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I have seen the previous post and I tried to follow the answer to solve the problem. But I encountered some problem and I can not get the expected answer $2a_{ij}$. I stuck at the final line since I thought when $k=j$ , we get $da_{ij}a_{ji}+a_{ij}(da_{jj})$.

$$(AA^T)ij=∑_{k=1}^{n}a_{ik}a_{jk}$$ \begin{align} d((AA^T)ij)&=d(∑_{k=1}^{n}a_{ik}a_{jk}) \\ &=∑_{k=1}^{n}[(da_{ik})a_{jk}+a_{ik}(da_{jk})] \\ \end{align}

Rodrigo de Azevedo
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PinYuan
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  • What are you differentiating with respect to? Are the matrix entries functions of different variables $x_l$? Or (as I guess) are you taking the derivative with respect to the coefficients $a_{ij}$ regarded as independent variables? – Dog_69 Dec 28 '19 at 12:08
  • I want to differentiate $AA^T$ w.r.t A. Just like you said, I want to take the derivative with respect to the coefficients $a_{ij}$ – PinYuan Dec 28 '19 at 12:13
  • In that case $$\frac{\partial (AA^t){km}}{\partial a{ij}} = \frac{\partial}{\partial a_{ij}}\sum_l a_{kl}a_{ml} = \sum_l \frac{\partial a_{kl}}{\partial a_{ij}}a_{ml} + a_{kl} \frac{\partial a_{ml}}{\partial a_{ij}}$$. But because the coefficients $a_{ij}$ are independent, $\frac{\partial a_{kl}}{\partial a_{ij}}=\delta {ik }\delta{lj}$. Therefore the last term above reads as $a_{mj}\delta_{ik }+a_{kj}\delta_{im}$. – Dog_69 Dec 28 '19 at 12:46

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