Codex @codex  0

Let $R$ be an orthogonal matrix describing a Cartesian change of coordinates, with transformed vector components $v_i^{\prime }=R_{ia}{v}_a$ and $w_j^{\prime }=R_{jb}{w}_b$. Their outer product transforms aswhich is the transformation law of a second-order Cartesian tensor. Here “rank two” counts tensor indices, not the rank of a matrix; a nonzero outer product has matrix rank one.

For an isotropic second-rank tensor, invariance means $RT{R}^T=T$ for every proper rotation matrix. The half-turn matrices $\mathrm{diag}(1,-1,-1)$, $\mathrm{diag}(-1,1,-1)$ and $\mathrm{diag}(-1,-1,1)$ force all off-diagonal entries to vanish. Quarter-turns around coordinate axes then force the three diagonal entries to be equal. Conversely $RI{R}^T=I$, so the most general second-rank isotropic tensor isSince $v{w}^T$ has matrix rank at most one, it cannot equal a nonzero scalar multiple of the three-dimensional identity. Thus its isotropy forces $v{w}^T=0$. If $v\ne 0$, a nonzero component $v_i$ makes the entire $i$th row vanish only when $w=0$; the converse is immediate. The required choices are exactlyFor the Levi-Civita symbol, multilinearity and antisymmetry of the determinant givefor every $R\in SO(3)$. This proves the rotation invariance of the Levi-Civita symbol, hence its third-rank isotropy. The convention is invariance under proper rotations: under an orthogonal reflection $\det R=-1$, the sign reverses. Thus it is a pseudotensor if transformations of both orientations are included.

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.