||
最后一题的解答见下: http://www.math.org.cn/forum.php?mod=viewthread&tid=37137&extra=&page=2 (陶哲轩小弟)
证明:实对称矩阵${A_n} =left[begin{matrix}
frac{1}{1}& frac{1}{2}& frac{1}{3}& cdots& frac{1}{n}\
frac{1}{2}& frac{1}{2}& frac{1}{3}& cdots& frac{1}{n}\
frac{1}{3}& frac{1}{3}& frac{1}{3}& cdots& frac{1}{n}\
vdots& vdots& vdots& & vdots\
frac{1}{n}& frac{1}{n}& frac{1}{n}& cdots& frac{1}{n}\
end{matrix}right]$的特征值都大于$0$,且小于等于$3+2sqrt{2}$.
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
[
0leqfrac{x^TB^TBx}{x^Tx}=frac{(Bx)^TBx}{x^Tx}leq 3+2sqrt{2},
]
其中$x=(x_1,x_2,cdots,x_n)^T$.等价于证明
[
0leqsum_{k=1}^{n-1}{frac{left(x_1+x_2+cdots +x_kright)^2}{kleft(k+1right)}}+frac{left(x_1+x_2+cdots +x_nright)^2}{n}leq (3+2sqrt{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_kright)geqleft(x_1+x_2+cdots +x_kright)^2.
]
不等式可以改写为
[
frac{left(x_1+x_2+cdots +x_kright)^2}{kleft(k+1right)}leqsum_{i=1}^k{frac{a_1+a_2+cdots +a_k}{kleft(k+1right)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_kright)^2}{kleft(k+1right)}}+frac{left(x_1+x_2+cdots +x_nright)^2}{n}leqsum_{k=1}^n{y_kx_{k}^{2}},
]
其中
[
y_k=sum_{i=k}^{n-1}{frac{a_1+a_2+cdots +a_i}{ileft(i+1right)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$y_kleq 3+2sqrt{2}$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+1right)sqrt{i}}}+frac{1}{sqrt{n}}right).
]
注意到
begin{align*}
2left(frac{1}{sqrt{i}}-frac{1}{sqrt{i+1}}right)&=2frac{sqrt{i+1}-sqrt{i}}{sqrt{i}cdotsqrt{i+1}}=2frac{1}{sqrt{i}cdotsqrt{i+1}left(sqrt{i+1}+sqrt{i}right)}\
&geq 2frac{1}{sqrt{i}cdotsqrt{i+1}left(sqrt{i+1}+sqrt{i+1}right)}=frac{1}{left(i+1right)sqrt{i}}.
end{align*}
因此
begin{align*}
y_k&leqfrac{1}{a_k}left[sum_{i=k}^{n-1}{2left(frac{1}{sqrt{i}}-frac{1}{sqrt{i+1}}right)}+frac{1}{sqrt{n}}right]=frac{1}{a_k}left[2left(frac{1}{sqrt{k}}-frac{1}{sqrt{n}}right)+frac{1}{sqrt{n}}right]\
&leqfrac{2}{a_ksqrt{k}}=frac{2}{left(sqrt{k}-sqrt{k-1}right)sqrt{k}}=frac{2left(sqrt{k}+sqrt{k-1}right)}{sqrt{k}}\
&=2left(1+sqrt{1-frac{1}{k}}right)<4<3+2sqrt{2}.
end{align*}
Archiver|手机版|科学网 ( 京ICP备07017567号-12 )
GMT+8, 2024-9-2 21:19
Powered by ScienceNet.cn
Copyright © 2007- 中国科学报社