Classifying Nondegenerate Stationary Points with the Hessian Matrix

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Hello! In the previous lesson, you built a quadratic approximation from the gradient and Hessian. At a stationary point, the gradient vanishes, so the quadratic term becomes the first term that can reveal the function’s local shape. Today you will use that term to distinguish local minima, local maxima, and saddle points.

This matters when a function represents potential energy: a stationary configuration need not sit at the bottom of a potential well. It may instead have directions in which the energy rises and others in which it falls.

What the Hessian must decide

Let $p$ be an interior point where a smooth function $f$ has zero gradient. For a small displacement $d$ , last lesson’s Taylor approximation becomes

f (p + d) = f (p) + \frac{1}{2} d^{T} H_{f} (p) d + o (∥ d ∥^{2}) .

The question is not merely whether an individual Hessian entry is positive or negative. It is whether the quadratic form $d^{T} H_{f} (p) d$ has the same sign for every nonzero displacement.

Watch How the Hessian Finds Minima, Maxima, and Saddle Points? from Practical stats for a visual introduction. In particular, notice why checking curvature along just the coordinate axes can miss a saddle.

How the Hessian Finds Minima, Maxima, and Saddle Points?

This short video connects stationary points, directional curvature, and the two-variable Hessian test.

Watch the explanation from the opening landscape example through the final discussion of eigenvalues. Focus on the determinant test beginning around 03:47, and on why the test first requires a zero gradient.

There are three decisive sign patterns:

If $d^{T} H d > 0$ in every nonzero direction, $H$ is positive definite and the stationary point is a strict local minimum.
If it is negative in every nonzero direction, $H$ is negative definite and the point is a strict local maximum.
If it is positive in some directions and negative in others, $H$ is indefinite and the point is a saddle.

“Local” means the comparison holds for all sufficiently nearby points; it says nothing by itself about the entire domain. The higher-order remainder cannot overturn a definite quadratic form sufficiently close to $p$ . When the quadratic form takes both signs, moving a small distance in the corresponding directions gives values both above and below $f (p)$ .

A stationary point is nondegenerate when its Hessian has no zero eigenvalues—or, equivalently, when its Hessian determinant is nonzero. For such a point, the three sign patterns above give a classification.

A practical test in two variables

For $f (x, y)$ , evaluate the Hessian at a stationary point $(a, b)$ and write

H_{f} (a, b) = (A B B C), A = f_{xx} (a, b), B = f_{x y} (a, b), C = f_{yy} (a, b) .

Its determinant is

D = det H_{f} (a, b) = A C - B^{2} .

The complete two-variable test is:

Hessian test	Classification at the stationary point
$D > 0$ and $A > 0$	Strict local minimum
$D > 0$ and $A < 0$	Strict local maximum
$D < 0$	Saddle point
$D = 0$	Second-derivative test inconclusive

Why does the mixed derivative enter as $B^{2}$ ? For a displacement $d = (h, k)^{T}$ , the quadratic form is

d^{T} H d = A h^{2} + 2 B hk + C k^{2} .

When $D > 0$ , $A$ cannot be zero. Completing the square gives

A h^{2} + 2 B hk + C k^{2} = A (h + \frac{B}{A} k)^{2} + \frac{D}{A} k^{2} .

Both terms have the sign of $A$ , proving the minimum and maximum cases. When $D < 0$ , the symmetric Hessian has one positive and one negative eigenvalue, so there are both uphill and downhill directions. Thus $f_{xx} > 0$ alone does not establish a minimum: it checks bending along only one line.

Classifying stationary points step by step

Consider a potential-like function with a coupling between coordinates:

V (x, y) = x^{2} + 2 x y + 3 y^{2} - 4 x - 8 y .

First locate its stationary points; the Hessian test must not be applied before this step.

V_{x} = 2 x + 2 y - 4, V_{y} = 2 x + 6 y - 8.

