= Solution
Define the <symmetric tensor> and <antisymmetric tensor> parts by $S=(T+T^T)/2$ and $A=(T-T^T)/2$. The transformed transpose satisfies $(T')^T=(MTM^T)^T=MT^TM^T$. Therefore
$$
S'=\frac12(T'+(T')^T)=M\frac12(T+T^T)M^T=MSM^T,
$$
and similarly
$$
A'=\frac12(T'-(T')^T)=M\frac12(T-T^T)M^T=MAM^T.
$$
Both obey the <tensor component transformation law>. Thus \b[the symmetric and antisymmetric parts are themselves second-order tensors]. Taking a transpose and forming linear combinations commute with this orthogonal change of frame.
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