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Suppose there are two symmetric matrices $A$ and $B$. Then can we write $$\frac{tr(AB)}{B}=A$$ Any help in this regard will be much appreciated.

The reason for question:

I am trying to understand an example from appendix A of convex optimization book (by Stephen Boyd). On page 642 it is written that $f(Z)\simeq f(X)+tr(X^{-1}(Z-X))$. Now the gradient of $f(X)$ is then shown to be $X^{-1}$. If I put $f(Z)$ in Equation A.4 then I have $$\frac{||f(Z)-f(X)-Df(X)(Z-X)||_{2}}{||Z-X||_2}=0$$ which results in $$tr(X^{-1}(Z-X))=Df(X)(Z-X)$$ Now if $tr(X^{-1}(Z-X))=X^{-1}(Z-X)$ then I can understand but I am not sure whether it is true or not. So any help in this regard will be much appreciated. Thanks in advance.

Frank Moses
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1 Answers1

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If $f:M_n (\mathbb R)\to\mathbb R $, then its derivative at $X $ is a linear map $Df(X):M_n (\mathbb R)\to\mathbb R $. What your text is saying is that in this case the map $Df (X) $ maps $$\tag1A \longmapsto\text {tr}(X^{-1}A) .$$

As every linear map $L:M_n (\mathbb R)\to\mathbb R $ is of the form $L (A)=\text {tr}\, (BA) $ for a matrix $B $, it is canonical to identify $L $ with $B $. With this in mind, is that we have the "equality" $$Df (X)=X^{-1}. $$

Martin Argerami
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  • Thank you for your answer. I can understand this if $X \in R^{n}$ but when $X \in S^{n}$ then I am unable to understand it. Can you provide detailed steps when $X \in S^{n}$. I will be very thankful to you – Frank Moses Jan 24 '18 at 03:34
  • And even when $X \in R^{n}$ then the derivative will be $X^{-1}$ . How to invert a vector? I do not have any idea. But anyway in my problem the $X$ can only be a symmetric matrix – Frank Moses Jan 24 '18 at 03:45
  • I wrote as if they were matrices, I'm editing that. Other than that, it is hard to tell because you provided no context. – Martin Argerami Jan 24 '18 at 04:45
  • Thank you for your comment. According to the last equation in my post $$tr(X^{-1}(Z-X))=Df(X)(Z-X)$$ (please tell if this is right or wrong). If this is right then $Df(X)(Z-X)=<Df(X),(Z-X)>$ is to be proved. Right? – Frank Moses Jan 24 '18 at 04:52
  • That's exactly what the left-hand-side means. There's nothing to prove. – Martin Argerami Jan 24 '18 at 05:00
  • Thanks for helping. Now the left hand side says $tr(X^{-1}(Z-X))=<X^{-1},Z-X>$ I can understand that. But on the right side how to prove that $Df(X)(Z-X)=<Df(X),Z-X>$. I hope I have clearly stated my problem. – Frank Moses Jan 24 '18 at 05:05