(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 $$\bex A=U\diag(\lm_1,\cdots,\lm_n)U^*, \eex$$ and thus $$\beex \bea W(A)&=\sed{x^*Ax;\sen{x}=1}\\ &=\sed{x^*U\diag(\lm_1,\cdots,\lm_n)U^*x;\sen{x}=1}\\ &=\sed{\sum_{i=1}^n \lm_i|y_i|^2; \sum_{i=1}^n |y_i|^2=1,\ y=U^*x}\\ &=\co\sed{\lm_1,\cdots,\lm_n}. \eea \eeex$$

(2). Let $u_1,\cdots,u_k$ be the first $k$ column vector of $U$, then $$\bex Au_i=\lm_iu_i,\quad 1\leq i\leq k. \eex$$ If $$\bex x=\sum_{i=1}^k x_iu_i,\quad \sen{x}=1\ra \sum_{i=1}^k |x_i|^2=1, \eex$$ then $$\beex \bea \sef{x,Ax}&=\sef{\sum_{i=1}^k x_iu_i,A\sum_{j=1}^k x_ju_j}\\ &=\sef{\sum_{i=1}^k x_iu_i,\sum_{j=1}^k\lm_j x_ju_j}\\ &=\sum_{i=1}^k |x_i|^2\lm_i\\ &\in \co\sed{\lm_1,\cdots,\lm_k}. \eea \eeex$$

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