4
$\begingroup$

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.

Improve this question
$\endgroup$
6
  • $\begingroup$ 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
    Commented Aug 4, 2013 at 4:42
  • $\begingroup$ This is from a communication paper ieeexplore.ieee.org/xpl/… $\endgroup$
    – Brian Lan
    Commented Aug 5, 2013 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
    Commented Aug 5, 2013 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
    Commented Aug 5, 2013 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
    Commented Aug 5, 2013 at 13:18

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .