8. 证明每个半正定矩阵都有唯一的半正定平方根, 即若 $A\geq 0$, 则存在唯一的 $B\geq 0$ 满足 $B^2=A$.

证明: 由 $A\geq 0$ 知存在酉阵 $U$, 使得 $$\bex U^*AU=\diag(\lm_1,\cdots,\lm_n),\quad \lm_i\geq 0. \eex$$ 取 $$\bex B=U\diag(\sqrt{\lm_1},\cdots,\sqrt{\lm_n})U^*, \eex$$ 则 $B\geq 0$, 且 $A=B^2$. 往证唯一性. 设 $C\geq 0$ 也适合 $A=C^2$, 则 $$\bex C=V\diag(\sqrt{\lm_1},\cdots,\sqrt{\lm_n})V^*. \eex$$ 于是 $$\bex U\diag(\lm_1,\cdots,\lm_n)U^*=B^2=A =C^2=V\diag(\lm_1,\cdots,\lm_n)V^*, \eex$$ $$\bex W\diag(\lm_1,\cdots,\lm_n)=\diag(\lm_1,\cdots,\lm_n)W,\quad W=V^*U. \eex$$ 故 $$\bex W=\diag(W_1,\cdots,W_s) \eex$$ 为准对角阵, 其中 $W_i$ 所对应的各 $\lm_j$ 相同. 如此, $$\bee\label{3_8_sqrt} W\diag(\sqrt{\lm_1},\cdots,\sqrt{\lm_n}) =\diag(\sqrt{\lm_1},\cdots,\sqrt{\lm_n})W, \eee$$ $$\beex \bea B&=U\diag(\sqrt{\lm_1},\cdots,\sqrt{\lm_n})U^*\\ &=VW \diag(\sqrt{\lm_1},\cdots,\sqrt{\lm_n})W^*V^*\\ &=V\diag(\sqrt{\lm_1},\cdots,\sqrt{\lm_n})V^*\quad\sex{\eqref{3_8_sqrt}}\\ &=C. \eea \eeex$$

[詹兴致矩阵论习题参考解答]习题3.8