Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 6

For a Cartesian change of components $x_i^{\prime }=R_{ij}{x}_j$, a rank-$n$ tensor transforms according towith repeated indices summed. An isotropic tensor has unchanged components under every proper rotation matrix.

For rank two write $T^{\prime }=RT{R}^T$. Invariance under quarter-turns implies invariance under their squares, the coordinate half-turnsEach off-diagonal entry changes sign under at least one of these, so all off-diagonal entries vanish. A quarter-turn about the third axis then interchanges the first two diagonal entries, making them equal; a quarter-turn about the first axis equates the other two. ThereforeConversely $R(\lambda I)R^T=\lambda I$ for every rotation. This proves the isotropic second-rank tensor classification without assuming that the original tensor was symmetric. Also $\lambda =T_{kk}/3$ is a scalar, since the trace is rotation-invariant.

Let $s=A_i{A}_i>0$, $q=A_i{T}_{ij}{A}_j$ and $\tau =T_{ii}$. Taking the trace of the proposed decomposition givessince $A_i{B}_i=0$ and $C_{ii}=0$. Contracting twice with $A$ gives $q=\alpha s+\beta s^2$. Solving these two scalar equations yieldsA single contraction gives $T_{ij}{A}_j=(\alpha +\beta s)A_i+s{B}_i$, and $\alpha +\beta s=q/s$. HenceFinally setThis is manifestly symmetric. The formulas give $A_i{B}_i=q/s-q/s=0$; substituting them into $C_{ij}{A}_j$ gives zero, and taking the trace gives $C_{ii}=0$. Since $C$ is symmetric, $A_i{C}_{ij}=0$ also holds.

A geometric form makes the axis decomposition of a symmetric matrix explicit. With the transverse orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/s$,Here $\alpha$ is half the transverse trace and $C$ is its symmetric traceless rank-two tensor part. Every term was forced by contractions, so the decomposition is unique.

The required contractions areSubstituting into the axis decomposition of a symmetric matrix givesIndeed $A^{T}B=-4+4=0$, $CA=0$ and $\mathrm{tr}C=-15+3+12=0$. Adding $25I+5A{A}^T+B{A}^T+A{B}^T+C$ reconstructs the given tensor.