Why the Two-Variable Determinant Test Works

Why the Determinant Appears in the Test

A second-derivative test for a function of two variables uses a short calculation involving the Hessian determinant. The calculation is useful, but its meaning is easier to remember when we see what the Hessian does to directions near a critical point.

The key idea is to choose orthonormal directions in which the Hessian’s quadratic form has no mixed term. In those coordinates, the form is a sum of two squares, each multiplied by an eigenvalue. The signs of those eigenvalues tell us whether the second-order change is positive, negative, or of both signs.

The Hessian and Its Quadratic Form

Let \(f\) have continuous second partial derivatives near a critical point \((a,b)\), so that \(\nabla f(a,b)=\mathbf{0}\). Its Hessian at that point is the symmetric matrix

The equality of the mixed partials follows from their continuity. For a displacement \(\mathbf{h}=(h_1,h_2)\), the Hessian’s quadratic form is

The second-order Taylor expansion at the critical point is

The linear term is absent because the gradient is zero. Thus, when the quadratic form is strictly positive in every nonzero direction, the leading change in \(f\) is positive; when it is strictly negative in every nonzero direction, the leading change is negative.

Diagonalizing a Symmetric Two-by-Two Matrix

We now show why orthonormal coordinates remove the mixed term. The eigenvalues of \(\mathbf{H}\) satisfy its characteristic equation:

The discriminant of this quadratic is \((A-C)^2+4B^2\), which is nonnegative. Therefore its eigenvalues are real. If they are distinct, corresponding eigenvectors are orthogonal: if \(\mathbf{H}\mathbf{e}_1=\lambda_1\mathbf{e}_1\) and \(\mathbf{H}\mathbf{e}_2=\lambda_2\mathbf{e}_2\), symmetry gives

Since \(\lambda_1\ne\lambda_2\), this implies \(\mathbf{e}_1^{\top}\mathbf{e}_2=0\). Normalize the eigenvectors to unit length. If the eigenvalues are repeated, the discriminant is zero, so \(A=C\) and \(B=0\); in that case \(\mathbf{H}=A\mathbf{I}\), and any orthonormal pair of directions works.

Let \(\mathbf{e}_1,\mathbf{e}_2\) be an orthonormal eigenvector pair with eigenvalues \(\lambda_1,\lambda_2\). Write a displacement in these directions as \(\mathbf{h}=u\mathbf{e}_1+v\mathbf{e}_2\). Orthogonality and the eigenvector equations give the diagonal form:

This expression reveals the directional behavior directly. If both eigenvalues are positive, every nonzero displacement has \(Q>0\). If both are negative, every nonzero displacement has \(Q<0\). If they have opposite signs, the form takes both positive and negative values along its eigenvector directions.

Deriving the Determinant Test

The characteristic polynomial also shows that the product of the eigenvalues is the constant term:

So the sign of \(D\) tells us whether the eigenvalues have the same sign or opposite signs, provided \(D\ne 0\). When \(D>0\), they have the same sign. In that case, \(A\) has that sign too: if \(\mathbf{w}\) is the unit vector in the first coordinate direction, then \(A=\mathbf{w}^{\top}\mathbf{H}\mathbf{w}\). Expressing \(\mathbf{w}\) in the orthonormal eigenvector directions makes this a sum of the eigenvalues weighted by nonnegative squared coordinates. If both eigenvalues are positive, this sum is positive; if both are negative, it is negative.

For the strict-extremum conclusions, the Taylor remainder matters as well as the quadratic form. For example, if both eigenvalues are positive, let \(m=\min(\lambda_1,\lambda_2)>0\). Then \(Q(\mathbf{h})\ge m\|\mathbf{h}\|^2\). The remainder is \(o(\|\mathbf{h}\|^2)\), so sufficiently close to \(\mathbf{0}\) its magnitude is less than \(\frac{m}{4}\|\mathbf{h}\|^2\). Consequently, the change in \(f\) is at least \(\frac{m}{2}\|\mathbf{h}\|^2-\frac{m}{4}\|\mathbf{h}\|^2>0\) for nonzero sufficiently small \(\mathbf{h}\). This proves a strict local minimum. Reversing the signs proves the strict local maximum case.

Worked Examples

A Common Mistake: Using \(D>0\) Alone

A positive determinant does not, by itself, distinguish a minimum from a maximum. For example, let \(r(x,y)=-\frac12(3x^2+2y^2)\). At the origin its Hessian is \(\begin{pmatrix}-3&0\\0&-2\end{pmatrix}\), and \(D=(-3)(-2)-0=6>0\). A mistaken conclusion would be: “The determinant is positive, so the origin is a minimum.” But the Hessian entries give \(r(x,y)=-\frac32x^2-y^2<0=r(0,0)\) for every nonzero \((x,y)\). The origin is a strict maximum. The correction is to use \(f_{xx}\) as well as \(D\): here \(r_{xx}(0,0)=-3<0\), which identifies the maximum case.

What the Test Does—and Does Not—Say

The determinant test is a compact way to read the eigenvalue signs without calculating eigenvectors in every application. For \(D>0\), the two eigenvalues have the same sign, and \(f_{xx}\) tells which sign. For \(D<0\), they have opposite signs, so the quadratic form has directions of positive and negative second-order change. If \(D=0\), at least one eigenvalue is zero, and this second-order argument does not settle the local behavior; higher-order information may matter.

The assumptions are important. The test applies at a critical point, requires continuous second partial derivatives near that point, and yields the strict-extremum conclusions when \(D>0\). It is the Taylor expansion, together with the eigenvalue signs, that connects the matrix calculation to the behavior of the function itself.