Why Small Errors Change a Classification
A Hessian test is a set of conditional conclusions, not a rule for labeling every point from a matrix alone. The point must first be stationary, the function must have the required derivatives, and the signs must be interpreted using the correct criterion for the dimension. Missing any of these conditions can turn a valid calculation into an invalid classification.
This tutorial focuses on three frequent mistakes: applying a Hessian test at a nonstationary point, calling every point with zero Hessian determinant a saddle, and using incorrect sign conditions for definiteness. The underlying idea is that the Hessian describes the quadratic part of the function near a stationary point; it does not replace the stationarity check or resolve every degenerate case.
Error 1: Classifying a Nonstationary Point from Its Hessian
The second derivative test classifies a stationary point, not an arbitrary point. For a twice continuously differentiable function, the standard conclusions are: a positive definite Hessian at a stationary point implies a strict local minimum; a negative definite Hessian implies a strict local maximum; and an indefinite Hessian implies a saddle point. If the Hessian is only semidefinite or otherwise fails these conditions, the test may not decide the classification.
The stationarity requirement is essential. Away from a stationary point, the linear term in the local change of the function need not vanish. That term can determine whether nearby values rise or fall, regardless of the curvature indicated by the Hessian.
Worked Example: A Positive Definite Hessian at a Nonstationary Point
Consider \(f(x,y)=x^2+y^2+3x\) at \((0,0)\). First compute the gradient:
The point is not stationary, so the Hessian test cannot classify it. The Hessian is
For any nonzero vector \((u,v)\), its quadratic form is \(2u^2+2v^2>0\), so this Hessian is positive definite. It would be a mistake to conclude that \((0,0)\) is a local minimum: the test's stationarity condition is not met.
Check using nearby points. The function value at the origin is \(f(0,0)=0\). Along the \(x\)-axis, \(f(t,0)=t^2+3t\). If \(0<t<3\), then \(t^2+3t>0\); for \(t=-\varepsilon\) with \(0<\varepsilon<3\), \(f(-\varepsilon,0)=\varepsilon^2-3\varepsilon<0\). There are arbitrarily close points with values both above and below \(0\). Thus \((0,0)\) is neither a local minimum nor a local maximum.
Error 2: Treating a Zero Determinant as a Saddle
For a function of two variables, let \(D=f_{xx}f_{yy}-f_{xy}^2\) at a stationary point. If \(D>0\), the sign of \(f_{xx}\) distinguishes a local minimum from a local maximum. If \(D<0\), the Hessian is indefinite and the point is a saddle. But when \(D=0\), this test is inconclusive. It does not say “saddle”; it says that the second derivative test does not decide.
Worked Example: Zero Determinant Can Accompany a Local Minimum
Let \(f(x,y)=x^2+y^4\). At the origin, \(f_x=2x\) and \(f_y=4y^3\), so \(\nabla f(0,0)=(0,0)\). The second derivatives are \(f_{xx}=2\), \(f_{xy}=0\), and \(f_{yy}=12y^2\). Therefore, at the origin,
The determinant test is inconclusive. To classify the point, examine the function itself: \(x^2+y^4\ge 0\) for all \((x,y)\), and equality holds only when \(x=0\) and \(y=0\). Hence the origin is a strict local minimum. This conclusion comes from the function values, not from interpreting \(D=0\) as a particular type of point.
Worked Example: The Same Zero Determinant Can Accompany a Saddle
Now let \(g(x,y)=x^2-y^4\). Its first derivatives are \(g_x=2x\) and \(g_y=-4y^3\), so the origin is stationary. Its Hessian at the origin is again
As in the previous example, the second derivative test alone gives no classification. Along the \(x\)-axis, \(g(t,0)=t^2>0\) for \(t\ne0\). Along the \(y\)-axis, \(g(0,t)=-t^4<0\) for \(t\ne0\). Both paths approach the origin, and their values have opposite signs relative to \(g(0,0)=0\). Therefore the origin is a saddle point.
The two examples have the same Hessian at the origin and the same zero determinant, but different classifications. That is why a zero determinant cannot, by itself, identify a saddle.
Why the Hessian Test Works at a Stationary Point
The test comes from the second-order Taylor expansion. For a twice continuously differentiable function near a point \(\mathbf{a}\), a small displacement \(\mathbf{h}\) gives
At a stationary point, \(\nabla f(\mathbf{a})=\mathbf{0}\), so the linear term vanishes. The quadratic form from the Hessian then controls the change in \(f\), provided it has a strict sign in every nonzero direction. Here is why that strict sign gives a local classification.
