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PropositionStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Trace is a linear functional on Mn(F)M_n(F)

Statement

For every natural nn, the map tr⁡:Mn(F)→F\operatorname{tr}:M_n(F)\to F is linear:

tr⁡(A+B)=tr⁡(A)+tr⁡(B),tr⁡(λA)=λtr⁡(A).\operatorname{tr}(A+B)=\operatorname{tr}(A)+\operatorname{tr}(B),\qquad \operatorname{tr}(\lambda A)=\lambda\operatorname{tr}(A).

Facts & Assumptions

Given: Matrices A,B∈Mn(F)A,B\in M_n(F) and a scalar λ∈F\lambda\in F.

[L1]

The trace is the finite sum of the diagonal entries (The trace tr⁡(A)\operatorname{tr}(A) as the sum of the diagonal entries).

Proof

technique · direct
1.1

Entrywise addition and the finite-sum law give tr⁡(A+B)=∑i<n(aii+bii)=∑i<naii+∑i<nbii=tr⁡(A)+tr⁡(B)\operatorname{tr}(A+B)=\sum_{i<n}(a_{ii}+b_{ii})=\sum_{i<n}a_{ii}+\sum_{i<n}b_{ii}=\operatorname{tr}(A)+\operatorname{tr}(B).

givenL1
2.1

Likewise, tr⁡(λA)=∑i<nλaii=λ∑i<naii=λtr⁡(A)\operatorname{tr}(\lambda A)=\sum_{i<n}\lambda a_{ii}=\lambda\sum_{i<n}a_{ii}=\lambda\operatorname{tr}(A).

step 1.1L1
3.1

If n=0n=0, every displayed sum is empty and has value 00, so trace is the zero linear functional on the zero matrix space.

step 2.1L1∎

Depends on

Used by

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Sources