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}


  • $\begingroup$ Looks like a nice inequality and it seems that the triangular nature of $R$ is essential. If all matrices are real, then writing $R=D_R+N$, where $D_R$ is the diagonal and $N$ is the strict upper triangle we see that $RR^T=|D_R|^2+Z$, where matrix $Z$ has only zero eigenvalues; this should suffice to prove the statement; in the complex case, some more work is needed... $\endgroup$ – Suvrit Aug 4 '13 at 4:42
  • $\begingroup$ This is from a communication paper ieeexplore.ieee.org/xpl/… $\endgroup$ – Brian Lan Aug 5 '13 at 2:42
  • $\begingroup$ Hi, suvrit. Thanks for your response. $\mathbf{Z}$ won't have zero eigenvalues. In fact, it won't be a positive semidefinite matrix, either. $\endgroup$ – Brian Lan Aug 5 '13 at 3:15
  • $\begingroup$ @Brian: I meant that in the case all matrices are real, then $Z = ND_R+D_RN+N^2$ has all zero eigenvalues (it is not symmetric however, so we are not talking about it being semidefinite, but having all zero eigenvalues is already nice; unfortunately, for complex matrices, this nice pattern breaks) $\endgroup$ – Suvrit Aug 5 '13 at 6:07
  • $\begingroup$ @suvrit: Let's take a simple example. Let $\mathbf{R} = \left[\begin{matrix}1 & 2\\ 0 & 1\end{matrix}\right]$. Then $\mathbf{Z} = \left[\begin{matrix}4 & 2\\ 2 & 0\end{matrix}\right]$ of which the eigenvalues are approximately $-0.8284$ and $4.8284$. $\endgroup$ – Brian Lan Aug 5 '13 at 13:18

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.