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I want to learn to find the determinant of an operator.

I am given an operator like $\Sigma _{\alpha\beta}=-k^2g_{\alpha\beta}+i\theta\epsilon_{\alpha\beta\gamma} k^\gamma$

$k^2=k^μk_μ$, $g^{αβ}$ is metric tensor, theta is a constant $ϵ$ is totally anti-symmetric tensor in 3 D and $k^γ$ is a contravariant 3 vector.

Determinant is found using

$Det A = \Pi \lambda$ or

$Det A = exp(Tr ln(A))$. In my case which one should I use any why?

I tried by expanding the explicit matrix form and found the determinant of matrix which gives $k^4(k^2-\theta^2)$ which does not match with given result.

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  • $\begingroup$ The word ``operator'' possibly means a linear mapping from one vector space to itself. Could you please tell us on what space you operator is defined and how it acts. $\endgroup$
    – Vladimir S Matveev
    May 10, 2015 at 8:41
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    $\begingroup$ You are not saying it explicitly (though you always should), but it appears that your $\Sigma$ is a 3x3 matrix. For any 3x3 matrix, its determinant can be computed using the formula $\det \Sigma = \frac{1}{3} \operatorname{tr} \Sigma^3 - \frac{1}{2} \operatorname{tr} \Sigma^2 \operatorname{tr} \Sigma + \frac{1}{6} \operatorname{tr}^3 \Sigma$, which does not require writing out the matrix explicitly. I'll leave the full computation as an exercise. $\endgroup$
    – Igor Khavkine
    May 10, 2015 at 10:25
  • $\begingroup$ @IgorKhavkine sorry i havnt seen such thing before. kindly tell me the way to solve 1st and 3rd term. I ll do the middle one. As i dnt knw what is $\Sigma^3$ and $tr^3$. $\endgroup$
    – Zohaib Aarfi
    May 11, 2015 at 6:29
  • $\begingroup$ See here for example. $\Sigma^k$ is the $k$-th matrix power and $\operatorname{tr}^k \Sigma = (\operatorname{tr} \Sigma)^k$. $\endgroup$
    – Igor Khavkine
    May 11, 2015 at 7:13
  • $\begingroup$ @IgorKhavkine thanx for your guidance kindly tell me some reference so that i could study this procedure in detail. M sure it ll help. $\endgroup$
    – Zohaib Aarfi
    May 11, 2015 at 11:21

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