2
$\begingroup$

Let $A$ and $B$ be two Hermitian matrices and let $D$ be a diagonal matrix.

Does there exist any inequality involving the trace for the diagonal entries $(D A D A D A B)_{i,i}$?

I am looking for something like

$$ |(D A D A D A B)_{i,i}| \leq | \text{Tr}\left(D^3\right)^{p} \text{Tr}\left(A^3\right)^{q} \text{Tr}\left(B\right)^{r} | $$

I was thinking about a tip like

$$|(D A D A D A B)_{i,i}| = |\text{Tr}(D A D A D A E_{i,i} B)| \leq | \text{Tr}\left(D^3A^3 E_{i,i}\right)^{p} \text{Tr}\left(B\right)^{q} |$$

where $E_{i,i}$ is the matrix with null entries everywhere except in position $(i,i)$, where the entry is equal to $1$. Does there exist any inequality like that?

$\endgroup$
  • 4
    $\begingroup$ That's clearly not working. If one of the traces on the RHS is zero, that doesn't make the LHS equal to zero. $\endgroup$ – Christian Remling Jul 20 '17 at 19:58
  • 2
    $\begingroup$ Hi Nawak. It looks like you have two accounts with the same name. If you would like to have them merged, there's a button "contact us" where you can ask an SE Community Manager to do this (site moderators can't do this, unfortunately). $\endgroup$ – Todd Trimble Jul 21 '17 at 13:12
1
$\begingroup$

Let $\rm A, B, D$ be $n \times n$ matrices. Using von Neumann's trace inequality [0],

$$\begin{array}{rl} | (\mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B)_{i,i} | = | \mathrm e_i^\top \mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B \, \mathrm e_i | &= | \,\,\mbox{tr} \left( \mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B \,\mathrm e_i \mathrm e_i^\top \right) | \\ &\leq \displaystyle\sum_{k=1}^n \sigma_k (\mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B) \, \sigma_k ( \mathrm e_i \mathrm e_i^\top ) \end{array}$$

Since $\mathrm e_i \mathrm e_i^\top$ is a rank-$1$ matrix, it has only one nonzero singular value, which is $1$. Hence,

$$| (\mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B)_{i,i} | \leq \sigma_{\max} (\mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B) = \| \mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B \|_2$$

and, since the spectral norm is submultiplicative,

$$| (\mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B)_{i,i} | \leq \| \mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B \|_2 \leq \| \mathrm A \|_2^3 \cdot \| \mathrm B \|_2 \cdot \| \mathrm D \|_2^3$$

Recall that $\| \rm M \|_2 \leq \| \rm M \|_{\text{F}}$. Since $\rm A$ and $\rm B$ are Hermitian,

$$| (\mathrm D \mathrm A \mathrm D \mathrm A \mathrm D \mathrm A \mathrm B)_{i,i} | \leq \| \mathrm A \|_{\text{F}}^3 \cdot \| \mathrm B \|_{\text{F}} \cdot \| \mathrm D \|_{\text{F}}^3 = \color{blue}{\left( \mbox{tr} (\mathrm A^2) \right)^{\frac 32} \cdot \left( \mbox{tr} (\mathrm B^2) \right)^{\frac 12} \cdot \left( \mbox{tr} (\mathrm D^* \mathrm D) \right)^{\frac 32}}$$


[0] L. Mirsky, A trace inequality of John von Neumann, Monatshefte für Mathematik, Volume 79, Issue 4, pp 303–306, December 1975.

$\endgroup$

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.