Recently, I have seen a matrix inequality but don't know how to prove it. The inequality goes as follows.


For an arbitrary $n\times n$ diagonal matrix $\mathbf{D}$ and an arbitrary upper-triangular matrix of the same size $\mathbf{R}$, we have
\begin{align}
\det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H) \geq \prod\_{i=1}^n(|\mathbf{D}\_{ii}|^2+|\mathbf{R}\_{ii}|^2).
\end{align}


Thanks.



Determinant inequality of square-product sum of diagonal matrix and upper-triangular matrix
I have found the proof by mathematical induction. Thanks to Sean Shih and Chu-Lan Kao for fruitful discussions.


For $n=1$, the inequality is obvious.


Suppose that for $n=m\in\mathcal{N}$, $\det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H)\geq \prod\_{i=1}^m(|\mathbf{D}\_{ii}|^2+|\mathbf{R}\_{ii}|^2)$ holds. 


For $n = m+1$, let $\mathbf{D}'=\left[\begin{matrix}\mathbf{D} & 0\\ 0 & d\end{matrix}\right]$ and $\mathbf{R}'=\left[\begin{matrix}\mathbf{R} & \mathbf{r}\\ \mathbf{0} & r\end{matrix}\right]$ where $\mathbf{D}$ is an $m\times m$ diagonal matrix, $\mathbf{R}$ is an $m\times m$ upper-triangular matrix, $\mathbf{r}$ is an $m \times 1$ vector, and $d$ and $r$ are scalars. Let $\mathbf{A} = \mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H$ be non-singular, or the inequality is obvious. Then
\begin{align}
&\det(\mathbf{D}'\mathbf{D}'^H+\mathbf{R}'\mathbf{R}'^H) \\
= &\det(\mathbf{A}+\mathbf{rr}^H)(|d|^2+|r|^2-|r|^2\mathbf{r}^2(\mathbf{A}+\mathbf{rr}^H)^{-1}\mathbf{r}^H)\\
=&\det(\mathbf{A})(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})(|d|^2+|r|^2-|r|^2(1-(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})^{-1}))\\
=&\det(\mathbf{A})(|d|^2+|r|^2)+\det(\mathbf{A})|d|^2\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r}\\
\geq &\det(\mathbf{A})(|d|^2+|r|^2) = \prod\_{i=1}^{m+1}(|\mathbf{D}'\_{ii}|^2+|\mathbf{R}'\_{ii}|^2)
\end{align}
where we use matrix inversion lemma and matrix determinant lemma.
By mathematical induction, the proof is completed.


