LINEAR-ALGEBRA · Interactive Practice — Unit 26 · Video 3

From Bowls to Saddles: How the Hessian Decides Where Functions Bottom Out

IKey Formulas

Concept	Formula	Meaning
Gradient zero	$\nabla f = 0$	Locates a critical point
Hessian (2D)	$H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{bmatrix}$	Matrix of second partials
Minimum test	$f_{xx} > 0$ AND $f_{xx}f_{yy} - f_{xy}^2 > 0$	$H$ is positive definite
Saddle test	$f_{xx}f_{yy} - f_{xy}^2 < 0$	$H$ is indefinite
Eigenvalue test	All eigenvalues of $H > 0$	Equivalent to positive definite

For a quadratic $f = a x^2 + 2b\, xy + c y^2$ , the Hessian is constant: $H = \begin{bmatrix} 2a & 2b \\ 2b & 2c \end{bmatrix}$

IIVisualization 1 — Bowl, Saddle, or Peak

The pivot $2a$ and the determinant of the Hessian decide whether the origin is a minimum, a maximum, or a saddle.

a 2 b 2 c 6

💡 Challenge: hold $a = 2$ and $c = 6$ fixed — how large must $b$ grow before the cross term overpowers the diagonal and the bowl collapses into a saddle?

IIIVisualization 2 — Eigenvectors as Principal Axes

The Hessian's eigenvectors are the principal axes of the level curves; each eigenvalue's sign is the curvature along its axis.

Hessian

💡 Every entry of $H = \begin{bmatrix} 2 & 6 \\ 6 & 7 \end{bmatrix}$ is positive, yet one eigenvalue is negative — the determinant, not the sign of the entries, decides definiteness.

IVQuiz Questions

Question 1

Consider $f(x, y) = 2x^2 + 12xy + 20y^2$ . The Hessian at the origin is $H = \begin{bmatrix} 4 & 12 \\ 12 & 40 \end{bmatrix}.$

Is the origin a minimum, a maximum, or a saddle?

✅ Correct! Pivot $4 > 0$ and $\det = 16 > 0$ , so $H$ is positive definite.

❌ Not quite. Check the pivot $f_{xx}$ and the determinant $f_{xx}f_{yy} - f_{xy}^2$ — both need to be positive for a minimum.

Show solution

Solution:

Apply both tests:

Upper-left pivot: $f_{xx} = 4 > 0$ .
Determinant: $\det H = 4 \cdot 40 - 12^2 = 160 - 144 = 16 > 0$ .

Both conditions are satisfied, so $H$ is positive definite. The origin is a minimum, and the surface is a bowl.

Question 2

A student claims: "The matrix $H = \begin{bmatrix} 2 & 6 \\ 6 & 7 \end{bmatrix}$ must be positive definite because every entry is positive."

True or False?

✅ Correct! $\det H = 14 - 36 = -22 < 0$ , so $H$ is indefinite (a saddle).

❌ Not quite. Compute $\det H = 2(7) - 6^2$ . Is it really positive? Positive entries do not guarantee positive definiteness.

Show solution

Solution:

Positive entries do not imply positive definite. Compute the determinant: $\det H = 2 \cdot 7 - 6^2 = 14 - 36 = -22 < 0.$

Since the determinant is negative, $H$ is indefinite — one eigenvalue is positive, the other negative. The corresponding surface is a saddle, not a bowl.

The student is confusing entries with pivots/eigenvalues. Always check the determinant condition.

Question 3

For the function $f(x, y) = 2x^2 + 12xy + 7y^2$ , compute the Hessian at the origin and classify the critical point.

What is the determinant of $H$ , and what type of critical point does the origin represent?

✅ Correct! $\det H = 56 - 144 = -88 < 0$ , so it's a saddle.

❌ Not quite. Be careful with the sign: $\det H = f_{xx}f_{yy} - f_{xy}^2 = 4(14) - 12^2$ . Is the cross term big enough to flip the sign?

Show solution

Solution:

Compute the second partials:

$f_{xx} = 4$
$f_{yy} = 14$
$f_{xy} = 12$

So $H = \begin{bmatrix} 4 & 12 \\ 12 & 14 \end{bmatrix}.$

Determinant: $\det H = 4 \cdot 14 - 12^2 = 56 - 144 = -88.$

Since $\det H < 0$ , $H$ is indefinite and the origin is a saddle point. The surface curves up along one eigenvector direction and down along another.

Question 4

You slice the graph $z = f(x, y)$ of a quadratic with positive definite Hessian by the horizontal plane $z = 1$ . What shape is the resulting curve in the $xy$ -plane?

✅ Correct! Positive definite → closed, bounded level sets → ellipses with axes along the eigenvectors of $H$ .

❌ Not quite. Positive definite means the surface curves up in every direction. What kind of horizontal cross-section does a bowl have?

Show solution

Solution:

A positive definite Hessian means the surface is a bowl curving upward in every direction. Slicing a bowl horizontally gives a closed curve — specifically, an ellipse.

The axes of the ellipse line up with the eigenvectors of $H$ .
The semi-axis lengths are $1/\sqrt{\lambda_i}$ where $\lambda_i$ are the eigenvalues.

By contrast:

An indefinite Hessian (saddle) slices to a hyperbola.
A degenerate Hessian can slice to parallel lines or a parabola.

The shape of the level set is the geometric signature of the matrix.

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