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Let $X$ and $L$ be real positive definite matrices.

$$\operatorname{Trace}(X^{-1}(X - e^{\log(X) - L})^2) \leq \operatorname{Trace}(XL^2)$$

where the exponential and the log are matrix exponential and log (defined as applying those functions to the eigenvalues)

I think I have for sanity check checked basic cases like when these are 1x1 or which basically implies the inequality for the case when X and L are diagonal which in turn implies it for the case when X and L commute.

I would love to have a simple proof if it is true. Even if this is true under some mild conditions on L (like spectral norm $\le$ constant) that is fine too.

Thanks

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  • $\begingroup$ It's false for $X=I_n,L=aI_n$ where $a<0$. $\endgroup$
    – user91684
    Sep 10, 2015 at 14:45
  • $\begingroup$ I am not sure about your example per say but there might be a slight error in an argument I had. Thanks anyways. $\endgroup$ Sep 10, 2015 at 16:46
  • $\begingroup$ I think the statement might hold for PSD matrices L. I have modified the statement accordingly. $\endgroup$ Sep 10, 2015 at 16:58
  • $\begingroup$ You don't need eigenvalues to define exponential and logarithm of a matrix; you can use power series: $$\exp X = \sum_{n=0}^\infty \frac{X^n}{n!}.$$ $\endgroup$ Sep 10, 2015 at 17:15
  • $\begingroup$ @MichaelHardy in what applications is the power series of a matrix more important to use instead of the eigenvalues of a matrix? aren't eigenvalues more analytically important than power series? $\endgroup$
    – develarist
    Dec 23, 2020 at 12:51

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