 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/17/a/ii/solution

Use Einstein notation, with repeated suffixes summed. The matrix multiplication and transpose give

C_{ij} = A_{ik} (A B)_{jk} = A_{ik} A_{j ℓ} B_{ℓ k} .

Interchange the free indices, then relabel the dummy indices

k, ℓ

C_{ji} = A_{jk} A_{i ℓ} B_{ℓ k} = A_{j ℓ} A_{ik} B_{k ℓ} = - A_{ik} A_{j ℓ} B_{ℓ k} = - C_{ij} .

The negative sign uses

B_{k ℓ} = - B_{ℓ k}

. Therefore $C$ is skew-symmetric. No orthogonality of

A

is needed for this part: matrix congruence preserves skew-symmetry.

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