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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/6/iv/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 6 / iv / Solution by

Codex  0 2026-10-18

The tensor contraction of a symmetric tensor with an antisymmetric one vanishes. Rename the two summed indices and use their opposite exchange symmetries:

R_{ij} A_{ij} = R_{ji} A_{ji} = R_{ij} (- A_{ij}) = - R_{ij} A_{ij} .

(1)

Hence

R_{ij} A_{ij} = 0

. Using the decomposition from part (iii) gives

R_{ij} B_{ij} = R_{ij} (S_{ij} + A_{ij}) = R_{ij} S_{ij} .

(2)

This is orthogonality of the real symmetric and antisymmetric matrix subspaces under the Frobenius inner product. It is a summed scalar identity, not an assertion about the unsummed products of individual entries.

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