MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 added 288 characters in body; edited body

I guess $\Delta_A$ denotes the derivative with respect to the elements of the matrix $A$ (more conventionally denoted by $\partial_{A}$).

To evaluate the derivative with respect to $A_{ij}$, write out the trace in terms of components and then use $\partial_{A_{ij}} A_{mn} = \delta_{im} \delta_{jn}$, $$\partial_{A_{ij}} \text{tr}(A B A^T C) = \partial_{A_{ij}}\sum_{mnkl} A_{mn} B_{nk} A_{lk} C_{lm}= \sum_{kl} B_{jk}A_{lk} C_{li} + \sum_{mn} A_{mn} B_{nj} C_{im}$$ $$= ( C^T A B^T+ C A B )_{ij}.$$ This is the component-wise version of your identity.

Note to the comment of Todd Trimble: the matrices $A,B$, and $C$ do not have to be necessarily square matrices. Their dimension just has to "match" ($A \in \mathbb{R}^{m \times n}$, $B\in \mathbb{R}^{n \times n}$, $C \in \mathbb{R}^{m \times m}$, with $m$ and $n$ arbitrary integers).

1

I guess $\Delta_A$ denotes the derivative with respect to the elements of the matrix $A$ (more conventionally denoted by $\partial_{A}$).

To evaluate the derivative with respect to $A_{ij}$, write out the trace in terms of components and then use $\partial_{A_{ij}} A_{mn} = \delta_{im} \delta_{jn}$, $$\partial_{A_{ij}} \text{tr}(A B A^T C) = \partial_{A_{ij}}\sum_{mnkl} A_{mn} B_{nk} A_{lk} C_{lm}= \sum_{kl} B_{jk}A_{lk} C_{li} + \sum_{mn} A_{mn} B_{nj} C_{im}$$ $$= ( C^T A B^T+ C A B )_{ij}.$$ This is the component-wise version of your identity.