Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F(X) = Tr(XBX^TA)$, where A,B are matrix.

What is the derivative $\frac{\partial F(X)}{\partial X}$?

Is it something like $A(XB^T+XB)$?

share|cite|improve this question
up vote 4 down vote accepted

Coming back to the definition of the differential, one looks for a linear functional $L$ such that $F(X+H)=F(X)+L(H)+o(\|H\|)$ when $H\to O$, and then one knows that $D_XF=L$. Here, $$ F(X+H)=F(X)+\mathrm{tr}(HBX^TA)+\mathrm{tr}(XBH^TA)+\mathrm{tr}(HBH^TA), $$ hence $$ D_XF:H\mapsto\mathrm{tr}(HBX^TA)+\mathrm{tr}(XBH^TA). $$ Since $\mathrm{tr}(U)=\mathrm{tr}(U^T)$ for every square matrix $U$ and $\mathrm{tr}(VW)=\mathrm{tr}(WV)$ for every matrices $V$ and $W$, this is also $$ (D_XF)(H)=\mathrm{tr}(H^TC),\qquad C=AXB+A^TXB^T. $$ One sees that $C=AX(B+B^T)$ when $A$ is symmetric, but not in general.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.