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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-3/4b/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 3 / 4B / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-05

The quotient theorem for Cartesian tensors here says: if contracting

T_{ij}

with every vector

v_{j}

gives the components

w_{i} = T_{ij} v_{j}

of a vector in every right-handed orthonormal basis, then

T

obeys the second-order tensor transformation law. Conversely that transformation law guarantees the contraction is a vector. One may equivalently test that

u_{i} T_{ij} v_{j}

is a scalar for every pair of vectors

u, v

By part (a),

v^{'} = R v

and

w^{'} = Rw

. The contraction property in the new basis says

T^{'} R v = w^{'} = Rw = RT v .

(1)

Since this holds for every test vector,

T^{'} R = RT

. Using

R^{- 1} = R^{T}

gives

T^{'} = RT R^{T}, T_{ij}^{'} = R_{ik} R_{j ℓ} T_{k ℓ} .

(2)

For the converse,

T^{'} v^{'} = RT R^{T} R v = RT v = Rw

, exactly the vector law. For the equivalent scalar formulation, invariance of

u^{T} T v

for every

u, v

gives

u^{T} R^{T} T^{'} R v = u^{T} T v

, forcing the same matrix identity.

The quantifier “every” matters: an arbitrary nonzero matrix can annihilate one fixed vector, so a single successful contraction cannot establish tensoriality. The stated bases test proper rotations; transformation under orthogonal reflections would require an additional hypothesis if that were wanted. “Second-order” counts tensor indices, rather than the matrix rank of a particular component matrix.

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