding eigenvector $v$ with $\|v\|=1$.

Then you can say that $$\rho(A) = |\lambda| = |\lambda v^*v| = \left| \frac{v^*Av}{v^*v}\right|\le \gamma(A)$$ (the latter inequality is true by definition of $\gamma(A)$).

On the other hand, $$\gamma(A)=\sup_{v\ne 0}\left|\frac{v^*Av}{v^*v}\right|= \sup_{v\ne 0} \frac{|v^*Av|}{ \|v\|^2} \le \|A\| $$ by the definition of $\|A\|$.

The last inequality should be done similarly.

ViktorStein
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TZakrevskiy
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