问题征解1 - Eufisky - The lost book
曾经的两道矩阵正定问题的解答
十个利用矩阵解决的经典题目

问题征解1

Eufisky posted @ 2017年1月02日 23:58 in 高等代数 with tags 征解题 北大 , 1864 阅读

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$3+2\sqrt{2}$.


证.感谢Veer的提示(其实是cholesky分解),我已经证明出来了.
首先注意到\[A_n=\left(\begin{matrix}b_1& b_2& b_3& \cdots& b_n\\0& b_2& b_3& \cdots& b_n\\\vdots&\ddots& b_3&\cdots&b_n\\\vdots& & \ddots& \ddots& \vdots\\0& \cdots& \cdots& 0& b_n\\\end{matrix}\right)\left(\begin{matrix}b_1&0& \cdots& \cdots& 0\\b_2& b_2& \ddots& & \vdots\\b_3& b_3& b_3& \ddots& \vdots\\\vdots& \vdots& \vdots& \ddots&0\\b_n& b_n& b_n& \cdots& b_n\\\end{matrix}\right)=B^TB,\]
其中
\[\left\{ \begin{array}{l}b_1^2 + b_2^2 + b_3^2 +  \cdots  + b_n^2 = 1,\\b_2^2 + b_3^2 +  \cdots  + b_n^2 = \frac{1}{2},\\\vdots \\b_n^2 = \frac{1}{n}.\end{array} \right.\]
解得$b_k=\frac1{\sqrt{k(k+1)}},k=1,2,\cdots,n-1$且$b_n=\frac1{\sqrt{n}}$.
 
由Rayleigh商定理可知只需证明
\[0\leq\frac{x^TB^TBx}{x^Tx}=\frac{(Bx)^TBx}{x^Tx}\leq 3+2\sqrt{2},\]
其中$x=(x_1,x_2,\cdots,x_n)^T$.等价于证明
\[0\leq\sum_{k=1}^{n-1}{\frac{\left(x_1+x_2+\cdots +x_k\right)^2}{k\left(k+1\right)}}+\frac{\left(x_1+x_2+\cdots +x_n\right)^2}{n}\leq (3+2\sqrt{2})\sum_{k=1}^n{x_{k}^{2}}.\]
 
左边的不等式是显然的.下面证明右边不等式.事实上,由Cauchy-Schwarz不等式可知
\[\left(\frac{x_{1}^{2}}{a_1}+\frac{x_{2}^{2}}{a_2}+\cdots +\frac{x_{k}^{2}}{a_k}\right)\left(a_1+a_2+\cdots +a_k\right)\geq\left(x_1+x_2+\cdots +x_k\right)^2.\]
不等式可以改写为
\[\frac{\left(x_1+x_2+\cdots +x_k\right)^2}{k\left(k+1\right)}\leq\sum_{i=1}^k{\frac{a_1+a_2+\cdots +a_k}{k\left(k+1\right)a_i}x_{i}^{2}}.\]
对于$k=1,2,\cdots,n-1$,构造类似的不等式,累加得
\[\sum_{k=1}^{n-1}{\frac{\left(x_1+x_2+\cdots +x_k\right)^2}{k\left(k+1\right)}}+\frac{\left(x_1+x_2+\cdots+x_n\right)^2}{n}\leq\sum_{k=1}^n{y_kx_{k}^{2}},\]
其中
\[y_k=\sum_{i=k}^{n-1}{\frac{a_1+a_2+\cdots +a_i}{i\left(i+1\right)a_k}}+\frac{a_1+a_2+\cdots +a_n}{na_k}.\]
只需证明数列$(a_1,a_2,\cdots,a_n)$满足$y_k\leq 3+2\sqrt{2}$即可.我们取$a_k=\sqrt{k}-\sqrt{k-1}$,则$a_1+a_2+\cdots+a_k=\sqrt{k}$.此时,我们有
\[y_k=\frac{1}{a_k}\left(\sum_{i=k}^{n-1}{\frac{1}{\left(i+1\right)\sqrt{i}}}+\frac{1}{\sqrt{n}}\right).\]
注意到
\begin{align*}2\left(\frac{1}{\sqrt{i}}-\frac{1}{\sqrt{i+1}}\right)&=2\frac{\sqrt{i+1}-\sqrt{i}}{\sqrt{i}\cdot\sqrt{i+1}}=2\frac{1}{\sqrt{i}\cdot\sqrt{i+1}\left(\sqrt{i+1}+\sqrt{i}\right)}\\&\geq 2\frac{1}{\sqrt{i}\cdot\sqrt{i+1}\left(\sqrt{i+1}+\sqrt{i+1}\right)}=\frac{1}{\left(i+1\right)\sqrt{i}}.\end{align*}
因此
\begin{align*}y_k&\leq\frac{1}{a_k}\left[\sum_{i=k}^{n-1}{2\left(\frac{1}{\sqrt{i}}-\frac{1}{\sqrt{i+1}}\right)}+\frac{1}{\sqrt{n}}\right]=\frac{1}{a_k}\left[2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{n}}\right)+\frac{1}{\sqrt{n}}\right]\\&\leq\frac{2}{a_k\sqrt{k}}=\frac{2}{\left(\sqrt{k}-\sqrt{k-1}\right)\sqrt{k}}=\frac{2\left(\sqrt{k}+\sqrt{k-1}\right)}{\sqrt{k}}\\&=2\left(1+\sqrt{1-\frac{1}{k}}\right)<4<3+2\sqrt{2}.\end{align*}

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 $\lambda>0$,

$$P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{X_k} - {m_k}} \right| \ge \lambda } \right) \le 2P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {{X_k} - {X'_k}} \right| \ge \lambda } \right).$$

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\[B_k=\left\{\underset{1\leq j\leq k-1}{\max}\left| X_j-X'_j\right|<\lambda ,\left| X_k-X'_k\right|\geq\lambda\right\},\]

\[P\left(\underset{1\leq k\leq n}{\max}\left| X_k-X'_k\right|\geq\lambda\right)=P\left(\bigcap_{k=1}^n{B_k}\right).\]

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*}

 
3.(高等代数)$F$为数域, $A,B,P\in M_n(F)$, $P$幂零且\[(A-B)P=P(A-B),\qquad BP-PB=2(A-B).\]

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.


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