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If an array is antisymmetric in every orthonormal basis and its contractions with every antisymmetric second-rank tensor are invariant scalars, it obeys the Cartesian second-rank tensor transformation law. The difference $D=R^{T}A\prime R-A$ is antisymmetric and orthogonal to every antisymmetric test matrix. Choosing the test matrix equal to $D$ gives $D:D=0$, hence $D=0$. The Frobenius inner product is nondegenerate on the antisymmetric subspace.

A Cartesian second-rank tensor invariant under every proper rotation preserving an unoriented axis $n$ has this form. Rotations about the axis give a planar block $aI+bJ$ and eliminate mixed axial-planar entries. A half-turn about a transverse axis reverses $J$, forcing $b=0$. If only rotations about the oriented axis are imposed, this planar antisymmetric term may survive.

The Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero. Thus invariant contractions of an array with all symmetric second-rank tensors test only its symmetric part; an arbitrary basis-dependent antisymmetric part is invisible. A nonzero antisymmetric matrix in one basis and the zero matrix in another passes all such tests with scalar zero but is not a Cartesian second-rank tensor.

A Cartesian second-rank tensor invariant under half-turns about all three coordinate axes is diagonal. Each half-turn has diagonal signs, and each off-diagonal entry changes sign under at least one of them. Invariance therefore kills all off-diagonal entries. It does not make the diagonal entries equal: that requires additional rotational invariance, as for an isotropic second-rank tensor.

Under a proper rotation matrix $R$, invariance of a Cartesian second-rank tensor reads $T=RT{R}^T$, equivalently $TR=RT$. The half-turn $R_z(\pi )=\mathrm{diag}(-1,-1,1)$ forces all entries mixing the $z$ direction with the $xy$ plane to vanish. Thus $T$ consists of a planar $2\times 2$ block $B$ and the entry $T_{33}=c$.

Commutation with the planar quarter-turn $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ gives $B=aI+bJ$. This block already commutes with every planar rotation. Now use the half-turn $R_x(\pi )=\mathrm{diag}(1,-1,-1)$: on the planar block its conjugation fixes $I$ and sends $J$ to $-J$, so invariance forces $b=0$. ThereforeThe formula uses the Kronecker delta and is sufficient as well as necessary: rotations preserving the unoriented $z$-axis fix both $I$ and $e_3{e}_3^T$. This is an axially invariant second-rank tensor with the planar antisymmetric part removed by the horizontal half-turn.

Use the Cartesian second-rank tensor transformation convention $T^{\prime }=RT{R}^T$, where $R$ is the orthogonal matrix converting components to the new orthonormal basis. The coordinate half-turns areFor a diagonal rotation $R=\mathrm{diag}(s_1,s_2,s_3)$, invariance gives $t_{ij}=s_i{s}_j{t}_{ij}$, with no summation in this equation. For each $i\ne j$, one of the listed half-turns makes $s_i{s}_j=-1$, so $t_{ij}=-t_{ij}=0$. ThusThis is coordinate half-turn invariance of a second-rank tensor. It does not force equal diagonal entries; invariance under all rotations would be the stronger isotropic second-rank tensor condition.

The first index of the printed Levi-Civita symbol is fixed at $1$. Thus the matrix in the first orthonormal basis isThe specified matrix in the other basis is $I$, with determinant one. If this array were a Cartesian second-rank tensor, an orthogonal basis change would give $A^{\prime }=RA{R}^T$, andThe determinants disagree, proving the array is not a tensor. The fixed index matters: the TeX aid drops it, but the original PDF clearly prints ${ϵ}_{1ij}$.

False. The counterexample just given is antisymmetric in every basis and has zero contraction with every symmetric second-rank tensor. It still fails the Cartesian second-rank tensor transformation law, because its value is nonzero in one basis and zero in another. Thus antisymmetry alone, together with symmetric tests, proves no tensor transformation law. The same blindness of symmetric contraction tests to antisymmetric arrays applies even when the array is assumed antisymmetric.

False. Tests against symmetric second-rank tensors see only the symmetric part $A_s=(A+A^T)/2$, because the Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero:An arbitrary basis-dependent antisymmetric part is invisible to these tests. For example, prescribein one basis, and $A^{\prime }=0$ in another rotated basis, choosing antisymmetric arrays in every remaining basis. Every contraction with any symmetric second-rank tensor is the invariant scalar zero. But an invertible rotation cannot transform this nonzero matrix to zero, so the array is not a Cartesian second-rank tensor. This is the blindness of symmetric contraction tests to antisymmetric arrays.

True. In all four tests, an array is understood to have components specified in each rotated orthonormal basis; being a scalar means that the contraction is invariant under those changes. Write the Frobenius inner product as $A:B={\sum }_{ij}{A}_{ij}{B}_{ij}$. Every test Cartesian second-rank tensor obeys $B^{\prime }=RB{R}^T$. The assumption saysChoosing the elementary matrices as $B$, or choosing $B=R^T{A}^{\prime }R-A$, gives $R^T{A}^{\prime }R=A$. Thereforewhich is the required Cartesian second-rank tensor transformation law. This is the scalar contraction test for a Cartesian tensor.

If an array specified in each orthonormal basis has invariant Frobenius inner product with every Cartesian second-rank tensor, it transforms as such a tensor. For a component change $R$, invariance says $(R^{T}A\prime R-A):B=0$ for every matrix $B$. Nondegeneracy of the Frobenius inner product forces the difference to vanish. Testing every tensor, rather than a single selected tensor, is essential.