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Suppose I have symmetric matrices $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times n}$ which are both positive definite. I am wondering if I one can bound ${\rm tr}\left(A - B \right)$ in the following way:

\begin{align*} {\rm tr}\left(A - B \right) & = {\rm tr}\left[ (A^{1/2} - B^{1/2})(A^{1/2} + B^{1/2}) \right] \\ & \leq {\rm tr}\left[ (A^{1/2} - B^{1/2})\right] f(A^{1/2} + B^{1/2}), \end{align*}

for some function $f$, e.g., spectral norm? Does such an inequality exist?

In full generality, I cannot say anything about the sign of the eigenvalues of $A^{1/2} - B^{1/2}$, so as far as I can tell, many of the standard inequalities do not apply.

Any insight would be very helpful.

Edit: An example showing that such an inequality will fail in certain cases was suggested by Darij Grinberg in the comments below.

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  • $\begingroup$ In this context, does positive definite imply symmetric? $\endgroup$ Nov 11, 2018 at 23:16
  • $\begingroup$ Yes - thank you. Edited to reflect A and B are symmetric. $\endgroup$
    – WazyMaze
    Nov 11, 2018 at 23:17
  • $\begingroup$ Your first inequality is bound to fail if $A = \operatorname{diag}(1, 16)$ and $B = \operatorname{diag}(4, 9)$, since the first factor on the RHS will just be $0$. $\endgroup$ Nov 12, 2018 at 1:28
  • $\begingroup$ Ahh... very good point. $\endgroup$
    – WazyMaze
    Nov 12, 2018 at 1:34
  • $\begingroup$ If you want to add that as a definitive answer, I'd be happy to accept it. $\endgroup$
    – WazyMaze
    Nov 12, 2018 at 2:49

1 Answer 1

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One way to get your inequality: let $e_i$ be an orthonormal eigenbasis of $A^{1/2} + B^{1/2}$. We have $$ {\rm tr}\left[ (A^{1/2} - B^{1/2})(A^{1/2} + B^{1/2}) \right] = \sum_{i=1}^n e_i^T(A^{1/2} - B^{1/2})(A^{1/2} + B^{1/2})e_i =\\ \sum_{i=1}^n \lambda_i e_i^T (A^{1/2} - B^{1/2}) e_i \leq \sum_{i=1}^n \|A^{1/2} + B^{1/2}\| \, e_i^T (A^{1/2} - B^{1/2}) e_i =\\ \|A^{1/2} + B^{1/2}\| \operatorname{tr}(A^{1/2} - B^{1/2}) $$


Another inequality that may interest you: because $A,B \mapsto \operatorname{tr}(AB)$ forms an inner product over the symmetric matrices, we may use the Cauchy-Schwarz inequality to conclude that $$ {\rm tr}\left[ (A^{1/2} - B^{1/2})(A^{1/2} + B^{1/2}) \right] \leq \sqrt{{\rm tr}[(A^{1/2} - B^{1/2})^2]} \sqrt{{\rm tr}[(A^{1/2} + B^{1/2})^2]} $$

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  • $\begingroup$ To clarify, here you use the $\|\cdot\|$ as the spectral norm? $\endgroup$
    – WazyMaze
    Nov 11, 2018 at 23:28
  • $\begingroup$ Could be an issue on my end, but in small scale experiments, the first inequality is frequently violated. In general, $e_i'(A^{1/2} - B^{1/2})e_i$ could be negative, so I'm not sure your $\leq$ argument holds. It would if $e_i'(A^{1/2} - B^{1/2})e_i$ was replaced with its absolute value. $\endgroup$
    – WazyMaze
    Nov 12, 2018 at 0:03
  • $\begingroup$ Ah, you're right. I'll leave the answer up in case you find it useful, but I probably won't find the time to fix it. $\endgroup$ Nov 12, 2018 at 0:17

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