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Exercise 510Linear algebra › Matrices and linear systemsL1Added 3 Nov 2026

Properties of the trace

#trace#cyclic property#similarity invariance

Statement

Show that the trace satisfies:
1. tr⁡(A+B)=tr⁡(A)+tr⁡(B)\operatorname{tr}(A+B) = \operatorname{tr}(A) + \operatorname{tr}(B);
2. tr⁡(AB)=tr⁡(BA)\operatorname{tr}(AB) = \operatorname{tr}(BA) for matrices of compatible sizes;
3. tr⁡(P−1AP)=tr⁡(A)\operatorname{tr}(P^{-1}AP) = \operatorname{tr}(A) for PP invertible.

Hint

For (2), write tr⁡(AB)=∑i(AB)ii=∑i,jaijbji\operatorname{tr}(AB) = \sum_i (AB)_{ii} = \sum_{i,j} a_{ij}b_{ji} and swap the order of summation.

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Solution

1. Immediate from the definition: tr⁡(A+B)=∑i(aii+bii)=∑iaii+∑ibii\operatorname{tr}(A+B) = \sum_i (a_{ii}+b_{ii}) = \sum_i a_{ii} + \sum_i b_{ii}.

2.
tr⁡(AB)=∑i∑jaijbji=∑j∑ibjiaij=tr⁡(BA).\operatorname{tr}(AB) = \sum_i \sum_j a_{ij}b_{ji} = \sum_j \sum_i b_{ji}a_{ij} = \operatorname{tr}(BA).

3. Using (2) with A=P−1AA = P^{-1}A and B=PB = P:
tr⁡(P−1AP)=tr⁡((P−1A)P)=tr⁡(P(P−1A))=tr⁡(A).\operatorname{tr}(P^{-1}AP) = \operatorname{tr}((P^{-1}A)P) = \operatorname{tr}(P(P^{-1}A)) = \operatorname{tr}(A).

Conclusion. The trace is linear, invariant under cyclic permutations, and invariant under similarity.