一道矩阵的行列式不等式

青青子衿

\(\Large\bf 定理\,\normalsize{\text{8-1-2}}\)：设\(\boldsymbol{M}\)与\(\boldsymbol{N}\)为任意\(m\times n\)矩阵，恒有不等式：
\[\large1-\sqrt{\displaystyle\det\left(\boldsymbol{I}+\boldsymbol{M}\boldsymbol{M}^H\right)-1}\cdot\sqrt{\displaystyle\det\left(\boldsymbol{I}+\boldsymbol{N}\boldsymbol{N}^H\right)-1}\leqslant\left|\det\left(\boldsymbol{I}+\boldsymbol{M}\boldsymbol{N}^H\right)\right|\]

orzweb111

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.

hbghlyj

math.stackexchange.com/questions/5003378

if $m\gt n$ you can just append columns of zeros to each of $M$ and $N$; this does not change $I + MM^*$ or $I+MN^*$ etc.  If $m\lt n$ I would probably try instead to work with $M^*M$, $N^*M$, $M^*N$ and $N^*N$ and then re-use the prior sentence's argument (i.e. appending columns of zeros to $M^*$ and $N^*$).

学习。不会