3. (Aronszajn) 设 $$\bex C=\sex{\ba{cc} A&X\\ X^*&B \ea} \eex$$ 为 Hermite 矩阵, $C\in M_n$, $A\in M_k$. 设 $A,B,C$ 的特征值分别为 $\al_1\geq \cdots\geq \al_k$, $\beta_1\geq \cdots \beta_{n-k}$, $\gamma_1\geq \cdots\geq \gamma_n$. 则对于满足 $i+j-1\leq n$ 的任意的 $i,j$, $$\bex \gamma_{i+j-1}+\gamma_n\leq \al_i+\beta_j. \eex$$

证明:

(1). 仅须证明在 $C$ 半正定的情形, $$\bee\label{3_3_goal} \gamma_{i+j-1}\leq \al_i+\beta_j. \eee$$ 事实上, 对一般的 $C$, $$\bex C-\gamma_nI =\sex{\ba{cc} A-\gamma_nI&X\\ X^*&B-\gamma_nI \ea} \eex$$ 半正定, 而由 \eqref{3_3_goal}, $$\bex \gamma_{i+j-1}-\gamma_n\leq (\al_i-\gamma-n)+(\beta_j-\gamma_n), \eex$$ $$\bex \gamma_{i+j-1}+\gamma_n\leq \al_i+\beta_j. \eex$$

(2). 当 $C$ 半正定时, 由 Courant-Fischer 极大极小刻画, $$\bex \exists\ S\subset \bbC^k,\ \dim S=k-i+1;\quad \exists\ T\subset \bbC^{n-k},\ \dim T=(n-k)-j+1 \eex$$ 使得 $$\bex \al_i=\max_{x\in S\atop \sen{x}=1}x^*Ax,\quad \beta_j=\max_{y\in T\atop \sen{y}=1} y^*By. \eex$$ 对 $$\bex z\in S\oplus T=\sed{\sex{\tilde x\atop \tilde y}\in \bbC^n;\ \tilde x\in S, \ \tilde y\in T},\quad \sen{z}=1 \eex$$ 可写成 $$\bex z=\sex{sx\atop ty};\quad s,t\in\bbC, |s|^2+|t|^2=1;\quad x\in S,\ \sen{x}=1;\quad y\in T,\ \sen{y}=1. \eex$$ 于是 $$\beex \bea 0\leq z^*Cz&=\sex{\bar sx^*,\bar t y^*} \sex{\ba{cc} A&X\\ X^*&B \ea}\sex{sx\atop ty}\\ &=|s|^2x^*Ax +\bar stx^*Xy +\bar tsy^*X^*x +|t|^2 y^*By\\ &\leq |s|^2\al_i +\bar st\delta +\bar ts\bar \delta +|t|^2\beta_j\quad\sex{\delta=x^*Xy}. \eea \eeex$$ 若 $t\neq 0$, 则 $$\bee\label{3_3_wst} 0\leq \frac{z^*Cz}{|t|^2} \leq \al_i w\bar w +\delta \bar w +\bar \delta w+\beta_j,\quad w=\frac{s}{t}. \eee$$ 令 $$\bex f(w)=\al_iw\bar w +\delta \bar w +\bar \delta w+\beta_j,\quad w\in \bbC, \eex$$ 则 $$\bee\label{3_3_w0} f(w)\geq 0,\quad \forall\ w\in \bbC. \eee$$ 而 $$\bex f\sex{-\frac{1}{\bar w}} =\frac{\al_i}{\bar ww} -\frac{\delta}{w} -\frac{\bar \delta}{\bar w} +\beta_j\geq 0, \eex$$ $$\bee\label{3_3_1z0} |w|^2f\sex{-\frac{1}{\bar w}} =\al_i-\delta \bar w -\bar \delta w +\beta_j w\bar w\geq 0. \eee$$ 将 \eqref{3_3_w0} 与 \eqref{3_3_1z0} 联立有 $$\bex 0\leq f(w)+|w|^2f\sex{-\frac{1}{\bar w}} =(\al_i+\beta_j)(1+w\bar w), \eex$$ $$\bex f(z)\leq (\al_i+\beta_j)(1+w \bar w), \eex$$ $$\bex f\sex{\frac{s}{t}}\leq (\al_i+\beta_j)\sex{1+\sev{\frac{s}{t}}^2}, \eex$$ $$\bee\label{3_3_final} |t|^2f\sex{\frac{s}{t}} \leq \al_i+\beta_j. \eee$$ 当 $t=0$ 时, \eqref{3_3_final} 也成立. 将 \eqref{3_3_final} 与 \eqref{3_3_wst} 联立有 $$\bex z^*Cz\leq |t|^2 f\sex{\frac{s}{t}} \leq \al_i+\beta_j,\quad \forall\ z\in S\oplus T: \sen{z}=1. \eex$$ 由 $$\beex \bea \dim S\oplus T &=\dim S+\dim T\\ &=(k-i+1)+[(n-k)-j+1]\\ &=n-(i+j-1)+1 \eea \eeex$$ 及 Courant-Fischer 极大极小刻画, $$\bex \gamma_{i+j-1}\leq \max_{z\in S\oplus T\atop \sen{z}=1} z^*Cz \leq \al_i+\beta_j. \eex$$

[詹兴致矩阵论习题参考解答]习题3.3