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Given the following definition:

definition for /delta

How to proof these two equations?

How to proof this?


enter image description here


Actually, there are two proofs preceding the two(I have no problem with the following two), they are: enter image description here

Maybe they are hints on solving the latter two.

I encounter this problem here(section 2.1 about page 8~page 9)

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By writing it out in index notation (personal preference), the first equation is simple application of the product rule (see Einstein notation)

\begin{align} \nabla_{A_{ij}} \delta_{kl} A_{km}B_{mn} A^T_{np} C_{pl} = \newline \nabla_{A_{ij}} A_{km}B_{mn} A_{pn} C_{pk} = \newline (\nabla_{A_{ij}} A_{km})B_{mn} A_{pn} C_{pk} + (\nabla_{A_{ij}}A_{pn}) A_{km}B_{mn}C_{pk} = \newline (\delta_{ik}\delta_{jm})B_{mn} A_{pn} C_{pk} + (\delta_{ip}\delta_{jn}) A_{km}B_{mn} C_{pk} = \newline B_{jn} A_{pn} C_{pi} + A_{km} B_{mj} C_{ik} = \newline C^T\cdot A\cdot B^T + C\cdot A \cdot B \end{align}

as for the second equation, I've only seen it derived by rewriting $|A|$ in terms of its eigenvalues and doing some tricks or Jacobi's formula. I don't think the two preceding equations give you much to work with here.

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Let $X=Y=A$. We have $(\ast): \textrm{tr} XBY^TC = \textrm{tr} CXBY^T = \textrm{tr} (CXBY^T)^T = \textrm{tr} Y(CXB)^T$. So $$ \begin{eqnarray*} \nabla_A \textrm{tr} ABA^TC &=& \nabla_X \textrm{tr} XBY^TC + \nabla_Y \textrm{tr} XBY^TC \quad\textrm{(by chain rule)}\\ &=& \nabla_X \textrm{tr} X(BY^TC) + \nabla_Y \textrm{tr} Y(CXB)^T\quad(\textrm{by }(\ast))\\ &=& (BY^TC)^T + CXB = C^TAB^T + CAB \end{eqnarray*} $$ and we get the first result. The second result is more straightforward. Recall that for any fixed $i$, by Laplace expansion, we have $\det A=\sum_j (-1)^{i+j}A_{ij}M_{ij}(A)$, where $M_{ij}(A)$ denotes the $(i,j)$-minor of $A$. Since the computation of $M_{ij}(A)$ does not involve $A_{ij}$, we have $\frac\partial{\partial A_{ij}}\det A=(-1)^{i+j}M_{ij}(A)=C_{ji}(A)$, where $C_{kl}(A)$ denotes the $(k,l)$-cofactor of $A$. Hence $\nabla_A\det(A)=\textrm{adj}(A)^T=(\det A)(A^{-1})^T$.

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thx, but can you explain what chain rule is? I think that is the obstacle for me to understand. – xiaohan2012 Sep 26 '11 at 16:05
Here chain rule means that for the function $g(X,Y)=\textrm{tr}XBY^TC$, we have $\frac{\partial g}{\partial a_{ij}} = \frac{\partial g}{\partial x_{ij}}\frac{dx_{ij}}{\partial a_{ij}} + \frac{\partial g}{\partial y_{ij}}\frac{dy_{ij}}{\partial a_{ij}}$. As $a_{ij}=x_{ij}=y_{ij}$, we get $\frac{\partial g}{\partial a_{ij}} = \frac{\partial g}{\partial x_{ij}} + \frac{\partial g}{\partial y_{ij}}$. This holds for each pair of $(i,j)$. Hence $\nabla_Ag=\nabla_Xg+\nabla_Yg$. – user1551 Sep 26 '11 at 18:29

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