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.
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" $$Df (X)=X^{-1}. $$