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Let $A$ and $B$ be two matrices with $\det(A)=\det(B)=1$. Does it follow that

$\sqrt{\mathrm{tr}(A^TB^TBA-I)}\le\sqrt{\mathrm{tr}(A^TA-I)}+\sqrt{\mathrm{tr}(B^TB-I)}$

I suspect that this can be shown using the singular value decomposition, but I've not been able to write a proof yet. Scaling arguments suggest that the determinant condition is really needed.

If needed, it may be assumed that all singular values of both $A$ and $B$ are positive.

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    $\begingroup$ Are the trace operators missing under the square roots on the right? (My guess would be that they are but I'll leave it to the OP to edit them in). $\endgroup$
    – fedja
    Nov 2, 2012 at 15:16
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    $\begingroup$ Anyway, looking at the sheer size, the claim seems fishy: let $A=B$ be the diagonal matrix with the entries $x,1/x$ on the diagonal. Then the LHS is about $x^2$ and the right hand side is about $2x$ as $x\to\infty$. $\endgroup$
    – fedja
    Nov 2, 2012 at 15:20
  • $\begingroup$ It seems that the question can be "fixed" but since the OP has not edited the question despite fedja's countex, I'm not sure if this fix would be of interest.... $\endgroup$
    – Suvrit
    Nov 6, 2012 at 18:05
  • $\begingroup$ @Suvrit: How would you fix this? I'm open to suggestions. $\endgroup$
    – Martijn
    Nov 7, 2012 at 8:08

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As per fedja's comment, the version of the inequality written in the question does not hold. A version that does hold is given below.

$\newcommand{\trace}{\mathrm{tr}}$ Define $\trace_n(X) := \frac{1}{n}\trace(X)$ be the normalized trace. Let $U, V$ be unitary matrices. Then, the following inequality can be shown to hold \begin{equation*} \sqrt{1-|\trace_n(UV)|^2} \le \sqrt{1-|\trace_n(U)|^2} + \sqrt{1-|\trace_n(V)|^2}, \end{equation*} with equality if and only if $U$ or $V$ is a unitary scalar matrix.

For an elementary proof, please see the paper: "A trace inequality for unitary matrices" by B.-Y. Wang and F. Zhang (AMM, 101(5), 1994).

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  • $\begingroup$ Thanks for posting this answer. Any chance that something can be done if $U$ or $V$ have an eigenvalue not on the unit circle? $\endgroup$
    – Martijn
    Nov 8, 2012 at 8:15
  • $\begingroup$ @Martijn: I had once obtained the above ineq. as a special case of something else; if i recall the context, i'll update the answer. some reasonable generalization is certainly possible because the above ineq. is really just the triangle ineq. in disguise. $\endgroup$
    – Suvrit
    Nov 8, 2012 at 17:04
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    $\begingroup$ Wang-Zhang's inequality was revisited in my little note: Remarks on Krein’s Inequality link.springer.com/article/10.1007/s00283-011-9270-z $\endgroup$
    – M. Lin
    Nov 4, 2015 at 10:20

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