= Solution
The <infinitesimal strain tensor> is the symmetric part of the <displacement gradient tensor>:
$$
\boxed{e_{ij}=\frac12\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right).}
$$
Its diagonal entries are infinitesimal normal extensions; twice an off-diagonal entry is the corresponding engineering shear <strain>. An infinitesimal rigid rotation has an antisymmetric <displacement gradient tensor> and therefore contributes no linearized <strain>.
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