曾经的两道矩阵正定问题的解答
设A∈Mn(R)对称正定, X∈Mn×m(R)且XTX=Im.证明XTA−1X−XTAX−1是半正定矩阵.
证.首先
(XTAXImImXTA−1X)=(XTAXXTXXTXXTA−1X)=(XT00XT)(AInInA−1)(X00X).
而
(In0−A−1In)(AInInA−1)(In−A−10In)=(A000).
所以(AInInA−1)半正定,所以(XTAXImImXTA−1X)半正定.
又(Im0−(XTAX)−1Im)(XTAXImImXTA−1X)(Im−(XTAX)−10Im)=(XTAX00XTA−1X−(XTAX)−1),
所以XTA−1X−XTAX−1是半正定矩阵.
设 xi>0,i=1,2,⋯, 证明矩阵 (ln(1+xi+xj)−ln(1+|xi−xj|)xi+xj−|xi−xj|)n×n是半正定矩阵.
证.(morrismodel)这是一道合成题, 我把它分解开来:
题1. 当0<x1<x2<⋯<xn时,
(min
正定.
证明. 由如下的矩阵恒等式得到:
\begin{align*}&\begin{pmatrix} a_1&\\ a_1&a_2\\ a_1&a_2&a_3\\ \cdots&\cdots&\cdots&\cdots\\ a_1&a_2&a_3&\cdots&a_n \end{pmatrix} \begin{pmatrix} a_1&a_1&a_1&\cdots&a_1\\ 0&a_2&a_2&\cdots &a_2\\ 0&0&a_3&\cdots&a_3\\ \cdots&\cdots&\cdots&\cdots\\ 0&0&0&\cdots&a_n \end{pmatrix}\\=& \begin{pmatrix} a_1^2&a_1^2&a_1^2&\cdots&a_1^2\\ a_1^2&a_1^2+a_2^2&a_1^2+a_2^2&\cdots &a_1^2+a_2^2\\ a_1^2&a_1^2+a_2^2&a_1^2+a_2^2+a_3^2&\cdots&a_1^2+a_2^2+a_3^2\\ \cdots&\cdots&\cdots&\cdots\\ a_1^2&a_1^2+a_2^2&a_1^2+a_2^2+a_3^2&\cdots&a_1^2+a_2^2+a_3^2+\cdots+a_n^2 \end{pmatrix}.\end{align*}
题2. 当正数x_1,x_2,\cdots,x_n两两不相等时,
\left(\min\{x_i,x_j\}\right)_{n\times n}= \begin{pmatrix} x_1&x_1&x_1&\cdots&x_1\\ x_1&x_2&x_2&\cdots &x_2\\ x_1&x_2&x_3&\cdots&x_3\\ \cdots&\cdots&\cdots&\cdots\\ x_1&x_2&x_3&\cdots&x_n \end{pmatrix}
正定.
证明. 对任何N阶置换\sigma\in S_N,
\begin{align*} &\sum_{i,j=1}^Na_{i}a_{j}\min\{x_{\sigma(i)},x_{\sigma(j)}\}=\sum_{i=1}^N\left(\sum_{j=1}^Na_{i}a_{j}\min\{x_{\sigma(i)},x_{\sigma(j)}\}\right)\\=&\sum_{i=1}^N\left(\sum_{j=1}^Na_{i}a_{\sigma^{-1}(j)}\min\{x_{\sigma(i)},x_{j}\}\right)=\sum_{j=1}^N\left(\sum_{i=1}^Na_{i}a_{\sigma^{-1}(j)}\min\{x_{\sigma(i)},x_{j}\}\right)\\=&\sum_{j=1}^N\left(\sum_{i=1}^Na_{\sigma^{-1}(i)}a_{\sigma^{-1}(j)}\min\{x_{i},x_{j}\}\right)=\sum_{i,j=1}^Na_{\sigma^{-1}(i)}a_{\sigma^{-1}(j)}\min\{x_{i},x_{j}\}. \end{align*}
再结合题1就得到结论.
题3. 当正数x_1,x_2,\cdots,x_n两两不相等时,
\left(e^{x_i+x_j-|x_i-x_j|}\right)_{n\times n}= \begin{pmatrix} e^{2x_1}&e^{2x_1}&e^{2x_1}&\cdots&e^{2x_1}\\ e^{2x_1}&e^{2x_2}&e^{2x_2}&\cdots &e^{2x_2}\\ e^{2x_1}&e^{2x_2}&e^{2x_3}&\cdots&e^{2x_3}\\ \cdots&\cdots&\cdots&\cdots\\ e^{2x_1}&e^{2x_2}&e^{2x_3}&\cdots&e^{2x_n} \end{pmatrix}
正定.
证明. 由题2得到.
题4. 当正数x_1,x_2,\cdots,x_n两两不相等且\theta\in[0,1]时,
\left(\frac{1}{1+\theta(x_i+x_j)+(1-\theta)|x_i-x_j|}\right)_{n\times n}
正定.
证明.
\begin{align*} &\sum_{i,j=1}^N\frac{a_ia_j}{1+\theta(x_i+x_j)+(1-\theta)|x_i-x_j|}\\=&\sum_{i,j=1}^Na_ia_j\int_0^\infty e^{-t(1+\theta(x_i+x_j)+(1-\theta)|x_i-x_j|)}dt\\ =&\int_0^\infty e^{-t}\left(\sum_{i,j=1}^N(a_ie^{-tx_i})(a_je^{-tx_j})e^{t(1-\theta)(x_i+x_j-|x_i-x_j|)}\right)dt. \end{align*}
再由题3得到结论.
题5. 当正数x_1,x_2,\cdots,x_n两两不相等时,
\left( {\frac{{\ln \left( {1 + {x_i} + {x_j}} \right) - \ln \left( {1 + \left| {{x_i} - {x_j}} \right|} \right)}}{{{x_i} + {x_j} - \left| {{x_i} - {x_j}} \right|}}} \right)_{n \times n}
正定.
证明.
{\frac{{\ln \left( {1 + {x_i} + {x_j}} \right) - \ln \left( {1 + \left| {{x_i} - {x_j}} \right|} \right)}}{{{x_i} + {x_j} - \left| {{x_i} - {x_j}}\right|}}}=\int_0^1 \frac{1}{1+\theta(x_i+x_j)+(1-\theta)|x_i-x_j|} d\theta.
再由题4得到结论.
回到原题, 没有假设正数x_i两两不等, 只需在题5的基础上摄动一下就能得到半正定性.
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