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Under a Cartesian frame change $B_i^{\prime }=R_{ik}{B}_k$, where $R$ is an orthogonal matrix. Orthogonality gives $B_k^{\prime }{B}_k^{\prime }=B_k{B}_k$ and $R_{ik}{R}_{j\ell }{\delta }_{k\ell }={\delta }_{ij}$. Thereforewhich is the tensor component transformation law. Also $T_{ij}=T_{ji}$ directly, proving that it is a symmetric Cartesian second-rank tensor for every choice of the two scalars.

For $\mathbf{B}\ne 0$, the direction of $\mathbf{B}$ is an eigenvector with eigenvalue $(\alpha +\beta )|\mathbf{B}{|}^2$. Every vector perpendicular to $\mathbf{B}$ is an eigenvector with eigenvalue $\beta |\mathbf{B}{|}^2$, a two-dimensional eigenspace. If $\alpha =0$, these eigenvalues coincide and every nonzero vector is an eigenvector. If $\mathbf{B}=0$, the tensor is zero and again every nonzero vector is an eigenvector, now with eigenvalue zero.

For the position-dependent field with zero divergence, the product rule givesContraction of the two Levi-Civita symbols in the curl and cross product givesConsequentlyThese constants give the magnetic part of the Maxwell stress tensor, with its physical prefactor suppressed. They are forced if the identity is to hold for every divergence-free field: $\mathbf{B}=(y,0,0)$ forces $2\beta =-1$; at the origin $\mathbf{B}=(y,1,0)$ then forces $\alpha =1$.

Apply the divergence theorem separately to each component:Since the whole vector field vanishes on $S$, all components of $T$ vanish there, and hence the vector volume integral is zero. The argument assumes a differentiable field and a boundary for which the divergence theorem applies.

Let $R_{ij}$ be the components of an orthogonal matrix relating two Cartesian frames by $v_i^{\prime }=R_{ij}{v}_j$. Repeated indices are summed from $1$ to $3$. The tensor component transformation law for a Cartesian second-rank tensor isFor a proper rotation, $\det R=1$; the transformation law itself also holds for orthogonal changes of Cartesian frame with determinant $-1$.

To find a cubically invariant second-rank tensor, first square each of the prescribed quarter-turns. For example the half-turn about the first axis is $S_x=\mathrm{diag}(1,-1,-1)$. Invariance under $S_{x}A{S}_x^{\mathsf{T}}$ forces $A_{12},A_{13},A_{21},A_{31}$ to vanish. The corresponding half-turns about the other axes eliminate the remaining off-diagonal components, even though symmetry of $A$ was never assumed.

Now write $A=\mathrm{diag}(a,b,c)$. A quarter-turn about the third axis interchanges $a,b$, so $a=b$; a quarter-turn about the first axis gives $b=c$. Conversely a scalar multiple of the identity matrix is invariant under every rotation. Thus the most general answer is

The coordinate transformation law for a affine connection has a homogeneous term with one upper and two lower index factors, and an inhomogeneous term involving second coordinate derivatives. The latter is the same for both Levi-Civita connections, so it cancels in their difference. Consequentlywhich is exactly the tensor component transformation law. Equivalently, $(\tilde{\nabla }-\nabla {)}_{X}Y$ is linear over functions in both $X$ and $Y$.

Write $V=T_p\mathcal{M}$ and $V^{\ast }$ for its dual space. A type-$(1,1)$ tensor is equivalently a linear map $A:V\to V$: the corresponding scalar-valued multilinear map is $(X,\eta )\mapsto \eta (A(X))$. Thus the composition definesIt is linear in $X$ because $A$ and $B$ are linear maps, and linear in the covector $\eta$ by evaluation. This proves directly that the composition is a tensor of type $(1,1)$.

