Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 6

Under an orthogonal Cartesian change of axes $Q$, the vector components transform as $v_i^{\prime }=Q_{ia}{v}_a$, while $v_k^{\prime }{v}_k^{\prime }=v_a{v}_a$ and ${\delta }_{ij}=Q_{ia}{Q}_{ja}$. Substitution givesThis is the rank-two Cartesian tensor transformation law. The nonzero assumption ensures that the denominator is positive. In matrix form $P=I-v{v}^T/(v^{T}v)$ is the orthogonal projection matrix onto the orthogonal complement of $v$.

Contract with $v_j$ and use its squared Euclidean norm:Thus $v$ lies in the kernel of this orthogonal projection.

The orthogonality condition is $v_j{u}_j=0$, soEvery vector in the orthogonal complement is fixed by the orthogonal projection.

The Cartesian tensor law is $T_{ik}^{\prime }=Q_{ia}{Q}_{kb}{T}_{ab}$. In every complete tensor contraction, summation over a transformed index contracts a pair of $Q$ factors to a Kronecker delta. In detail,Thus all three are invariant under rotations. They are $\mathrm{tr}T$, $\mathrm{tr}(T^2)$ and $\mathrm{tr}(T^3)$. For a real symmetric tensor, orthogonal diagonalization givesExpanding ${\prod }_i(\lambda -{\lambda }_i)$ and using Newton's identities to recover the characteristic polynomial from traces of powers yieldsComparison identifies $\boxed{c_2=-1,c_1=\frac{1}{2},c_0=-\frac{1}{6}}$.