[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.9

(1). When $A$ is normal, the set $W(A)$ is the convex hull of the eigenvalues of $A$. For nonnormal matrices, $W(A)$ may be bigger than the convex hull of its eigenvalues. For Hermitian operators, the first statement says that $W(A)$ is the close interval whose endpoints are the smallest and the largest eigenvalues of $A$.

 

(2). If a unit vector $x$ belongs to the linear span of the eigenspaces corresponding to eigenvalues $lm_1,cdots,lm_k$ of a normal operator $A$, then $sef{x,Ax}$ lies in the convex hull of $lm_1,cdots,lm_k$. (This fact will be used frequently in Chapter III.)

 

Solution.

 

(1). When $A$ is normal, by the spectral theorem, there exists a unitary $U$ such that $$ex A=Udiag(lm_1,cdots,lm_n)U^*, eex$$ and thus $$eex ea W(A)&=sed{x^*Ax;sen{x}=1}\ &=sed{x^*Udiag(lm_1,cdots,lm_n)U^*x;sen{x}=1}\ &=sed{sum_{i=1}^n lm_i|y_i|^2; sum_{i=1}^n |y_i|^2=1, y=U^*x}\ &=cosed{lm_1,cdots,lm_n}. eea eeex$$

 

(2). Let $u_1,cdots,u_k$ be the first $k$ column vector of $U$, then $$ex Au_i=lm_iu_i,quad 1leq ileq k. eex$$ If $$ex x=sum_{i=1}^k x_iu_i,quad sen{x}=1 a sum_{i=1}^k |x_i|^2=1, eex$$ then $$eex ea sef{x,Ax}&=sef{sum_{i=1}^k x_iu_i,Asum_{j=1}^k x_ju_j}\ &=sef{sum_{i=1}^k x_iu_i,sum_{j=1}^klm_j x_ju_j}\ &=sum_{i=1}^k |x_i|^2lm_i\ &in cosed{lm_1,cdots,lm_k}. eea eeex$$

原文地址:https://www.cnblogs.com/zhangzujin/p/4106628.html