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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.

First separate the symmetric tensor and antisymmetric tensor parts:The trace of the antisymmetric part vanishes. In three dimensions, remove one third of the trace from the symmetric part, and identify the remaining antisymmetric part with an axial vector. The scalar, symmetric-traceless and axial tensor decomposition isHere $S_{ji}=S_{ij}$ and $S_{ii}=P_{ii}-P_{kk}=0$. The contraction of two Levi-Civita symbols verifies the reconstruction:ThereforeThe pieces contain $1$, $5$ and $3$ independent components respectively, accounting for all nine entries. Taking the trace, then the traceless symmetric part, then the antisymmetric part shows that the decomposition is unique.

Contract the given isotropic fourth-rank tensor with the strain array. The stress tensor isWrite $p_{ij}=p{\delta }_{ij}+s_{ij}+{ϵ}_{ijk}{a}_k$, where $s_{ij}$ is symmetric and traceless. Then $p_{kk}=3p$ and $p_{ji}=p{\delta }_{ij}+s_{ij}-{ϵ}_{ijk}{a}_k$. Substitution givesBy uniqueness of the scalar, symmetric-traceless and axial tensor decomposition,This is isotropic elasticity on tensor components: the three tensor sectors do not mix. If the strain is symmetric, $a_k=0$, hence $A_k=0$ and $\boxed{P_{ij}=P_{ji}}$. No equality between $\mu$ and $\nu$ is needed for that conclusion.

An isotropic tensor has unchanged components under every rotation of the Cartesian axes. Under proper rotations in three dimensions, rank one has no nonzero isotropic tensor, rank two has $\boxed{T_{ij}=A{\delta }_{ij}}$, and rank three has $\boxed{T_{ijk}=B{ϵ}_{ijk}}$. Here $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. If invariance under reflections is also required for a true tensor, the rank-three case vanishes; $ϵ$ is then a pseudotensor.

The velocity gradient tensor for the given field isIts symmetric part, the strain-rate tensor if $\mathbf{u}$ is interpreted as a velocity, isThe antisymmetric part has the specified ordering $D_{ik}-e_{ik}={ϵ}_{ijk}{\omega }_j$, withThe index ordering matters for this sign; it agrees with a local rotation velocity $\mathbf{\omega }\times \mathbf{x}$. This is the axial part of the scalar, symmetric-traceless and axial tensor decomposition, retaining the trace inside the symmetric part here.

At the specified point, $e=\left(\begin{array}{ccc}9& 6& 0\\ 6& 0& 0\\ 0& 0& 4\end{array}\right)$. The upper block has characteristic polynomial ${\lambda }^2-9\lambda -36=(\lambda -12)(\lambda +3)$. Thus the principal values and unit eigenvectors areThese mutually orthogonal directions are the principal strain rates' axes for this local field; reversing an eigenvector does not change its axis.

A two-index Cartesian tensor has components transforming as $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$ under an orthogonal change of orthonormal basis. The tensor order here counts indices, rather than the matrix rank. Its scalar, symmetric-traceless and axial tensor decomposition follows by settingThen $T_{ij}=Y{\delta }_{ij}+{\Omega }_{ij}+S_{ij}$, with $\Omega$ antisymmetric and $S$ symmetric and traceless. There are three independent antisymmetric components and five independent symmetric traceless components: three off-diagonal entries and two independent diagonal ones in the latter case.

For the symmetric tensor relative to a nonzero vector $A$, put $q=A_i{A}_i>0$, $s=A_i{T}_{ij}{A}_j$, and $t=T_{ii}$. The symmetric tensor decomposition relative to an axis givesIn particular $A_i{B}_i=s/q-s/q=0$. Define the orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/q$ onto the plane perpendicular to $A$, and setThis tensor is symmetric and kills $A$. Its trace is $t-s/q-2\alpha =0$. To verify the full decomposition, set $u=A/\sqrt{q}$ and split $T$ into its longitudinal component $(u^{T}Tu)u{u}^T$, two mixed components, and $PTP$. The mixed terms are exactly $B{A}^T+A{B}^T$, and $\alpha +\beta q=s/q$ gives the longitudinal coefficient. Thus all requested constraints and the reconstruction hold.

Write $A_{ij}={\partial }_j{u}_i$. The trace fixes the isotropic part, and splitting the remaining matrix into its symmetric and antisymmetric parts gives the scalar, symmetric-traceless and axial tensor decomposition:Here $s_{ij}=s_{ji}$ and $s_{ii}=0$. The antisymmetric part is $a_{ij}=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Its associated axial vector isIndeed the contraction identity ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$ gives ${ϵ}_{ijk}{\omega }_k=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Adding the three parts reconstructs $A_{ij}$.

The minus sign is required by the index order $A_{ij}={\partial }_j{u}_i$. The symmetric gradient and curl decomposition written instead for ${\partial }_i{u}_j$ has the opposite axial sign. The divergence supplies $3p$, the trace-free symmetric part describes shape change, and the antisymmetric part describes local rotation.

The scalar, symmetric-traceless and axial tensor decomposition fixes the scalar part by the trace:Subtracting this isotropic part from the symmetric tensor in part (c) givesIts trace is zero. The antisymmetric tensor part isAdding $PI+\hat{S}+\hat{A}$ reconstructs every entry of $F$.

Subtracting a scalar multiple of the identity preserves every eigenvector: if $Sv=\lambda v$, then $(S-PI)v=(\lambda -P)v$. Thus the principal axes are the same as in part (c), with shifted principal valuesThe two tensors are coaxial symmetric tensors. At $r=0$ both vanish, so neither specifies unique axes.

Take the trace first: the antisymmetric tensor has zero diagonal and $X$ is traceless, so $T_{ii}=3t$. Taking symmetric and antisymmetric parts then gives the scalar, symmetric-traceless and axial tensor decomposition:The Kronecker delta satisfies ${\delta }_{ii}=3$, which verifies $X_{ii}=0$. Substitution verifies the original decomposition and also proves its uniqueness.