If \(H_f(\mathbf{a})\) is positive definite, its smallest eigenvalue \(\lambda\) is positive. Therefore \(\mathbf{h}^{\top}H_f(\mathbf{a})\mathbf{h}\ge\lambda\|\mathbf{h}\|^2\) for every \(\mathbf{h}\).
Since \(r(\mathbf{h})/\|\mathbf{h}\|^2\to0\), for sufficiently small nonzero \(\mathbf{h}\) we have \(|r(\mathbf{h})|<(\lambda/4)\|\mathbf{h}\|^2\).
Using stationarity and the expansion, \(f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})\ge(\lambda/2)\|\mathbf{h}\|^2-(\lambda/4)\|\mathbf{h}\|^2=(\lambda/4)\|\mathbf{h}\|^2>0\). Thus \(\mathbf{a}\) is a strict local minimum.
If the Hessian is negative definite, apply the same argument to \(-f\) to obtain a strict local maximum. If it is indefinite, there are directions in which the quadratic form is positive and directions in which it is negative; the Taylor expansion then gives function values both above and below \(f(\mathbf{a})\) arbitrarily near \(\mathbf{a}\), so \(\mathbf{a}\) is a saddle.
This also explains why a semidefinite Hessian may be inconclusive: its quadratic form can vanish in some nonzero directions, where higher-order terms may determine what happens. The two examples with zero determinant illustrate that possibility.
Error 3: Using the Wrong Definiteness Signs
For a symmetric \(3\times3\) Hessian, Sylvester’s criterion checks the leading principal minors. Write \(\Delta_1\) for the upper-left entry, \(\Delta_2\) for the determinant of the upper-left \(2\times2\) block, and \(\Delta_3\) for the determinant of the full matrix. The required sign patterns are:
| Classification | Required signs |
|---|---|
| Positive definite | \(\Delta_1>0,\ \Delta_2>0,\ \Delta_3>0\) |
| Negative definite | \(\Delta_1<0,\ \Delta_2>0,\ \Delta_3<0\) |
A common error is to require all three leading principal minors to be negative for negative definiteness. The middle sign must instead be positive. Another error is to decide definiteness from the full determinant alone: a positive determinant does not imply a positive definite Hessian in three variables.
Worked Example: A Positive Determinant Does Not Mean Positive Definite
Consider \(F(x,y,z)=\frac12(-x^2-y^2+z^2)\) at the origin. Its gradient is \(\nabla F=(-x,-y,z)\), which vanishes at the origin. Its Hessian is
The leading principal minors are \(\Delta_1=-1\), \(\Delta_2=(-1)(-1)-0^2=1\), and \(\Delta_3=(-1)(-1)(1)=1\). The full determinant is positive, but the positive-definite sign pattern fails because \(\Delta_1<0\). The negative-definite pattern also fails because \(\Delta_3>0\), not \(\Delta_3<0\).
Direct checks confirm the classification. The function value at the origin is \(0\). Along the \(x\)-axis, \(F(t,0,0)=-t^2/2<0\) for \(t\ne0\); along the \(z\)-axis, \(F(0,0,t)=t^2/2>0\) for \(t\ne0\). Thus the origin is a saddle point, despite the positive full determinant.
A Practical Checklist
Before assigning a local classification, check each condition in order. This prevents a correct computation from being used to support a conclusion it cannot justify.
- Confirm that the point is stationary by evaluating the gradient there.
- Confirm that the function has continuous second partial derivatives near the point when using the Hessian test.
- For two variables, use the determinant test only with its stated conditions; treat \(D=0\) as inconclusive.
- For three variables, apply the leading-principal-minor sign patterns, not a remembered but incorrect variation.
- If the test is inconclusive, examine the function’s local behavior using appropriate directions or higher-order terms.
Common mistake and correction: “The Hessian determinant is zero, so the point is a saddle.” This is not a valid inference. A zero determinant means the standard determinant test does not settle the classification. Check the function itself, as in the examples \(x^2+y^4\) and \(x^2-y^4\), which have the same Hessian at the origin but different local behavior.
Check Your Understanding
For each question, identify which conditions are needed before drawing a conclusion.
- Why is a positive definite Hessian not enough to classify a point if the gradient there is nonzero?
- For a stationary point of a twice continuously differentiable function of two variables, what does \(D=0\) tell you—and what does it not tell you?
- What are the leading-principal-minor sign conditions for a negative definite \(3\times3\) Hessian?
- Can a positive determinant alone establish that a symmetric \(3\times3\) Hessian is positive definite? Explain briefly.
- Give two directions or paths that could help determine the behavior of a point when the Hessian test is inconclusive.