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Define $f(X) = \operatorname{tr}(MXX^T)$. If $M$ is a positive semi-definite matrix, can we prove that $f$ is convex?

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$$ s\cdot f(X)+(1-s)\cdot f(Y)=f(sX+(1-s)Y)\color{red}{+s(1-s)\cdot f(X-Y)} $$

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  • $\begingroup$ And...what's next? $\endgroup$ – Mathematics Lover Mar 23 '12 at 15:07
  • $\begingroup$ It's already over, for anybody knowing the meaning of the words in your question. $\endgroup$ – Did Mar 23 '12 at 15:16
  • $\begingroup$ Er, yes, I understand from the above that f is convex, I just don't quite understand how we can get this equality... $\endgroup$ – Mathematics Lover Mar 23 '12 at 16:33
  • $\begingroup$ $M(sX+(1-s)Y)(sX+(1-s)Y)^T=\ldots$ and the operator trace is linear hence $f(sX+(1-s)Y)=\ldots$ $\endgroup$ – Did Mar 23 '12 at 18:02

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