每週問題 November 28, 2016

Let $A=[a_{ij}]$ be an $n\times n$ Hermitian and positive semidefinite matrix and $B=[b_{ij}]$ with the property $b_{ij}=1/a_{ij}$. Show that $B$ is positive semidefinite if and only if $\hbox{rank}A=1$.

$\displaystyle \begin{vmatrix} a_{ii}&a_{ij}\\ a_{ji}&a_{jj} \end{vmatrix}=a_{ii}a_{jj}-a_{ij}\overline{a}_{ij}=\Vert\mathbf{p}_i\Vert^2\Vert\mathbf{p}_j\Vert^2-\vert\mathbf{p}_i^\ast\mathbf{p}_j\vert^2\ge 0$

$\displaystyle \begin{vmatrix} b_{ii}&b_{ij}\\ b_{ji}&b_{jj} \end{vmatrix}=(a_{ii}a_{jj})^{-1}-(a_{ij}\overline{a}_{ij})^{-1}=\Vert\mathbf{p}_i\Vert^{-2}\Vert\mathbf{p}_j\Vert^{-2}-\vert\mathbf{p}_i^\ast\mathbf{p}_j\vert^{-2}\ge 0$

\displaystyle \begin{aligned} \mathbf{x}^\ast B\mathbf{x}&= \sum_{i=1}^n\sum_{j=1}^n\overline{x}_ia_{ij}^{-1}x_j=\sum_{i=1}^n\sum_{j=1}^n\frac{\overline{x}_ix_j}{v_i\overline{v}_j}\\ &=\left(\sum_{i=1}^n\frac{\overline{x}_i}{v_i}\right)\overline{\left(\sum_{j=1}^n\frac{\overline{x}_j}{{v_j}}\right)}=\left|\sum_{i=1}^n\frac{\overline{x}_i}{v_i}\right|^2\ge 0. \end{aligned}

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