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I have this equation, $$\frac{\partial\phi_i}{\partial \mu_j}\mu_j = T_{i,j}\frac{\partial\phi_j}{\partial \mu_k}\mu_k + g_i$$

Could I use identity matrix to change the indices so that I can rewrite it as,

$$I_{i,j}\frac{\partial\phi_j}{\partial \mu_k}\mu_k = T_{i,j}\frac{\partial\phi_j}{\partial \mu_k}\mu_k + g_i$$

$$(I_{i,j}-T_{i,j})\frac{\partial\phi_j}{\partial \mu_k}\mu_k = g_i$$

Can such a change of indices be performed? If no, then how do I solve the problem for $\mu_i$. Also $i,j,k$ all vary from 1 to N

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    $\begingroup$ Yes, this is OK. (But the identity matrix is usually denoted $\delta_{ij}$ in Einstein notation.) $\endgroup$ Commented Feb 3 at 19:54
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    $\begingroup$ You have a linear system for $v$, where $v_j := (\partial\phi_j/\partial\mu_k)\mu_k$. $\endgroup$ Commented Feb 3 at 19:55
  • $\begingroup$ Thanks.. Is there a simple way to understand why this can be done $\endgroup$ Commented Feb 3 at 19:56

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Since Einstein notation is nothing but an implied sum over twice appearing indices, yes you can, although the more common notion would be a Kronecker delta

$$a_i = \sum \delta_{ij} a_j = \delta_{ij}a_j$$

Also you can rename twice appearing indices into something else.

$$a_i b_i = \sum \limits_i a_i b_i = \sum \limits_j a_j b_j = a_j b_j$$

So all your steps are correct. When in doubt, just put the summation back in.

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