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:
(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 H | Every direction… | Verdict |
|---|---|---|
| both positive | curves upward | minimum |
| both negative | curves downward | maximum |
| opposite signs | disagree | saddle |
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 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: