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[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.3.1

更新时间:2022-09-06 15:47:09

Let $A=A_1\oplus A_2$. Show that

 

(1). $W(A)$ is the convex hull of $W(A_1)$ and $W(A_2)$; i.e., the smallest convex set containing $W(A_1)\cup W(A_2)$.

 

(2). $$\beex \bea \sen{A}&=\max\sed{\sen{A_1},\sen{A_2}},\\ \spr(A)&=\max\sed{\spr(A_1),\spr(A_2)},\\ w(A)&=\max\sed{w(A_1),w(A_2)}. \eea \eeex$$

 

Solution.

 

(1). We have $$\beex \bea W(A)&=\sed{x^*Ax;\sen{x}=1}\\ &=\sed{y^*A_1y+z^*A_2z;\sen{y}^2+\sen{z}^2=1}\\ &\supset W(A_1)\cup W(A_2), \eea \eeex$$ and $$\bex W(A)=\sed{\sen{y}^2 \sex{\frac{y}{\sen{y}}}^*A_1\frac{y}{\sen{y}} +\sen{z}^2 \sex{\frac{z}{\sen{z}}}^*A_2\frac{z}{\sen{z}}; \sen{y}^2+\sen{z}^2=1} \eex$$ is contained in any convex set containing $W(A_1)\cup W(A_2)$.

 

(2). $$\beex \bea \sen{Ax}^2&=\sen{\sex{A_1y\atop A_2z}}^2\quad\sex{x=\sex{y\atop z}}\\ &=\sen{A_1y}^2+\sen{A_2z}^2\\ &\leq \sen{A_1}^2\sen{y}^2+\sen{A_2}^2\sen{z}^2\\ &\leq \max\sed{\sen{A_1},\sen{A_2}}^2 \sex{\sen{y}^2+\sen{z}^2}\\ &=\max\sed{\sen{A_1},\sen{A_2}}^2 \sen{x}^2. \eea \eeex$$ $$\beex \bea &\quad Ax=\lm x\quad\sex{x\neq 0}\\ &\ra A_1y=\lm y,\quad A_2z=\lm z\quad\sex{x=\sex{y\atop z}}\\ &\ra |\lm|\leq\sedd{\ba{ll} \spr(A_1),&y\neq 0\\ \spr(A_2),&z\neq 0 \ea}\\ &\ra |\lm|\leq \max\sed{\spr(A_1),\spr(A_2)};\\ &\quad A_1y=\lm y\quad\sex{y\neq 0}\\ &\ra A\sex{y\atop 0}=\lm \sex{y\atop 0}\\ &\ra |\lm|\leq \spr(A);\\ &\quad A_2z=\lm z\quad\sex{z\neq 0}\\ &\ra |\lm|\leq \spr(A). \eea \eeex$$ $$\beex \bea w(A)&=\sup_{\sen{x}=1}\sev{\sef{x,Ax}}\\ &=\sup_{\sen{y}^2+\sen{z}^2=1} \sev{\sef{y,A_1y}+\sev{z,A_2z}}\\ &\leq \sup_{\sen{y}^2+\sen{z}^2=1} \sez{ \sen{y}^2w(A_1)+\sen{z}^2w(A_2) }\\ &\leq \max\sed{w(A_1),w(A_2)};\\ w(A_1)&=\sup_{\sen{y}=1}\sen{\sef{y,A_1y}}\\ &=\sup_{\sen{\sex{y\atop 0}}=1} \sev{\sef{\sex{y\atop 0},A\sex{y\atop 0}}}\\ &\leq w(A),\\ w(A_2)&\leq w(A). \eea \eeex$$