Choose a basis $e_{\alpha }$ of the tangent space and its dual basis $e^{\beta }$. With the paper's component order $A(e_{\alpha })=A_{\alpha }{}^{\gamma }{e}_{\gamma }$,Equivalently, in the usual output-first matrix notation, $C^{\beta }{}_{\alpha }=B^{\beta }{}_{\gamma }{A}^{\gamma }{}_{\alpha }$: the matrix product is $BA$, not $AB$. This composition of mixed tensors contracts one output index with one input index. Under the tensor component transformation law, the two Jacobian factors at that contracted index cancel, leaving exactly one factor for each free index.

In coordinates $x^{\mu }$, write the type-$(1,1)$ tensor asFor a covector $\eta ={\eta }_{\mu }d{x}^{\mu }$ and vector $V=V^{\nu }{\partial }_{\nu }$,Since this must equal $\eta (V)={\eta }_{\nu }{V}^{\nu }$ for every $\eta$ and $V$, the components areThus $T$ is the identity tensor.

The tensor component transformation law givesThe components are therefore the same Kronecker delta in every coordinate system.

In Cartesian coordinates a rank-$r$ tensor is an array whose components transform under a change of orthonormal axes by one rotation matrix factor per index:An isotropic tensor has unchanged components under every proper rotation. The Levi-Civita symbol obeysfor $Q\in SO(3)$, proving rotation invariance of the Levi-Civita symbol. Under an orthogonal transformation of determinant $-1$ its sign reverses; including such transformations requires the pseudotensor distinction. Here isotropy is with respect to proper rotations.

Contract the determinant identity with ${ϵ}_{ijk}$. Since ${ϵ}_{ijk}{ϵ}_{ijk}=3!=6$, one obtainsEvery index on the right is contracted, so the tensor component transformation law leaves no free rotation factor: the determinant is a scalar. Equivalently, $A^{\prime }=QA{Q}^T$ gives $\det A^{\prime }=(\det Q{)}^2\det A=\det A$.

For the matrix inverse, defineThis is the cofactor matrix. To verify the inverse formula rather than just recognize it, contract with $A_{ib}$ and use the determinant identity again:Here the contraction of two Levi-Civita symbols gives ${ϵ}_{ajk}{ϵ}_{ijk}=2{\delta }_{ai}$. Therefore $A{C}^T=(\det A)I$ and, when $\det A\ne 0$,These are the Levi-Civita formulas for a three-dimensional determinant and inverse.

Now differentiate the cubic determinant formula with respect to the individual entry $A_{ab}$. The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. ThusSubstituting the inverse expression proves the entrywise derivative of the determinant:The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.

For the time-dependent Jacobian matrix, the stated smooth expansion impliesApplying the chain rule and the entrywise determinant derivative gives Jacobi's formula near $t=0$:At zero, $C=I$, so ${{\partial }_t\det A|}_0={\sum }_a{\partial }_a{u}_a=\nabla \cdot u$. A Taylor expansion therefore yieldsThe Jacobian determinant measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.

Let $Q$ be the orthogonal matrix taking old components to new ones, so $x_i^{\prime }=Q_{ip}{x}_p$. With repeated indices summed, the tensor component transformation law isAn isotropic tensor has the same components after every proper rotation. Since $Q_{ip}{Q}_{jp}={\delta }_{ij}$, the Kronecker delta is invariant. Applying this identity to each factor shows that $a{\delta }_{ij}$ and every linear combination $b{\delta }_{ij}{\delta }_{kl}+c{\delta }_{ik}{\delta }_{jl}+d{\delta }_{il}{\delta }_{jk}$ are invariant as well.

To show that an isotropic second-rank tensor must be scalar, start with any matrix $t$, without assuming symmetry. Invariance under quarter-turns about axes 1 and 2 implies invariance under their squares, the half-turnsThe first forces $t_{12},t_{21},t_{13},t_{31}$ to vanish. The second also forces $t_{23},t_{32}$ to vanish. The matrix is therefore diagonal. A quarter-turn about axis 1 interchanges its 22 and 33 entries; one about axis 2 interchanges its 11 and 33 entries. All three must agree, giving $t=aI$.

