[詹兴致矩阵论习题参考解答]习题1.4

4. 证明数值半径 $w(\cdot)$ 和谱范数 $\sen{\cdot}_\infty$ 满足如下关系:

$$\bex \frac{1}{2}\sen{A}_{\infty} \leq w(A)\leq \sen{A}_\infty,\quad A\in M_n. \eex$$

 

 

证明: (1). 当 $\sen{x}_2=\sen{y}_2=1$ 时,

$$\beex \bea \sev{\sef{y,Ax}} &=\frac{1}{4}\sev{ \sum_{k=0}^3 \sef{y+i^kx,A(y+^kx)}}\\ &\leq \frac{1}{4}w(A)\sen{y+i^k x}_2\\ &\leq \frac{1}{4}\sum_{k=0}^3 w(A)\sex{ \sen{y}_2+\sen{x}_2}\\ &\leq 2w(A). \eea \eeex$$ 故 $$\bex \sen{Ax}_2 =\max_{\sen{y}_2=1}|\sef{y,Ax}| \leq 2w(A),\quad \sen{x}_2=1, \eex$$ $$\bex \sen{A}_\infty =\max_{\sen{x}_2=1}\sen{Ax}_2 \leq 2w(A). \eex$$ (2). 由 $$\bex |\sef{Ax,x}| \leq \sen{Ax}_2\sen{x}_2 \leq \sen{A}_\infty \sen{x}_2^2 \eex$$ 即知 $$\bex w(A)\leq \sen{A}_\infty. \eex$$