# Well known matrix inequality?

I suspect that the following matrix inequality is well known, but I can't find a reference or proof:

Given $n \times n$ symmetric matrices $A,B$ such that $I_n \leq A,B$, is the following true? $${Tr}\big[\big( \log A^{\frac{1}{2}} B A^{\frac{1}{2}}\big)^p\big] \geq {Tr}\big[\big( \log B \big)^p\big] + {Tr}\big[\big( \log A\big)^p\big].$$

For clarity, for a continuous function $f$ and symmetric matrix $C$, $f(C)$ is simply equal to $U {diag}(f(\lambda_1),...,f(\lambda_n))U^t$, where $U$ is a unitary matrix diagonalizing $C$, i.e., $C = U diag(\lambda_1,...,\lambda_n)U^t$.

UPDATE: after the discussion with amakelov below, I edited the statement to be more aesthetic.

• just a little note: $BA^{-1}$ is not necessarily symmetric - was that the intention? If not, maybe you can replace it by something like $A^{-1/2}BA^{-1/2}$ (of course this still works fine but you need to define the log without appealing to the diagonalization) – amakelov May 12 '18 at 22:24
• What you write is worth considering. Unless I am mistaken, $BA^{-1}$ and $A^{-1/2}BA^{-1/2}$ seem to have the same eigenvalues. This is the main reason behind my formulation. The other was the fact that I wanted to have a statement similar to the one with real numbers. But I do understand the logic behind working with "only" symmetric matrices. Now that I write this, your point of view seems more appealing. Edited. – Hammerhead May 12 '18 at 22:38
• Seems to be trivial, denote by a=log A and by b=logB and use the first inequality and the additivity of the trace. – Alan May 13 '18 at 5:33
• @Alan, if you think you have a solution, please write it out. I will gladly accept it. Saying that it is trivial and leaving a few superficial comments does not seem to serve any purpose. – Hammerhead May 13 '18 at 5:51

The desired inequality follows from an majorisation argument :

According to Bhatia Matrix Analysis Corollary III.4.6 we have :

$log \lambda(AB) \succ log \lambda^\downarrow(A) + log \lambda^\uparrow(B)$.

Using Corollary II.3.4 from the same book, we conclude that
$(log \lambda(AB))^p \succ_w (log \lambda^\downarrow(A) + log \lambda^\uparrow(B))^p \succ_w (log \lambda^\downarrow(A))^p + (log \lambda^\uparrow(B))^p$ , since $x \mapsto x^p$ is convex and monoton increasing for $x \ge 0$ .

From this it follows the trace inequality.

• Very Nice answer! – Mahdi May 13 '18 at 17:16
• I used Corollary II.3.4 in Bathia and applied $x \mapsto x^p$ componentwise ("induced map") . – jjcale May 13 '18 at 18:45