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I am familiar with the Cauchy_Schwartz inequality

$$|Tr[A^*B]|^2 \le |Tr[A^*A]| |Tr[B^*B]|$$

where $*$ denotes the conjugate-transpose operation. I am wondering if there is a similar inequality for $Tr[ABCD]$? Of course, one could ask the same question for the product of $n$ such matrices, but I am interested only in four!

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  • $\begingroup$ Hello :) Yes, you just have to write the entries of $A$ (and $B$) in a vector to get the ordinary cauchy schwarz inequality. We get $(\sum_{i, j=1}^na_{ij}b_{ij})^2\leq (\sum_{i, j=1}^na_{ij}^2)(\sum_{i, j=1}^nb_{ij}^2)$ $\endgroup$
    – Jochen
    Commented 2 days ago
  • $\begingroup$ Thanks @Jochen, could you expand your comment as answer? $\endgroup$
    – Mike
    Commented 2 days ago
  • $\begingroup$ I don't know if I'd call it "similar" necessarily, but one natural extension of Cauchy-Schwarz to more vectors is positive-(semi)definiteness of the Gram matrix. It is not difficult to prove that the Gram matrix is positive-semidefinite, and positive definite if and only if the vectors used to form it are linearly independent. If using two vectors, then you get precisely Cauchy-Schwarz, complete with equality condition. Of course, a $4 \times 4$ determinant will yield a fairly messy inequality involving inner products! $\endgroup$ Commented 2 days ago

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Replacing $A$ by $A^*$ in the classical Cauchy-Schwarz inequality, you obtain that for all $A,B$, $$ |\mathrm{tr}(AB)|^2 \leqslant \mathrm{tr}(A^*A)\mathrm{tr}(B^*B). $$ You can use this inequality twice to obtain a new one with $4$ matrices, \begin{align*} |\mathrm{tr}(ABCD)|^4 & \leqslant \mathrm{tr}((AB)^*AB)^2\mathrm{tr}((CD)^*CD)^2\\ & = \mathrm{tr}(A^*ABB^*)^2\mathrm{tr}(C^*CDD^*)^2\\ & \leqslant \mathrm{tr}((A^*A)^*A^*A)\mathrm{tr}((BB^*)^*BB^*)\mathrm{tr}((C^*C)^*C^*C)\mathrm{tr}((DD^*)^*DD^*)\\ & = \mathrm{tr}\left((A^*A)^2\right)\mathrm{tr}\left((B^*B)^2\right)\mathrm{tr}\left((C^*C)^2\right)\mathrm{tr}\left((D^*D)^2\right). \end{align*}

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  • $\begingroup$ No, take $A = B = I_n$, $C$ an invertible matrix with $\mathrm{tr}(C) = 0$ and $D = C^{-1}$. Then $\mathrm{tr}(ABCD) = n$ and $\mathrm{tr}(A^*BC) = 0$ $\endgroup$
    – Cactus
    Commented yesterday

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