问题征解1
1.(高等代数)证明:实对称矩阵${A_n} = \left[ {\begin{array}{*{20}{c}}{\frac{1}{1}}&{\frac{1}{2}}&{\frac{1}{3}}& \cdots &{\frac{1}{n}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{3}}& \cdots &{\frac{1}{n}}\\{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}& \cdots &{\frac{1}{n}}\\\vdots & \vdots & \vdots &{}& \vdots \\{\frac{1}{n}}&{\frac{1}{n}}&{\frac{1}{n}}& \cdots &{\frac{1}{n}}\end{array}} \right]$的特征值都大于$0$,且小于等于$3+2\sqrt{2}$.
2.(高等概率论)Let the stochastic processes $\{X_k,1\leq k\leq n\}$ and $\{X'_k,1\leq k\leq n\}$ be independent of one another and have the same joint distributions. If $m_k$ is a median of $X_k,1\leq k\leq n$. Prove: for $\lambda>0$,
证.(DH)令$S_{k}^{\ast}=\underset{1\leq j\leq k}{\max}\left| X_j-m_j\right|$且$S_{0}^{\ast}=0$.记\[A_1=\left\{S_{1}^{\ast}\geq\lambda\right\},\quad A_k=\left\{S_{k-1}^{\ast}<\lambda ,\left| X_k-m_k\right|\geq\lambda\right\},\]
则有$A_1\cup\cdots\cup A_n=\left\{\underset{1\leq k\leq n}{\max}\left| X_k-m_k\right|\geq\lambda\right\}$且$A_i\cap A_j=\emptyset,i\neq j$.并记
\[A_k^+=\left\{S_{k-1}^{\ast}<\lambda , X_k-m_k\geq\lambda\right\},\quad A_k=\left\{S_{k-1}^{\ast}<\lambda , X_k-m_k\leq -\lambda\right\},\]则有$A_k^+\cap A_k^-=\emptyset,A_k^+\cup A_k^-=A_k$,且$\forall 1\leq i<j\leq n, A_i^+\cap A_j^+=A_i^-\cap A_j^-=\emptyset$.又因为$A_i\cap A_j=\emptyset$,则$\forall 1\leq i<j\leq n,A_i^+\cap A_j^-=\emptyset$.再令$M_{k}^{+}=\left\{m_k-X'_k\geq 0\right\},M_{k}^{-}=\left\{m_k-X'_k\leq 0\right\}$且\[B_k=\left\{\underset{1\leq j\leq k-1}{\max}\left| X_j-X'_j\right|<\lambda ,\left| X_k-X'_k\right|\geq\lambda\right\},\]
则
\[P\left(\underset{1\leq k\leq n}{\max}\left| X_k-X'_k\right|\geq\lambda\right)=P\left(\bigcap_{k=1}^n{B_k}\right).\]
仿照$A_k^+,A_k^-$,构造$B_k^+,B_k^-$,亦有$B_i^+\cap B_j^-=\emptyset,\forall 1\leq i,j \leq n$成立.因此
\[P\left(\underset{1\leq k\leq n}{\max}\left| X_k-X'_k\right|\geq\lambda\right)=P\left(\bigcap_{k=1}^n{B_k^+}\right)+P\left(\bigcap_{k=1}^n{B_k^-}\right).\]
因为$X_k,X'_k$同分布,所以$m_k=m'_k$且$X_k-m_k+m_k-X'_k=0$,所以$B_k^+\supset A_k^+\cap M_k^+,B_k^-\supset A_k^-\cap M_k^-$.又因为$\left\{X_k\right\}$与$\left\{X'_k\right\}$独立,因此
\begin{align*}P\left(\underset{1\leq k\leq n}{\max}\left| X_k-X'_k\right|\geq\lambda\right)&\geq P\left(\bigcap_{k=1}^n{\left(A_{k}^{+}\cap M_{k}^{+}\right)}\right)+P\left(\bigcap_{k=1}^n{\left(A_{k}^{-}\cap M_{k}^{-}\right)}\right)\\&=\sum_{k=1}^n{P\left(A_{k}^{+}\right)\cdot P\left(M_{k}^{+}\right)}+\sum_{k=1}^n{P\left(A_{k}^{-}\right)\cdot P\left(M_{k}^{-}\right)}\\&=\frac{1}{2}\sum_{k=1}^n{\left[P\left(A_{k}^{+}\right)+P\left(A_{k}^{-}\right)\right]}=\frac{1}{2}P\left(\bigcap_{k=1}^n{\left(A_{k}^{+}\cap A_{k}^{-}\right)}\right)\\&=\frac{1}{2}P\left(\bigcap_{k=1}^n{A_k}\right)=\frac{1}{2}P\left\{\underset{1\leq k\leq n}{\max}\left| X_k-m_k\right|\geq\lambda\right\}.\end{align*}
求一个可逆矩阵$Q$使得$AQ=QB$.
证.(SUCCEME)记$H=A-B,G=P/2$,则我们有$HP=PH,BG-GB=H$,只需取一个可逆矩阵$Q$使得$(B+H)Q=QB$.
事实上,我们可取\[Q = {e^{ - G}} = 1 - G + \frac{{{G^2}}}{{2!}} - \frac{{{G^3}}}{{3!}} + \cdots \]
由$P$幂零可知上面的和为有限和.由$e^G\cdot e^{-G}=I$可知$Q$可逆.又由
\begin{align*}B{G^n} - {G^n}B&= \left( {B{G^n} - GB{G^{n - 1}}} \right) + \left( {GB{G^{n - 1}} - {G^2}B{G^{n - 1}}} \right) + \cdots + \left( {{G^{n - 1}}BG - {G^n}B} \right)\\&= H{G^{n - 1}} + GH{G^{n - 2}} + \cdots + {G^{n - 1}}H = nH{G^{n - 1}},\end{align*}
可知
\[B{G^n} - {G^n}B=nH{G^{n - 1}}.\]
因此
\begin{align*}\left( {B + H} \right)Q - QB &= \left( {B + H} \right)\left( {1 - G + \frac{{{G^2}}}{{2!}} - \frac{{{G^3}}}{{3!}} + \cdots } \right) - \left( {1 - G + \frac{{{G^2}}}{{2!}} - \frac{{{G^3}}}{{3!}} + \cdots } \right)B\\&= \sum\limits_{n \ge 0} {\left[ {\left( {B + H} \right)\frac{{{{\left( { - 1} \right)}^n}{G^n}}}{{n!}} - \frac{{{{\left( { - 1} \right)}^n}{G^n}}}{{n!}}B} \right]} = \sum\limits_{n \ge 0} {\frac{{{{\left( { - 1} \right)}^n}}}{{n!}}\left( {B{G^n} - {G^n}B + H{G^n}} \right)} \\&= \sum\limits_{n \ge 0} {\frac{{{{\left( { - 1} \right)}^n}}}{{n!}}\left( {nH{G^{n - 1}} + H{G^n}} \right)} = \sum\limits_{n \ge 0} {{{\left( { - 1} \right)}^n}\left( {\frac{{H{G^{n - 1}}}}{{\left( {n - 1} \right)!}} + \frac{{H{G^n}}}{{n!}}} \right)} = 0.\end{align*}
这里约定$n=-1$时, $HP^n/n!=0$.所以此时$\left( {B + H} \right)Q = QB$成立.