= Solution
A Cartesian <Cartesian second-rank tensor> has components transforming under a constant orthogonal change of axes as
$$
\boxed{T'_{ij}=p_{ia}p_{jb}T_{ab},\qquad T'=PTP^T.}
$$
Here $\mathbf X'=P\mathbf X$ and $\mathbf u'=P\mathbf u$. By the <chain rule> and orthogonality,
$$
\frac{\partial u_i'}{\partial X_j'}
=p_{ia}\frac{\partial u_a}{\partial X_b}\frac{\partial X_b}{\partial X_j'}
=p_{ia}p_{jb}\frac{\partial u_a}{\partial X_b}.
$$
Thus the <displacement> <gradient> is a <Cartesian second-rank tensor>. Its symmetric part transforms in the same way:
$$
e'_{ij}=\frac12p_{ia}p_{jb}
\left(\frac{\partial u_a}{\partial X_b}+\frac{\partial u_b}{\partial X_a}\right)
=p_{ia}p_{jb}e_{ab}.
$$
Hence \b[the <infinitesimal strain tensor> is also second-rank]. This is a passive coordinate rotation, under which both reference and spatial coordinates rotate; it does not assert finite-<deformation> frame indifference of the linearized strain formula.
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