Derivative of $tr(A)$, $tr(A^2)$ , $tr(A^3)$ with respect to $A$ in indicial notation

Question

How can I show (prove) the three following equations in indicial notation:

$$\frac{\partial (trA)}{\partial A}=I \\ \frac{\partial (trA^2)}{\partial A}=2A^T \\ \frac{\partial (trA^3)}{\partial A}=3(A^2)^T $$

A is a second-order tensor.

Thank you

Answer 1

$$ \frac{\partial}{\partial A_{ab}} \operatorname{tr}{(A^n)} = \frac{\partial}{\partial A_{ab}} ( A_{i_1 i_2} A_{i_2 i_3} \dotsm A_{i_{n-2}i_1} ) \\ = \delta_{ai_1} \delta_{bi_2} A_{i_2 i_3} \dotsm A_{i_n i_1} + A_{i_1 i_2} \delta_{ai_2} \delta_{bi_{3}} \dotsm A_{i_n i_1} + \dotsb + A_{i_1 i_2} \dotsc A_{i_{n-1}i_n} \delta_{ai_n} \delta_{b i_1} \\ = A_{b i_3} \dotsm A_{i_n a} + A_{i_1 a} A_{bi_4} \dotsm A_{i_n i_1} + \dotsb + A_{b i_2} \dotsc A_{i_{n-1}a} $$ Now, we can relabel the indices: because all the produands are the same, we end up with $$ n A_{b i_3} A_{i_3 i_4} \dotsm A_{i_n a} = [n(A^{n-1})^T]_{ab} . $$

An easier way to see this is to write out the calculation using a small perturbation $t B$: $$ \operatorname{tr}{((A+tB)^n)} = \operatorname{tr}{(A^n + nA^{n-1} tB + \frac{n(n-1)}{2}A^{n-2} tB A + \frac{n(n-1)(n-2)}{6}A^{n-3} tB A^2 + \dotsb + ntBA^{n-1} + O(t^2)} \\ = \operatorname{tr}{(A^n)} + t\operatorname{tr}{n(BA^{n-1})} + O(t^2), $$ using that the trace is invariant under cyclic permutations. Dividing by $t$ and taking the limit $t \to 0$ gives the directional derivative $\operatorname{tr}{n(BA^{n-1})}$, and one can again write this out in index notation to get the result.

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For $n=1$: $$ \frac{\partial}{\partial A_{ab}} \operatorname{tr}{(A)} = \frac{\partial}{\partial A_{ab}} ( A_{i_1 i_1}) \\ = \delta_{ai_1} \delta_{bi_1} = \delta_{ab} = [I]_{ab} $$ For $n=2$: $$ \frac{\partial}{\partial A_{ab}} \operatorname{tr}{(A^n)} = \frac{\partial}{\partial A_{ab}} ( A_{i_1 i_2} A_{i_2 i_1} ) \\ = \delta_{ai_1} \delta_{bi_2} A_{i_2 i_1} + A_{i_1 i_2} \delta_{ai_2} \delta_{bi_{1}} \\ = A_{ba} + A_{ba} = [2A^T]_{ab} $$ For $n=3$: $$ \frac{\partial}{\partial A_{ab}} \operatorname{tr}{(A^3)} = \frac{\partial}{\partial A_{ab}} ( A_{i_1 i_2} A_{i_2 i_3} A_{i_3 i_1} ) \\ = \delta_{ai_1} \delta_{bi_2} A_{i_2 i_3} A_{i_3 i_1} + A_{i_1 i_2} \delta_{ai_2} \delta_{bi_3} A_{i_3 i_1} + A_{i_1 i_2} A_{i_2 i_3} \delta_{ai_3} \delta_{b i_1} \\ = A_{b i_3} A_{i_3 a} + A_{i_1 a} A_{b i_1} + A_{b i_2} A_{i_2 a} = [3(A^2)^T]_{ab} $$