# Trace inequality with matrix square-roots

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.

• In this context, does positive definite imply symmetric? Nov 11, 2018 at 23:16
• Yes - thank you. Edited to reflect A and B are symmetric. Nov 11, 2018 at 23:17
• 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 $0$. Nov 12, 2018 at 1:28
• Ahh... very good point. Nov 12, 2018 at 1:34
• If you want to add that as a definitive answer, I'd be happy to accept it. Nov 12, 2018 at 2:49

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 $${\rm tr}\left[ (A^{1/2} - B^{1/2})(A^{1/2} + B^{1/2}) \right] \leq \sqrt{{\rm tr}[(A^{1/2} - B^{1/2})^2]} \sqrt{{\rm tr}[(A^{1/2} + B^{1/2})^2]}$$
• To clarify, here you use the $\|\cdot\|$ as the spectral norm? Nov 11, 2018 at 23:28
• 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. Nov 12, 2018 at 0:03