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Hi I need some insight into the following proof provided by my book. The aim is to prove that the trace of a tensor matrix is invariant wrt rotation of the axes, using index notation.

Consider the tensor $X_{ij}$. The trace is the sum over diagonal elements i.e. $Tr(\chi)=X_{ii}$, where again we use the summation convention. In the new frame of reference, the trace is $X'_{ii}$. We now evaluate $$X_{ii}'=R_{ij}R_{ik}X_{jk} = \delta_{jk}X_{jk} = X_{jj} = X_{ii}$$ so the trace is invariant.

Now I understand every passage, apart from the first one. $$X_{ii}'=R_{ij}R_{ik}X_{jk}$$ how do we know this is true? Where does this come from? Normally we write $$X' = RXR'$$ where $R$ is the rotation matrix.

Can you help me understanding?

Euler_Salter
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1 Answers1

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We have: $$ X_{ii}' = (RXR')_{ii} = (RX)_{ik}R_{ki}' = R_{ij}X_{jk}R_{ki}' = R_{ij}X_{jk}R_{ik} = R_{ij}R_{ik}X_{jk} $$ In the usual notation, we might write $$ \operatorname{Trace}(RXR') = \operatorname{Trace}[(RX)R'] = \operatorname{Trace}[R'(RX)] = \operatorname{Trace}[(R'R)X] = \operatorname{Trace}(X) $$

Ben Grossmann
  • 225,327