Trace, Symmetric and Skew-Symmetric Matrices
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Direct answer
The trace of a square matrix is the sum of its diagonal entries, and it obeys two results beyond the obvious linearity: tr(AB) = tr(BA), so trace is cyclic — tr(ABC) = tr(BCA) = tr(CAB) — and consequently tr(AB − BA) = 0. A matrix is symmetric when Aᵀ = A and skew-symmetric when Aᵀ = −A; skew-symmetric matrices have zero diagonal, hence trace zero, and every odd-order skew-symmetric matrix has determinant zero. Every square matrix decomposes uniquely as a symmetric part plus a skew-symmetric part: A = ½(A + Aᵀ) + ½(A − Aᵀ).
What you must remember
- Linearity: tr(A + B) = trA + trB, tr(kA) = k·trA, and tr(Iₙ) = n — the trace of the identity is the order.
- Cyclic property: tr(AB) = tr(BA) always, extended to tr(ABC) = tr(BCA) = tr(CAB); the order may rotate but not reverse — tr(ABC) ≠ tr(ACB) in general.
- Commutator corollary: tr(AB − BA) = 0, so no matrix X satisfies AX − XA = I: taking traces gives 0 = n, a contradiction — a favourite one-line Advanced proof.
- Decomposition: A = ½(A + Aᵀ) + ½(A − Aᵀ); the first term is symmetric, the second skew-symmetric, and the split is unique because a matrix that is both must be zero.
- Skew facts: diagonal entries of a skew-symmetric matrix are zero ⇒ trace 0; |A| = |Aᵀ| = |−A| = (−1)ⁿ|A| forces |A| = 0 for odd n.
- Products: AAᵀ and AᵀA are always symmetric; AB is symmetric iff AB = BA; (AB)ᵀ = BᵀAᵀ reverses the order.
- Eigenvalue framing (Advanced): trace equals the sum of the characteristic roots and determinant their product — useful for quick checks on 2×2 and 3×3 matrices.
One pair of matrices, every property checked
Take A = [[1, 2], [3, 4]] and B = [[0, 1], [1, 0]], so tr(A) = 5. Compute AB = [[2, 1], [4, 3]] and BA = [[3, 4], [1, 2]]: the products differ, yet tr(AB) = 5 = tr(BA) — the cyclic property surviving non-commutativity in one glance. The commutator AB − BA = [[−1, −3], [3, 1]] has trace −1 + 1 = 0, confirming tr(AB − BA) = 0 with explicit numbers. Now decompose A: ½(A + Aᵀ) = [[1, 5/2], [5/2, 4]] is symmetric and ½(A − Aᵀ) = [[0, −1/2], [1/2, 0]] is skew-symmetric with trace 0; adding them returns A, and the zero diagonal of the skew piece is visible rather than quoted. Finally, its determinant: |½(A − Aᵀ)| = 0×0 − (−1/2)(1/2) = 1/4 — nonzero, as it must be, since order 2 is even; the vanishing determinant is an odd-order privilege.
Where students slip
The commonest written falsehood is tr(AB) = tr(A)·tr(B) — trace is additive over sums, not multiplicative over products; with the A, B above, tr(A)tr(B) = 0 while tr(AB) = 5. Second, students assume AB is symmetric whenever A and B are: symmetry of the product requires commutation, since (AB)ᵀ = BᵀAᵀ equals AB only when AB = BA. Third, the zero diagonal of skew-symmetric matrices is a consequence, not the definition — defining skew-symmetry as "diagonal zero and off-diagonal pairs opposite" works for checks but fails in proofs, where Aᵀ = −A does the work: putting i = j in aᵢᵢ = −aᵢᵢ is what forces the diagonal to vanish. Examiners phrase questions precisely to catch which version a student carries.
Frequently asked questions
Is tr(AB) = tr(BA) always true?
Yes, for any two square matrices of the same order, even when A and B do not commute — trace is invariant under cyclic rotation of products.
Why does an odd-order skew-symmetric matrix have zero determinant?
|A| = |Aᵀ| = |−A| = (−1)ⁿ|A|; for odd n this reads |A| = −|A|, forcing |A| = 0.
How is a matrix split into symmetric and skew parts?
A = ½(A + Aᵀ) + ½(A − Aᵀ); the first bracket is symmetric, the second skew-symmetric, and the decomposition is unique.
When is the product of two symmetric matrices symmetric?
Exactly when the factors commute: (AB)ᵀ = BᵀAᵀ equals AB iff AB = BA.
What is the trace of a commutator AB − BA?
Zero — every diagonal contribution cancels, which is why no equation of the form AX − XA = I can hold.