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Let $A$ be such that $\operatorname{det}A > 0$ and $B$ be symmetric positive definite matrix. Is it then true that $$f(A, B) = \operatorname{tr}(ABA^{\top})$$ is a jointly convex function in $(A, B)$?

I'm familiar with the paper of Elliott Lieb but I'm not sure what can be said about the simple case above. Perhaps, no convexity/concavity result can be obtained in this case? And if so, can someone explain in detail?

sleepingrabbit
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    A good first shot is to try with scalars. This would reduce to $f(A,B)=A^2B$ on $A>0, B>0$. A quick look at $f(t,1-t) = t^2-t^3$ on $[0,1]$ will show that it is not convex. – copper.hat Sep 18 '20 at 16:16
  • Yup, @copper.hat 's counterexample is a good one. In general when you've got that sort of nonlinearity (in your case, cubic, or even quadratic) on your inputs, instances of joint convexity are hard to come by. This is discussed at-length in the blind deconvolution literature as well. – Zim Sep 18 '20 at 16:24
  • @Zim Blind deconvolution is a new one for me! – copper.hat Sep 18 '20 at 16:29

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