# Inequality on diagonal entries of a matrix product

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?

• That's clearly not working. If one of the traces on the RHS is zero, that doesn't make the LHS equal to zero. – Christian Remling Jul 20 '17 at 19:58
• 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). – Todd Trimble Jul 21 '17 at 13:12

## 1 Answer

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.