A rotation preserves the ball $V_R$, $|x{|}^2$ and the volume element. Changing integration variables by that rotation shows that $A$ and $B$ are isotropic tensors. Write $A_{ij}=a{\delta }_{ij}$. Taking its tensor contraction givesso $a=4\pi R^7/21$. Every permutation of the indices of $B$ leaves its integrand unchanged. Its complete symmetry therefore makes all three isotropic coefficients equal, say $d$:Double contraction gives $15d=B_{iikk}={\int }_{V_R}|x{|}^{4}dV=4\pi R^7/7$. Thus the isotropic fourth moment of a ball and its radial contraction are

Put $d=4\pi R^7/105$, so that $a=5d$. Contract $B$ with the prescribed unit vector twice:This has the spectrum of a scalar matrix plus a rank-one projector. In the direction $n$, the outer product $n{n}^T$ acts as the identity; on the plane perpendicular to $n$, it acts as zero. ThereforeAny nonzero multiple of $n$ is a parallel eigenvector, and any nonzero vector in the perpendicular plane is a transverse eigenvector. The transverse eigenvalue has multiplicity two; an orthonormal basis of that plane completes a set of principal axes.

A Cartesian tensor of order $n$ is a multilinear geometric object whose components under an orthogonal change of axes obey the tensor component transformation law,An isotropic tensor has the same components under every proper rotation. For the Kronecker delta,Thus ${\delta }_{ij}$ is isotropic, even under improper orthogonal transformations.

To prove multiplicativity of the determinant from the given formula, first consider $C_{pqr}={ϵ}_{ijk}{A}_{ip}{A}_{jq}{A}_{kr}$. Exchanging any two of $p,q,r$ changes its sign, so it is proportional to the Levi-Civita symbol. Its $123$ component is the determinant, as follows by expanding the given double-epsilon expression. ThereforeSubstitute $(AB{)}_{il}=A_{ip}{B}_{pl}$ into that same determinant formula:HenceNo invertibility assumption was used, so this identity includes singular matrices.

Now differentiate the double-epsilon formula for a differentiable $A(t)$. Its three differentiated terms are equal after renaming the paired dummy indices: the simultaneous swaps in both epsilon symbols cancel their signs. Thuswhere $C_{il}$ is the cofactor of entry $A_{il}$. For an invertible matrix, the adjugate matrix identity gives $C_{il}=(\det A)(A^{-1}{)}_{li}$. Consequently the Jacobi determinant derivative formula isThe order of the factors inside the trace follows from the reversed cofactor indices; no commutativity of $A$ and $\dot{A}$ is assumed. The cofactor version is also Jacobi's formula and remains valid when $A$ is singular, although the inverse version then cannot be used.

Define the outer product $T_{ij}=u_i{v}_j$. Transforming each vector givesThus $u_i{v}_j$ is a rank-two tensor, since it obeys the two-index tensor component transformation law. This is the component form of the tensor product of two vectors; its two indices are independent, not contracted.

For a constant orthogonal Cartesian coordinate change $x_i^{\prime }=Q_{ij}{x}_j$, the tensor component transformation law givesRepeated indices are summed. Each index of an ordinary Cartesian tensor contributes one factor of $Q$. A pseudotensor would have an additional determinant factor under an orientation-reversing transformation; the rules here concern ordinary tensors.

Let $B$ be the rank-two tensor and $v$ the vector. The components of their tensor product are $T_{ijk}=B_{ij}{v}_k$. Applying the transformation rules givesThe product therefore obeys the three-index tensor component transformation law, proving that $B_{ij}{v}_k$ is a rank-three tensor. Tensor-product order adds because no pair of indices is summed out.

Under the same Cartesian change, a vector field has $u_i^{\prime }(x^{\prime })=Q_{ip}{u}_p(x)$ with $x=Q^T{x}^{\prime }$. The chain rule gives $\partial /\partial x_j^{\prime }=Q_{jq}\partial /\partial x_q$. Since $Q$ is constant,This is exactly the order-two tensor component transformation law. Thus the Cartesian vector gradient is a Cartesian second-rank tensor, as in the velocity gradient tensor when $u$ is a velocity field.

