[QUE/ME-02022] ME-PROBLEM

Consider the right hand side of the identity (a) to be proved and write all terms in the sum. There will be four terms as \(i\) and \(j\) take values 1 and 2. This gives \begin{eqnarray} \frac{1}{2}S_{ij}\big(T_{ij}+T_{ji}\big) &=&\frac{1}{2}\sum_{i=1}^2\sum_{j=1}^2S_{ij}\big(T_{ij}+T_{ji}\big)\\ &=& \frac{1}{2}\sum_{i=1}^2\Big( S_{i1}\big(T_{i1}+T_{1i}\big)+ S_{i2}\big(T_{i2}+T_{2i}\big)\Big)\\ &=&\frac{1}{2} \Big( S_{11}\big(T_{11}+T_{11}\big)+ S_{12}\big(T_{12}+T_{21}\big)\Big)\\ && +\frac{1}{2} \Big( S_{21}\big(T_{21}+T_{12}\big)+ S_{22}\big(T_{22}+T_{22}\big)\Big)\\ &=& S_{11}T_{11} +S_{12}T_{12} + S_{21}T_{21}+ S_{22}T_{22}\\ &=& \sum_{i=1}^2 \sum_{j=1}^2 S_{ij}T_{ij} = S_{ij}T_{ij} \end{eqnarray} Note:- This approach will not be useful for more complicated expressions.

\begin{eqnarray} S_{ij}T_{ij} &=& S_{ji}T_{ij} \qquad \HighLight{Used symmetry property of S} \\ &=&S_{nm}T_{mn} \qquad \HighLight{renamed dummy indices}\\ &=& S_{ij}T_{ji} \qquad \HighLight{renamed dummy indices again} \label{EQ02} \end{eqnarray} Therefore the right hand side of (a) becomes \begin{eqnarray} \frac{1}{2}S_{ij}\big(T_{ij}+T_{ji}\big) &=&\frac{1}{2}S_{ij}T_{ij} +\frac{1}{2} S_{ij}T_{ji} \qquad \text{\HighLight{Now use \eqRef{EQ02} in second term}}\nonumber\\ &=& \frac{1}{2}S_{ij}T_{ij} +\frac{1}{2} S_{ij}T_{ij} = S_{ij}T_{ij}. \end{eqnarray}

\begin{eqnarray}\nonumber S_{ij}T_{ij} &=& S_{ji}T_{ji} \qquad \HighLight{Change Dummy indices}\\\nonumber S_{ij}T_{ij} &=& S_{ji}T_{ji} \qquad \HighLight{Use symmetry of S}\\ \text{add} \qquad S_{ij}T_{ij} &=& \frac{1}{2}S_{ij}\big(T_{ij}+T_{ji}\big)\nonumber \end{eqnarray}