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Show $$ (Q^3)_{ij}-\frac{1}{2}Tr(Q^2)Q_{ij}-\det(Q)\delta_{ij}=0 $$ where Q is a real symmetric traceless tensor. $δ_{ij}$ is Kronecker delta symbol which is 1 if $i=j$ or $0$ otherwise. We can possibly use eigenvalues & eigenvectors of Q to help us with this, I am not sure how to go about it though. Also given is $$ Q_{ij}=\epsilon_{ij}-\frac{1}{3}Tr(\epsilon)\delta_{ij}$$

where $\epsilon$ is the totally anti-symmetric tensor-Levi-Civita symbol. This post recently got taken off because it lacked explanation. This is a math problem from Russian math books by Demidovich, not sure what the problem is, please do not delete the post if you cannot solve it. It is just a tensor property that we should all know, I am trying :)

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  • $\begingroup$ Is there any restriction on the size of $Q$? That is, on the range(s) of the indices $i$ and $j$? Also, what is $\epsilon_{ij}$? $\endgroup$ – Robert Lewis Dec 15 '13 at 8:26
  • $\begingroup$ Oh, I think I get it; your second equation implies the size of $Q$ is $3$! That is, when coupled with the fact that $Q$ is traceless! $\endgroup$ – Robert Lewis Dec 15 '13 at 8:29
  • $\begingroup$ Yes 3 is correct! Sorry for not being clear about that $\endgroup$ – Jeff Faraci Dec 15 '13 at 8:31
  • $\begingroup$ I do not understand: is $\epsilon$ the Levi-Civita symbol? If yes, then its trace is equal to $0$...if not, what it is? A given matrix? $\endgroup$ – Avitus Dec 15 '13 at 8:40
  • $\begingroup$ @Avitus Yes it is the totally anti-symmetric tensor levi civita symbol $\endgroup$ – Jeff Faraci Dec 15 '13 at 8:41
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Since $Q$ is a $3\times3$ matrix, Cayley-Hamilton theorem says that $Q^3-tQ^2+sQ-rI=0$, for some specific scalar $(r,s,t)$. For example, $t$ is the trace of $Q$, hence $t=0$ in the present case, and $r$ is the determinant of $Q$. Hence the identity to prove holds as soon as $$s=-\tfrac12\mathrm{tr}(Q^2). $$ The characteristic polynomial $x^3-tx^2+sx-r$ is the determinant of $xI-Q$ hence $s$ is the derivative at $x=0$ of this determinant, that is, $$ s=\sum_i\det(Q^{(i,i)}), $$ where $Q^{(i,i)}$ is the $2\times2$ matrix obtained by cancelling the $i$th column and the $i$th row of $Q$. Thus, $$ s=\tfrac12\sum_{i\ne j}(Q_{ii}Q_{jj}-Q_{ij}Q_{ji}). $$ Furthermore, the trace $t$ is zero hence $$ t^2=0=\sum_iQ_{ii}^2+\sum_{i\ne j}Q_{ii}Q_{jj}. $$ Finally, $$ \mathrm{tr}(Q^2)=\sum_i(Q^2)_{ii}=\sum_{i,j}Q_{ij}Q_{ji}=\sum_{i}Q_{ii}^2+\sum_{i\ne j}Q_{ij}Q_{ji}. $$ Massaging the three last displayed identities above yields the desired one.

All this only uses that $Q$ is a $3\times 3$ traceless matrix.

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  • $\begingroup$ Very nice, Thank you. $\endgroup$ – Jeff Faraci Dec 15 '13 at 17:59

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