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What is the matrix form of $\epsilon_{ijk} A_{jk}$?

user157226
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1 Answers1

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$\epsilon_{ijk} A_{jk}$ has only one free index (the $i$), assuming that repeated indices are summed. Hence, it is a vector. I assume you are working in 3D.

Define the vector $v_i$

$$v_i = \epsilon_{ifg} A_{fg}$$

and calculate its components by using the fact that, for example, $\epsilon_{112}=\epsilon_{133}=0$ (i.e. the $\epsilon$ is zero when you have the same index appearing twice):

$$ v_1 = \epsilon_{1fg} A_{fg} = \epsilon_{123} A_{23} + \epsilon_{132} A_{32} $$

You know that $\epsilon_{123}=1$ and $\epsilon_{132}=-1$. Therefore,

$$ v_1 = \epsilon_{1fg} A_{fg} = A_{23} - A_{32} $$

Similarly for the other two components $v_2$ and $v_1$.

Quillo
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    Thus, it would read $\begin{bmatrix} A_{23}-A_{32}\ A_{31}-A_{13}\ A_{12}-A_{21} \end{bmatrix}$ – user157226 Sep 17 '21 at 16:32
  • @user157226 I confirm that the signs in your solution are correct: the trick is $v_a = A_{bc}-A_{cb}$ where $abc$ is an even permutation of $123$ (hence, $acb$ an odd permutation of $123$, and this is why $A_{cb}$ has the minus). – Quillo Sep 17 '21 at 16:36