# Trace of non-commutable matrices

Let $M_1$ and $M_2$ be two symmetric $d\times d$ matrices. What is the relationship between $tr(M_1M_2M_1M_2)$ and $tr(M_1^2 M_2^2 )$?

P.S. I tried a few examples and found $$tr(M_1M_2M_1M_2) \le tr(M_1^2 M_2^2 )$$ seems always true. Is there a theorem?

• No there isn't. Try matrix units $M_1 = E_{12}, M_2 = E_{21} \in \mathbb M_2$. Then the LHS is 1 and the RHS is 0. – Chris Ramsey Nov 7 '17 at 20:34
• Oh. Sorry, I should have added symmetric condition! I corrected the statement. – Koltchinskii Nov 7 '17 at 20:38

Your conjecture is a special case of the following result which essentially follows from the Lieb-Thirring inequality.

Let $A$ and $B$ be Hermitian matrices. Then, for every positive integer $p$ we have \begin{equation*} |\text{tr}(AB)^{2p}| \le \text{tr}A^{2p}B^{2p} \end{equation*}

• Wikipedia insists that something like this be called Araki–Lieb–Thirring, not just Lieb–Thirring. Its result by the latter name seems unrelated, and its result by the former name requires that the matrices be positive definite, as well as considering $B A B$ in place of $A B$. Is there some other version that drops these differences? – LSpice Nov 7 '17 at 21:25
• So the proof is $tr(ABAB) = tr( (A^{1/2} B A^{1/2})^2) \leq tr(A^{1/2} B A^{1/2})^2 = tr(AB)^2 \overset{ALT}{\leq} tr(A^2 B^2)$, yes? – Johannes Hahn Nov 7 '17 at 21:35
• @LSpice Araki proved a more general inequality (see link.springer.com/article/10.1007%2FBF01045887); here the simpler Lieb-Thirring result from 1976 suffices. – Suvrit Nov 7 '17 at 23:53
• "Lieb-Thirring inequality" normally means: bound the $\ell^p$ norm of the negative eigenvalues of a Schrodinger operator by the $L^q$ norm of the potential, though it's very possible of course that they proved the matrix inequality above somewhere in these works. – Christian Remling Nov 7 '17 at 23:55
• @ChristianRemling I'm just using the version apparent from Araki's paper, who cites Lieb-Thirring as having proved the simpler inequality. But I agree, outside of matrix inequality land, dropping Araki's name may confuse some. – Suvrit Nov 7 '17 at 23:58

Not to take anything away from Suvrit's answer, but this is actually much simpler. First, we can assume $M_1$ is diagonal. Call it $diag(x_1, \dotsc, x_i).$ Then the difference between the LHS and the RHS is

$$\sum_{i> j} a_{ij}^2 (x_i - x_j)^2,$$ where the $a_{ij}$ are the entries of $M_2.$

• So it seems we only need to assume one of them is symmetric. – Koltchinskii Nov 8 '17 at 6:18
• Another proof is $0 \leq \text{tr}[A, B]^{t}[A, B] = 2\text{tr}(A^{2}B^{2}) - 2\text{tr}(ABAB)$, for symmetric $A$ and $B$. – Dan Fox Nov 8 '17 at 8:38
• @Koltchinskii I was assuming both of them were symmetric... – Igor Rivin Nov 8 '17 at 13:27