The restriction to Cartesian transformations matters. In a general position-dependent basis, differentiating the basis contributes extra terms, so the tensorial derivative is the covariant derivative, not an arbitrary array of partial derivatives.

For a constant orthogonal Cartesian change of coordinates $x_i^{\prime }=Q_{ij}{x}_j$, the tensor component transformation law for order $n$ isRepeated indices are summed. In Cartesian orthonormal coordinates the metric identifies upper and lower indices, so the same orthogonal matrix acts on every index.

An isotropic tensor has the same numerical components after every proper orthogonal change of axes: $Q^{\otimes n}T=T$ for every $Q$ with $Q^{T}Q=I$ and $\det Q=1$. For example, ${\delta }_{ij}$ is isotropic because $Q_{ik}{Q}_{jk}={\delta }_{ij}$. Some conventions require invariance under all orthogonal transformations, including reflections; this stronger convention must be distinguished for odd-order tensors, notably the Levi-Civita symbol, which is invariant under proper rotations but changes sign under its ordinary three-index transformation for a reflection.

Preserving Cartesian lengths and Euclidean inner products requires an orthogonal matrix:Its determinant is $\pm 1$. Relative to an originally right-handed frame, $\det M=+1$ gives a right-handed frame and $\det M=-1$ a left-handed frame. More generally, this sign determines whether the two frames have the same or opposite orientation.

The tensor component transformation law isFor a tensor field, the components are compared at the same physical point, whose coordinates satisfy $x=M^T{x}^{\prime }$.

Define the symmetric tensor and antisymmetric tensor parts by $S=(T+T^T)/2$ and $A=(T-T^T)/2$. The transformed transpose satisfies $(T^{\prime }{)}^T=(MT{M}^T{)}^T=M{T}^T{M}^T$. Thereforeand similarlyBoth obey the tensor component transformation law. Thus the symmetric and antisymmetric parts are themselves second-order tensors. Taking a transpose and forming linear combinations commute with this orthogonal change of frame.

A second-order Cartesian tensor assigns components in each orthonormal coordinate system according toHere $R$ is the matrix taking old vector components to new ones, and repeated indices are summed. This is the tensor component transformation law; an arbitrary array of nine numbers is not a tensor unless its components in different frames obey this rule. No symmetry $T_{ij}=T_{ji}$ is implied by being a second-order tensor.

Let $R$ be an orthogonal change of axes. Since $A$ is a second-order tensor, $A_{ij}^{\prime }=R_{ik}{R}_{j\ell }{A}_{k\ell }$. The scalar-contraction assumption givesfor every tensor $A$. In one fixed frame its components can be any array: in particular, choose one component at a time to be one, all others zero. Equality of the coefficients then forcesMultiplying by $R$ and $R^T$ yieldsThis is exactly the tensor component transformation law. It is the full-contraction version of the quotient theorem for Cartesian tensors. The words “for every” are essential: testing a single $A$, such as the zero tensor or the identity, would not determine all components.

In orthonormal Cartesian coordinates, a rank-$n$ tensor has components transforming under a change of axes $R$ asRepeated indices are summed. This is the tensor component transformation law; for rank one it is precisely the transformation law of a vector.

Use the vector triple product to write the double cross product asBoth dot products are invariant under an orthogonal change of axes, while $u^{\prime }=Ru$ and ${\Omega }^{\prime }=R\Omega$. ThereforeHence $W$ is a vector. This argument also avoids the sign issue for an individual cross product under an orientation-reversing orthogonal transformation: the double product has two such factors, whose signs cancel.

For a rotation matrix $R\in SO(3)$, the Cartesian tensor component transformation law isAn isotropic tensor is unchanged under every such rotation, so its components satisfy $T_{ijkl}=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}$ for every $R\in SO(3)$. The repeated indices are summed by the Einstein summation convention.