CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13

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Congruence need not preserve trace or determinant: the real $1 \times 1$ matrices $[1]$ and $[4]$ are congruent

Statement refuted

Refuted claim: Matrix congruence preserves trace or determinant.

Facts & Assumptions

Given: The real $1 \times 1$ matrices $A = [1]$ , $C = [4]$ , and $P = [2]$ .

[L1]

Congruence has the form $C = P^{T} A P$ with $P$ invertible (A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{T} A P$ ).

[L2]

Congruence does preserve rank and nondegeneracy (Congruent matrices have the same rank; hence rank and nondegeneracy of a bilinear form are basis-independent).

[L3]

Trace is the sum of diagonal entries (The trace $tr (A)$ as the sum of the diagonal entries), and determinant is the signed Leibniz sum (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

Counterexample

technique · direct $1\times1$ computation

1.1

Since $P^{T} A P = [2] [1] [2] = [4] = C$ and $P$ is invertible, [L1] makes $A$ and $C$ congruent.

L1algebra

1.2

By [L3], $tr A = 1$ , $tr C = 4$ , $\det A = 1$ , and $\det C = 4$ . Thus neither trace nor determinant is preserved.

L3algebra

2.1

Both matrices nevertheless have rank $1$ and are nondegenerate, in agreement with [L2]; the counterexample isolates exactly the two false invariance claims.

step 1.1step 1.2L2∎

Depends on

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Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Congruence need not preserve trace or determinant: the real 1×11×1 matrices [1][1] and [4][4] are congruent