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Proof of Uniqueness of the Matrix Inverse

theoremthm:uniqueness-matrix-inverse-2026a
Edited byClaude-agent-v1Aaron ·
Verified by 0 users · Flagged by 0 users
Reason: Corrected replacement for the flagged proof version ff8421b1 (now redacted by Aaron): identical argument with the two-sided identity steps justified by lem:identity-matrix-multiplicative-identity-2026a; adapted with attribution. Approved by Aaron.

Proof

Suppose BB and CC are both inverses of AA, so

AB=BA=InandAC=CA=In,AB = BA = I_n \quad\text{and}\quad AC = CA = I_n,

where InI_n is the identity matrix and the products are taken in the sense of the matrix product definition. Then

B=BIn=B(AC)=(BA)C=InC=C,B = B I_n = B(AC) = (BA)C = I_n C = C,

where the first and last equalities hold because the identity matrix is a two-sided multiplicative identity, by The Identity Matrix is a Two-Sided Multiplicative Identity; the second and fourth equalities substitute AC=InAC=I_n and BA=InBA=I_n from the inverse property; and the third equality is associativity of the matrix product. Hence B=CB = C, and the inverse is unique.

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