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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Similar matrices have the same trace

Statement

If square matrices AA and BB over FF are similar, then tr⁡(A)=tr⁡(B)\operatorname{tr}(A)=\operatorname{tr}(B).

Facts & Assumptions

Given: Similar matrices A,B∈Mn(F)A,B\in M_n(F).

[L1]

Similarity means B=P−1APB=P^{-1}AP for an invertible matrix PP (Similar matrices: B=P−1APB=P^{-1}AP for an invertible PP).

[L2]

For conformable rectangular matrices X,YX,Y, one has tr⁡(XY)=tr⁡(YX)\operatorname{tr}(XY)=\operatorname{tr}(YX) (For A∈Mm×n(F)A\in M_{m\times n}(F) and B∈Mn×m(F)B\in M_{n\times m}(F), tr⁡(AB)=tr⁡(BA)\operatorname{tr}(AB)=\operatorname{tr}(BA)).

[L3]

Matrix multiplication is associative and unital (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

Proof

technique · direct
1.1

Choose invertible PP with B=P−1APB=P^{-1}AP as in [L1].

givenL1
2.1

Apply [L2] to X=P−1AX=P^{-1}A and Y=PY=P to obtain tr⁡(B)=tr⁡(P(P−1A))\operatorname{tr}(B)=\operatorname{tr}(P(P^{-1}A)).

step 1.1L1L2
3.1

By associativity and unitality from [L3], P(P−1A)=(PP−1)A=InA=AP(P^{-1}A)=(PP^{-1})A=I_nA=A, so tr⁡(B)=tr⁡(A)\operatorname{tr}(B)=\operatorname{tr}(A).

step 2.1L1L2L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 16 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources