张祖锦常用结论13两个正定矩阵的 Hadamard 积还是正定的

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# 张祖锦常用结论两个正定矩阵的 Hadamard 积还是正定的 设矩阵 $\displaystyle A=(a\_\{ij\})\_\{n\times n\}, B=(b\_\{ij\})\_\{n\times n\}, C=(c\_\{ij\})\_\{n\times n\}$, 其中 $\displaystyle c\_\{ij\}=a\_\{ij\}b\_\{ij\}, i,j=1,2,\cdots,n$. 如果 $\displaystyle A, B$ 均为正定矩阵, 则 $\displaystyle C$ 也是正定矩阵. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle A$ 正定知存在可逆阵 $\displaystyle T$ 使得 $\displaystyle A=T^\mathrm\{T\} T$. 于是对 $\displaystyle \forall\ 0\neq x\in\mathbb\{R\}^n$, \begin\{aligned\} x^\mathrm\{T\} Cx=&\sum\_\{i,j\}c\_\{ij\}x\_ix\_j =\sum\_\{i,j\}a\_\{ij\}b\_\{ij\}x\_ix\_j =\sum\_\{i,j,k\} t\_\{ki\}t\_\{kj\}b\_\{ij\}x\_ix\_j\\\\ \stackrel\{y\_\{ki\}=t\_\{ki\}x\_i\}\{=\}&\sum\_k \left\[ \sum\_\{i,j\}b\_\{ij\}y\_\{ki\}y\_\{kj\}\right\] \stackrel\{y\_k=(y\_\{k1\},\cdots,y\_\{kn\})^\mathrm\{T\}\}\{=\}\sum\_k y\_k^\mathrm\{T\} B y\_k\\\\ > &0\left((y\_1,\cdots,y\_n)=T\mathrm\{diag\}(x\_1,\cdots,x\_n)\Rightarrow \mbox\{至少某\}y\_l\neq 0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/