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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02
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The two-variable Hessian determinant test

Statement

Let a be a critical point of a C2 function of two variables, and put A=fxx(a), B=fxy(a), C=fyy(a) and Δ=AC−B2. If Δ>0, then A>0 gives a strict local minimum and A<0 a strict local maximum. If Δ<0, there is neither. If Δ=0, this test gives no conclusion.

Facts & Assumptions

Given: a is a critical point of a C2 scalar field f on an open subset of R2.

[L1]
[L2]

The second-derivative test classifies a critical point from definiteness or indefiniteness of its Hessian (The multivariable second-derivative test by definiteness of the Hessian).

Proof

technique · calculation
1.1

By [L1], the Hessian quadratic form is Q(s,t)=As2+2Bst+Ct2. If A≠0, completing the square gives Q=A(s+(B/A)t)2+(Δ/A)t2.

L1algebra
2.1

If Δ>0, the two coefficients in step 1.1 have the sign of A, so Q is positive definite for A>0 and negative definite for A<0.

step 1.1algebra
2.2

If Δ<0 and A≠0, then Q(1,0)=A while Q(−B/A,1)=Δ/A, which have opposite signs; hence Q is indefinite. If A=0, then −B2=Δ<0, so B≠0, and Q(1,t)=2Bt+Ct2 has both signs for sufficiently small positive and negative t.

step 1.1algebra
3.1

Apply [L2] to steps 2.1 and 2.2. When Δ=0 and A≠0, step 1.1 makes Q=A(s+(B/A)t)2, so it is semidefinite but not definite; when A=0, then B=0 and Q=Ct2, with the same conclusion (including Q=0). Thus this is the inconclusive case of [L2].

L2step 1.1step 2.1step 2.2algebra∎

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