Setting both derivatives to zero gives $x + y = 2$ and $x + 3 y = 4$ , hence the sole stationary point is $(1, 1)$ . Next compute

H_{V} = (2226), D = (2) (6) - 2^{2} = 8.

Since $D > 0$ and $V_{xx} = 2 > 0$ , $(1, 1)$ is a strict local minimum. Here the function itself is quadratic, so its Taylor approximation is exact. With $x = 1 + h$ and $y = 1 + k$ ,

V (1 + h, 1 + k) = - 6 + h^{2} + 2 hk + 3 k^{2} = - 6 + (h + k)^{2} + 2 k^{2} .

The final expression makes the result visible without a test: every nonzero displacement raises $V$ . For this particular quadratic, the minimum is also global; the Hessian test alone ordinarily establishes only a local result.

For a maximum, take $g (x, y) = 5 - (x - 1)^{2} - 2 (y + 2)^{2}$ . Its gradient vanishes at $(1, - 2)$ , and its Hessian is

H_{g} = (- 2 0 0 - 4) .

Here $D = 8 > 0$ but $g_{xx} = - 2 < 0$ , so $(1, - 2)$ is a strict local maximum.

Finally, consider $s (x, y) = x^{2} + 4 x y + y^{2}$ . The gradient vanishes at the origin, while

H_{s} (0, 0) = (2442), D = 4 - 16 = - 12.

The origin is a saddle despite both coordinate-axis second derivatives being positive. Along $y = x$ , $s (t, t) = 6 t^{2} > 0$ ; along $y = - x$ , $s (t, - t) = - 2 t^{2} < 0$ . The downhill direction lies between the coordinate axes.

The following simpler saddle has its opposing directions aligned with those axes.

The surface $f(x,y)=x^2-y^2$ rises away from the origin along the $x$-axis and falls along the $y$-axis. Its stationary origin has an indefinite Hessian.

What nondegeneracy rules out

Read the subsection “Minima, maxima, and saddle points” in MIT OpenCourseWare’s Second Derivatives, Bilinear Maps, and Hessian Matrices. It puts the directional-sign test into general matrix language. Its notation for a second derivative evaluated on two displacements corresponds here to the Hessian quadratic form $d^{T} H d$ .

Second Derivatives, Bilinear Maps, and Hessian Matrices

Read the classification discussion from MIT OpenCourseWare to connect the two-variable determinant rule to Hessian eigenvalues in any number of variables.

Find Section 12.4.3, “Minima, maxima, and saddle points,” around pp. 93–94, after the discussion of approximating Hessians. Begin where the author returns to the quadratic approximation and read the classification through the end of the subsection. Focus on the signs of the Hessian eigenvalues; you can skip the preceding material on computational methods.

In any number of variables, a symmetric Hessian has real eigenvalues. All positive means a strict local minimum, all negative a strict local maximum, and a mixture of positive and negative a saddle. The convenient $D$ -and- $f_{xx}$ table is specific to two variables: in higher dimensions, a determinant alone does not tell you all the eigenvalue signs.

A zero eigenvalue makes a stationary point degenerate, and the second-order test may leave out the decisive behavior. For example,

f_{+} (x, y) = x^{4} + y^{2}, f_{-} (x, y) = - x^{4} + y^{2}

have the same Hessian at the origin,

H (0, 0) = (0002) .

Yet the origin is a minimum of $f_{+}$ and a saddle of $f_{-}$ . The fourth-order term, invisible to the Hessian there, decides what happens along the $x$ -axis. Thus $D = 0$ means inconclusive, not “saddle” or “no extremum.”

Looking ahead

To classify a nondegenerate stationary point, first solve $\nabla f = 0$ , then evaluate the Hessian there and determine the signs of its quadratic form. In two variables, its determinant and $f_{xx}$ provide a compact test; the mixed derivative is essential because important directions need not align with the axes. The result describes local, unconstrained behavior.

Next, you will study constrained extrema. When motion is restricted to a curve or surface, only displacements that respect that constraint are allowed, so the unconstrained stationary-point test is no longer the whole story.