Divergence of traceless matrix

[jouvelot]

Assume that $\partial M_{ab}/\partial \hat{n}_c$ is completely symmetric in $a, b$ and $c$. Then, it is stated in the book I read that the divergence of the traceless part of $M$ is proportional to the gradient of the trace of $M$. More precisely,
$$ \partial /\partial \hat{n}_a (M_{ab} - \delta_{ab} {\rm Tr} (M)/2) = \partial ({\rm Tr} (M)/2)/\partial \hat{n}_b .$$ Can anyone provide some hints on why this is true, please?
Thanks in advance.
Pierre

 

[pasmith]

The trace of M is M_{aa}. By symmetry, <br /> \frac{\partial M_{ab}}{\partial \hat{n}_a} = \frac{\partial M_{aa}}{\partial \hat{n}_b}<br />

 

[jouvelot]

Thanks. This was pretty simple; I should have gotten that ;)

Bye,

Pierre