Showing a trace inequality for positive semidefinite matrices.

I'm trying to show the following inequality (although I'm not sure if it's true or not): Let $$A$$ and $$B$$ be positive semidefinite matrices. Then I want to show $$\mathrm{tr}((A+B)^{-1}) \leq \textrm{tr}(A^{-1}).$$ Since the eigenvalues of a positive semidefinite matrix are positive, and trace is just sum of eigenvalues, it's clear that $$\textrm{tr}(A)\leq\textrm{tr}(A+B).$$ What's not clear is if I can invert both sides and flip the inequality. Any help would be greatly appreciated.

2 Answers

Yes, the inequality you're trying to prove holds. One proof is as follows.

Note that $$A^{-1/2}BA^{-1/2}$$ is PSD. It follows that $$I + A^{-1/2}BA^{-1/2}$$ is PSD with eigenvalues greater than or equal to $$1$$. It follows that $$(I + A^{-1/2}BA^{-1/2})^{-1}$$ is PSD with eigenvalues less than or equal to $$1$$. So, $$I - (I + A^{-1/2}BA^{-1/2})^{-1}$$ is positive semidefinite. On the other hand, $$A^{-1/2}(I - (I + A^{-1/2}BA^{-1/2})^{-1})A^{-1/2} = \\ A^{-1/2}(I - [A^{-1/2}(A + B)A^{-1/2}]^{-1})A^{-1/2} = \\ A^{-1/2}(I - A^{1/2}(A + B)^{-1}A^{1/2})A^{-1/2} = \\ A^{-1} - (A + B)^{-1}.$$ So, $$A^{-1} - (A + B)^{-1}$$ is PSD. So, the trace of $$A^{-1} - (A + B)^{-1}$$ is non-negative.

Thus, the trace of $$A^{-1}$$ is greater than or equal to that of $$(A + B)^{-1}$$, as desired.

If $$C\le D$$ then $$D^{-1}\le C^{-1}$$. A Hint since your result follows from that directly.

• I agree that my result follows directly, but I'm having trouble showing this. It would suffice to show $$\lambda_1 + \dots + \lambda_n\leq \nu_1+\dots + \nu_n\implies \lambda_1^{-1} + \dots + \lambda_n^{-1} \geq\nu_1^{-1} + \dots + \nu_n^{-1},$$ where $\lambda_i,\nu_i$ all positive, but it's not clear to me how one would show this.
– Phil
Commented Oct 23, 2019 at 15:56
• Indeed i know proofs for the inequality in matrices but there are not trivial. Any where it should be found the other answer Maybe proves it also. Besides if you want your particular result it maybe different since it is weaker... Commented Oct 23, 2019 at 20:55
• Any chance you could provide a reference to one such non-trivial proof?
– Phil
Commented Oct 23, 2019 at 20:56
• i ll check if you want a proof for both things. Your trace inequality i don't see it in general like trivial. Commented Oct 23, 2019 at 21:01
• @Phil Regarding the inequality, the usual (direct) proof is as follows: $$C \leq D \iff\\ D^{-1/2}CD^{-1/2} \leq I \iff\\ D^{1/2}C^{-1}D^{1/2} \geq I \iff\\ C^{-1} \geq D^{-1}$$ Here $\leq$ is the Loewner ordering. Horn and Johnson's Matrix Analysis should be a sufficient reference if you want to look it up. Commented Oct 27, 2019 at 22:30