一道矩阵的行列式不等式
\(\Large\bf 定理\,\normalsize{\text{8-1-2}}\):设\(\boldsymbol{M}\)与\(\boldsymbol{N}\)为任意\(m\times n\)矩阵,恒有不等式:\[\large1-\sqrt{\displaystyle\det\left(\boldsymbol{I}+\boldsymbol{M}\boldsymbol{M}^H\right)-1}\cdot\sqrt{\displaystyle\det\left(\boldsymbol{I}+\boldsymbol{N}\boldsymbol{N}^H\right)-1}\leqslant\left|\det\left(\boldsymbol{I}+\boldsymbol{M}\boldsymbol{N}^H\right)\right|\]
{:dizzy:} You asked this two years ago, I am not sure if you are still interested now.
Here is a proof. Clearly
$$\begin{pmatrix} I+MM^* & I+MN^*\\ I+NM^* & I+NN^*
\end{pmatrix}\ge \begin{pmatrix} I & I\\ I & I
\end{pmatrix},$$ where $A\ge B$ means $A-B$ is positive semidefinite.
This gives $$\begin{pmatrix} \det(I+MM^*) & \det(I+MN^*)\\ \det(I+NM^*) & \det(I+NN^*)
\end{pmatrix}\ge \begin{pmatrix} 1 & 1\\ 1 & 1
\end{pmatrix},$$
in other words, the matrix $ \begin{pmatrix} \det(I+MM^*)-1 & \det(I+MN^*)-1\\ \det(I+NM^*)-1 & \det(I+NN^*)-1
\end{pmatrix}$ is positive semidefinite. Taking determinant gives the desired result.
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