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The rank-four law $c_{ijkl}=\alpha {\delta }_{ij}{\delta }_{kl}+\beta {\delta }_{ik}{\delta }_{jl}+\gamma {\delta }_{il}{\delta }_{jk}$ sends a strain array to $P_{ij}=\alpha {\delta }_{ij}{T}_{kk}+\beta T_{ij}+\gamma T_{ji}$. Applying the scalar, symmetric-traceless and axial tensor decomposition gives the displayed independent actions on the three components. The inverse requires all relevant coefficients nonzero. Restricting to a symmetric second-rank tensor eliminates the axial part and gives $P_{ij}=\alpha {\delta }_{ij}{T}_{kk}+(\beta +\gamma )T_{ij}$. Without that symmetry restriction the two-parameter form is not generally valid.

For a Cartesian second-rank tensor, writeThe first term is a symmetric second-rank tensor and the second is an antisymmetric second-rank tensor. Taking the trace of the symmetric term and subtracting it gives the scalar, symmetric-traceless and axial tensor decomposition:Here $S_{ij}=S_{ji}$ and $S_{ii}=0$. To check reconstruction of the antisymmetric part, contract the Levi-Civita symbols:Thus $P_{ij}=P{\delta }_{ij}+S_{ij}+{ϵ}_{ijk}{A}_k$, and the displayed contractions also prove uniqueness. The vector associated with the antisymmetric part is an axial vector under reflections, and an ordinary vector under proper rotations.

For the specified isotropic tensor law, contract the Kronecker deltas first:Write $T_{ij}=T{\delta }_{ij}+W_{ij}+{ϵ}_{ijk}{V}_k$, where $T=T_{kk}/3$ and $W$ is symmetric and traceless. Transposition preserves its scalar and symmetric-traceless parts and reverses its antisymmetric part. ThereforeUniqueness of the decomposition gives the isotropic elasticity on tensor components relations. Provided their respective coefficients are nonzero, the requested inverses areIf one coefficient vanishes, that strain component has zero stress response: the corresponding stress component must be zero for a solution to exist, and that strain component is not uniquely recoverable. Isotropy by itself does not guarantee invertibility.

If $T_{ij}$ is symmetric then $V=0$, hence $A=0$ and the stress is symmetric. Equivalently, direct transposition gives $P_{ij}-P_{ji}=(\beta -\gamma )(T_{ij}-T_{ji})=0$. Restricting to this symmetric-strain case givesThis final two-parameter expression uses the symmetry assumption from the preceding request. For a general nonsymmetric array, the transpose term cannot be absorbed into the same coefficient: it acts with $\beta +\gamma$ on the symmetric-traceless part and with $\beta -\gamma$ on the antisymmetric part. The stated coefficient $\mu$ is the coefficient of $T_{ij}$ itself; if one instead writes the conventional elastic law with $2\mu$ multiplying the strain, that shear-modulus convention is half this coefficient.