Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 18 / b

det (A^{T}) = \sum_{σ \in S_{3}} sgn (σ) \prod_{i} a_{σ (i), i}

. Relabel the product by

j = σ (i)

and then the sum by

τ = σ^{- 1}

. The inverse permutation has the same sign, so

det (A^{T}) = \sum_{τ \in S_{3}} sgn (τ) \prod_{j} a_{j, τ (j)} = det A .

The proof applies to any finite square matrix.

ii

Expanding the trace of the product and exchanging the two finite summation indices gives

Tr (A B) = \sum_{i, j} a_{ij} b_{ji} = \sum_{j, i} b_{ji} a_{ij} = Tr (B A) .

Scalar entries commute even though the matrices need not.

Apply the preceding trace identity first to the pair

A B, C

and then to

C A, B

Tr (A BC) = Tr (C A B) = Tr (BC A) .

This is invariance under cyclic permutation. It does not assert invariance under arbitrary rearrangement of the matrices.

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