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In three-dimensional Euclidean space, contracting an antisymmetric second-rank tensor with the Levi-Civita symbol gives an axial vector. The epsilon contraction identity proves the inverse formula: $\frac{1}{2}{ϵ}_{kij}{ϵ}_{klm}{A}_{lm}=\frac{1}{2}(A_{ij}-A_{ji})=A_{ij}$. With this index convention,Thus $A$ is minus the cross-product matrix for $a\times x$. Proper rotations transform $a$ as a vector; an orientation-reversing orthogonal change adds a determinant sign. Stating the index convention prevents a sign ambiguity when converting an antisymmetric tensor into a cross-product operator.

For a fixed nonzero real vector $b$ in three dimensions, a real linear map commuting with every proper rotation about that axis has the displayed form, where $[b{]}_{\times }x=b\times x$ is the cross-product matrix. On the axis it acts by $a+c\Vert b{\Vert }^2$. On the perpendicular plane its matrix in an oriented orthonormal basis is $\left(\begin{array}{cc}a& -d\Vert b\Vert \\ d\Vert b\Vert & a\end{array}\right)$. Its eigenvalues are therefore $a+c\Vert b{\Vert }^2$ and $a\pm id\Vert b\Vert$, andTo see the converse, commuting with all rotations preserves their common fixed axis and its perpendicular plane; there can be no nonzero coupling vector fixed by every planar rotation. A planar matrix commuting with every rotation is a scalar identity plus a scalar right-angle rotation. Thus the block description is exhaustive. If $d\ne 0$, only the axial eigenvalue is real; if $d=0$, the perpendicular plane also consists of real eigenvectors.

The cross-product matrix $A\mathbf{x}=\mathbf{n}\times \mathbf{x}$ obeys $A\mathbf{n}=0$ and $A^2=\mathbf{n}{\mathbf{n}}^T-|\mathbf{n}{|}^{2}I$, by the vector triple product. On the orthogonal complement of a nonzero $\mathbf{n}$, this is $-|\mathbf{n}{|}^{2}I$; hence the two remaining complex eigenvalues are $\pm i|\mathbf{n}|$. In an orthonormal basis adapted to $\mathbf{n}$, $A$ is a planar quarter-turn multiplied by $|\mathbf{n}|$, with zero action along its axis.

Expanding the cross product component by component gives its cross-product matrix:It is a real skew-symmetric matrix and annihilates $\mathbf{n}$. By the vector triple product,so $A^2=\mathbf{n}{\mathbf{n}}^T-I$ for a unit vector $\mathbf{n}$. Choose a unit vector $\mathbf{u}\perp \mathbf{n}$ and put $\mathbf{v}=\mathbf{n}\times \mathbf{u}$. Then $(\mathbf{u},\mathbf{v},\mathbf{n})$ is an orthonormal basis and $A\mathbf{u}=\mathbf{v}$, $A\mathbf{v}=-\mathbf{u}$, $A\mathbf{n}=0$. Thus the characteristic polynomial is $\lambda ({\lambda }^2+1)$ and the cross-product matrix spectrum isOver the complex numbers, corresponding eigenvectors are $\mathbf{n}$, $\mathbf{u}-i\mathbf{v}$, and $\mathbf{u}+i\mathbf{v}$. Only the zero eigenvalue has real eigenvectors.

For the specified third coordinate axis, direct multiplication givesIts zero eigenspace is $\mathrm{span}\{(0,0,1)\}$, and its $-1$ eigenspace is the whole plane $\mathrm{span}\{(1,0,0),(0,1,0)\}$. Hence the eigenvalues of $A^2$ are $-1,-1,0$, with every nonzero vector in the appropriate eigenspace an eigenvector.

For $a=(a_1,a_2,a_3{)}^T$, the cross-product matrix isIts diagonal is zero, so $\mathrm{tr}A=0$. Also $Aa=a\times a=0$, and $a\ne 0$, so $A$ has a nontrivial kernel. Hence

The antisymmetric contribution is the negative of twice the cross-product matrix for $\mathbf{a}$:The cross-product matrix spectrum is $0,\pm i|\mathbf{a}|$, so $M$ has eigenvalues $1,1\pm 2i|\mathbf{a}|$. Multiplication givesDirect expansion of the displayed matrix yields the same real positive determinant.

The unique symmetric and antisymmetric decomposition is $S=(T+T^T)/2$, $A=(T-T^T)/2$. Both pieces are Cartesian tensors: transposing $T^{\prime }=RT{R}^T$ gives $(T^{\prime }{)}^T=R{T}^T{R}^T$, so $S^{\prime }=RS{R}^T$ and $A^{\prime }=RA{R}^T$. Hence this is a decomposition into a symmetric tensor and an antisymmetric second-rank tensor, rather than just a decomposition of numerical arrays.

For the axial vector of an antisymmetric second-rank tensor, contract the two Levi-Civita symbols:ThusWith this convention $Ax=x\times a$, so the displayed matrix is the negative of the cross-product matrix for $a\times x$.