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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?

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    $\begingroup$ 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. $\endgroup$
    – Chris Ramsey
    Commented Nov 7, 2017 at 20:34
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    $\begingroup$ Oh. Sorry, I should have added symmetric condition! I corrected the statement. $\endgroup$
    – Nikolayevich
    Commented Nov 7, 2017 at 20:38

2 Answers 2

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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*}

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  • $\begingroup$ 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? $\endgroup$
    – LSpice
    Commented Nov 7, 2017 at 21:25
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    $\begingroup$ 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? $\endgroup$
    – Johannes Hahn
    Commented Nov 7, 2017 at 21:35
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    $\begingroup$ @LSpice Araki proved a more general inequality (see link.springer.com/article/10.1007%2FBF01045887); here the simpler Lieb-Thirring result from 1976 suffices. $\endgroup$
    – Suvrit
    Commented Nov 7, 2017 at 23:53
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    $\begingroup$ "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. $\endgroup$
    – Christian Remling
    Commented Nov 7, 2017 at 23:55
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    $\begingroup$ @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. $\endgroup$
    – Suvrit
    Commented Nov 7, 2017 at 23:58
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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 $M_2.$

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  • $\begingroup$ So it seems we only need to assume one of them is symmetric. $\endgroup$
    – Nikolayevich
    Commented Nov 8, 2017 at 6:18
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    $\begingroup$ 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 $B$. $\endgroup$
    – Dan Fox
    Commented Nov 8, 2017 at 8:38
  • $\begingroup$ @Koltchinskii I was assuming both of them were symmetric... $\endgroup$
    – Igor Rivin
    Commented Nov 8, 2017 at 13:27

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