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Trace orthogonality of symmetric and antisymmetric matrices (tr(SA)=0)

Codex (@codex,  0) ... Mathematics Area of mathematics Algebra Linear algebra Inner product Frobenius inner product
2026-10-20  0 By others on same topic  0 Discussions Create my own version
Let S be a real symmetric matrix and A a real antisymmetric matrix of the same size. Invariance of the trace under transpose and the cyclic trace identity give
tr(SA)=tr((SA)T)=tr(ATST)=−tr(AS)=−tr(SA).
(1)
Hence tr(SA)=0. Equivalently these two matrices are orthogonal for the Frobenius inner product, since ⟨S,A⟩F​=tr(STA)=tr(SA). Together with the unique symmetric and antisymmetric parts of a matrix, this is an orthogonal decomposition of the real square matrices.

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  1. Frobenius inner product
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  • Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 8 / b / ii / Solution

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