Back to courseLesson 13 of 13

The Hessian: eigenvalues decide

What you'll learn

Classify any critical point with the second-derivative test — and see it's the eigenvalue story from module 3 all over again.

One question is left: standing at a flat spot, is it a summit, a pit, or a pass? The answer welds this course's two halves — linear algebra and surfaces — into a single test.

Curvature comes as a matrix

In one variable, the second derivative settles it: f″ positive means a smile (minimum). In two variables there are four second partials — how the x-slope changes with x, with y, and likewise for the y-slope — and they arrange into the Hessian matrix:

H=(fxxfxyfyxfyy)H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix}

(The mixed partials f_xy and f_yx agree for any smooth f, so H is symmetric.) The Hessian is the landscape's curvature, written as a 2×2 matrix — which means everything from module 3 applies to it.

Eigenvalues decide

Near a critical point the surface is approximately a pure quadratic bowl or saddle, bent by H. The eigenvectors of H are the principal directions of the bend, and the eigenvalues are the curvatures along them:

Eigenvalues of HEvery direction…Verdict
both positivecurves upwardminimum
both negativecurves downwardmaximum
opposite signsdisagreesaddle

That's the whole classification. A saddle is a Hessian whose eigenvalues argue.

The shortcut you'll actually use

You rarely need the eigenvalues themselves — module 3 taught that their product is the determinant. Opposite signs means negative product, so:

D=detH=fxxfyyfxy2D = \det H = f_{xx} f_{yy} - f_{xy}^2

D negative → saddle. D positive → a true extremum (both eigenvalues share a sign — read which from f_xx: positive = minimum, negative = maximum). D = 0: the test abstains; the terrain is too flat to call.

Finish last lesson's example: f = x³ − 3x + y² has f_xx = 6x, f_yy = 2, f_xy = 0. At (1, 0): D = 12 > 0 with f_xx > 0 — minimum. At (−1, 0): D = −12 — saddle. The poke-test impressions, now proved.

The full circle

Step back and look at what you used: vectors to describe directions, the gradient to find flat spots, a matrix to hold the curvature, determinants and eigenvalues to read its character. Every module of this course is load-bearing in this one test — and it's the same test running inside the surface explorer's critical-point readout, and inside every optimizer that checks whether it found a true minimum or just a pass in the fog.

That's higher dimensions: not harder math — the same math, holding hands.

Check your understanding

Question 1 of 2

The Hessian at a critical point has eigenvalues 4 and −2. The point is: