2020-10-05 矩阵A+B之最小特征值 ≥ 正定矩阵A的最小特征值+正定矩阵B的最小特征值

定理: $A,B,C$: 均正定 $C=A+B$ ${\lambda }_{\min }(C)⩾{\lambda }_{\min }(A)+{\lambda }_{\min }(B)$ ${\lambda }_{\max }(C)⩽{\lambda }_{\max }(A)+{\lambda }_{\max }(B)$

证明: $x^{T}Ax⩾{\lambda }_{\min }(A)x^{T}x$. Thus, $x^T(A-{\lambda }_{\min }(A)E)x⩾0$, that is, $A-{\lambda }_{\min }(A)E$ is non-negative definite.

因此，有 $A-{\lambda }_{\min }(A)E⩾0$ $B-{\lambda }_{\min }(B)E⩾0$ $A-{\lambda }_{\min }(A)E+B-{\lambda }_{\min }(B)E⩾0$

显然， $Cx=\lambda (C)x$，所有 $(A-{\lambda }_{\min }(A)E+B-{\lambda }_{\min }(B)E)x=\left(\lambda (C)-{\lambda }_{\min }(A)-{\lambda }_{\min }(B)\right)x$ 也就是说，$\lambda (C)-{\lambda }_{\min }(A)-{\lambda }_{\min }(B)$是$A-{\lambda }_{\min }(A)E+B-{\lambda }_{\min }(B)E$的特征值

因为$A-{\lambda }_{\min }(A)E+B-{\lambda }_{\min }(B)E⩾0$，是非负定阵，因此$\lambda (C)-{\lambda }_{\min }(A)-{\lambda }_{\min }(B)⩾0$

所以 $\lambda (C)⩾{\lambda }_{\min }(A)+{\lambda }_{\min }(B)$ ${\lambda }_{\min }(C)⩾{\lambda }_{\min }(A)+{\lambda }_{\min }(B)$

同理 ${\lambda }_{\max }(C)⩽{\lambda }_{\max }(A)+{\lambda }_{\max }(B)$