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 :)

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.