[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.10

(1). The numerical radius defines a norm on $\scrL(\scrH)$.

(2). $w(UAU^*)=w(A)$ for all $U\in \U(n)$.

(3). $w(A)\leq \sen{A}\leq 2w(A)$ for all $A$.

(4). $w(A)=\sen{A}$ if (but not only if) $A$ is normal.

Solution.

(1). We only need to show that $$\beex \bea w(A)=0&\ra \sef{x,Ax}=0,\quad \forall\ x:\sen{x}=1\\ &\ra \sef{y,Ax}=\frac{1}{4} \sum_{k=0}^3 i^k\sef{x+i^ky,A(x+i^ky)}=0,\quad\forall\ x,y:\sen{x}=\sen{y}=1\\ &\ra Ax=0,\quad \forall\ x:\sen{x}=1\\ &\ra A=0. \eea \eeex$$

(2). $$\beex \bea w(UAU^*)&=\sup_{\sen{x}=1}\sev{\sef{x,UAU^*}}\\ &=\sup_{\sen{x}=1}\sev{(U^*x)^*A(U^*x)}\\ &=\sup_{\sen{y}=1}\sev{y^*Ay}\quad\sex{y=U^*x}\\ &=w(A). \eea \eeex$$

(3). $$\beex \bea w(A)&=\sup_{\sen{x}=1}\sev{\sef{x,Ax}}\\ &\leq \sup_{\sen{x}=1} \sex{\sen{x}\cdot \sen{Ax}}\\ &=\sup_{\sen{x}=1}\sen{Ax}\\ &=\sen{A};\\ \sen{A}&=\sup_{\sen{x}=\sen{y}=1}\sev{\sef{y,Ax}}\\ &=\sup_{\sen{x}=\sen{y}=1} \sev{\frac{1}{4}\sum_{k=0}^3 i^k\sef{y+i^kx,A(y+i^kx)}}\\ &\leq \sup_{\sen{x}=\sen{y}=1} \frac{1}{4}\sum_{k=0}^3 \sev{\sef{y+i^kx,A(y+i^kx)}}\\ &\leq \sup_{\sen{x}=\sen{y}=1} \frac{1}{4}\sum_{k=0}^3 \sen{y+i^kx}^2\cdot w(A)\\ &=\sup_{\sen{x}=\sen{y}=1} \frac{1}{4}\cdot 4\sex{\sen{x}^2+\sen{y}^2} \cdot w(A)\\ &=2w(A). \eea \eeex$$

(4). If $A$ is normal, then by the spectral theorem, there exists a unitary $U$ such that $$\bex A=U\diag(\lm_1,\cdots,\lm_n)U^*, \eex$$ and hence $$\beex \bea \sen{Ax}^2&=\sef{Ax,Ax}\\ &=x^*A^*Ax\\ &=Ux^*\diag(|\lm_1|^2,\cdots,|\lm_n|^2)U^*x\\ &=\sum_{i=1}^n |\lm_i|^2|y_i|^2\quad\sex{y=U^*x}\\ &\leq \max_i\sen{\lm_i}^2\sen{y}^2\\ &\leq w(A)\sen{x}^2. \eea \eeex$$

时间: 2024-02-15 16:48:46

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