Quadratic forms — positive definite, negative definite, indefinite: Worked examples

We lean on two tools throughout — the eigenvalue test (via the Spectral Theorem for Symmetric Matrices) and Sylvester's determinant test. We close with the Hessian Matrix / Second Derivative Test payoff.

The scenario matrix

Every $2 \times 2$ symmetric quadratic form lands in exactly one row below. The columns tell you what the two tests should say, so you can check yourself.

#	Cell (what makes it special)	Eigenvalue signs	Sylvester $D_{1}, D_{2}$	Type
A	Both dials up	$+, +$	$+, +$	PD
B	Both dials down	$-, -$	$-, +$	ND
C	One up, one down	$+, -$	any, $-$	Indefinite
D	One dial flat (up branch)	$+, 0$	$+, 0$	PSD (not PD)
E	One dial flat (down branch)	$-, 0$	$-, 0$	NSD (not ND)
F	Both dials flat — degenerate zero machine	$0, 0$	$0, 0$	Both PSD & NSD (only zero output)
G	Limiting family — a parameter tips PD → PSD → indefinite	crosses $0$	$D_{2}$ crosses $0$	transitions
H	Trap: " $D_{1} > 0$ so it's PD" (Sylvester misread)	—	$+, -$	Indefinite (not PD)
I	Word problem — real Hessian, minimize a cost	$+, +$	$+, +$	PD → minimum
J	Exam twist — $3 \times 3$ , mixed	mixed	—	Indefinite

The examples below hit every one of A–J.

Example 1 — Cell A (Positive definite)

Example 2 — Cell B (Negative definite)

Example 3 — Cell C (Indefinite saddle) — with figure

Figure — Quadratic forms — positive definite, negative definite, indefinite

Look at the saddle: the pale-yellow ridge (going up) is the $λ = 3$ eigendirection, the chalk-blue valley (going down) is the $λ = - 1$ direction. Standing at the center, you go up one way and down the other — that is what "indefinite" looks like.

Example 4 — Cell D (PSD, not PD)

Example 5 — Cell E (NSD, not ND)

Example 6 — Cell F (Degenerate zero machine)

Example 7 — Cell G (Limiting family: one parameter tips the type) — with figure

Worked example Cell G: watch definiteness change continuously

A parameterized machine: $Q_{t} (x, y) = x^{2} + 2 t x y + y^{2}$ for a real dial $t$ . Find the values of $t$ where the type changes.

Forecast: At $t = 0$ it's $x^{2} + y^{2}$ (PD). For huge $t$ the cross term dominates (indefinite, like Example 3). Guess: PD near $t = 0$ , and it tips somewhere.

Build $A$ : $A = (1 t t 1)$ .
Track $D_{2}$ . $D_{1} = 1 > 0$ (fixed), $D_{2} = 1 - t^{2}$ . This is the switch. Why this step? $D_{1}$ never changes sign, so the whole story lives in the sign of $D_{2} = 1 - t^{2}$ .
Read the three regimes:
- $∣ t ∣ < 1$ : $D_{2} > 0$ with $D_{1} > 0$ ⇒ PD.
- $∣ t ∣ = 1$ : $D_{2} = 0$ ⇒ borderline PSD (this is Example 4's matrix when $t = - 1$ , and $(x + y)^{2}$ when $t = 1$ ).
- $∣ t ∣ > 1$ : $D_{2} < 0$ ⇒ indefinite.
Eigenvalue confirmation. $λ = 1 \pm t$ . Both positive $⟺ - 1 < t < 1$ ; one hits zero at $t = \pm 1$ ; opposite signs when $∣ t ∣ > 1$ . Same story.

Verify: at $t = \frac{1}{2}$ , $λ = 1.5, 0.5$ (PD) ✓; at $t = 1$ , $λ = 2, 0$ (PSD) ✓; at $t = 2$ , $λ = 3, - 1$ (indefinite) ✓.

The figure plots the two eigenvalues $1 + t$ (pale yellow) and $1 - t$ (chalk blue) against the dial $t$ . PD is the band where both curves sit above the axis; at $t = \pm 1$ one curve touches zero (PSD); outside that, one curve crosses below → indefinite. Definiteness is literally "are both curves above the line?"

Example 8 — Cell H (The Sylvester trap)

Worked example Cell H: "

$D_{1} > 0$ , so it must be PD" — resist it Classify $Q (x, y) = x^{2} + 6 x y + y^{2}$ .

Forecast: A tempting-but-wrong reflex: " $D_{1} = 1 > 0$ , positive-looking squares, call it PD." Force yourself to compute $D_{2}$ before deciding.

Build $A$ : diagonals $1, 1$ ; off-diagonal $6/2 = 3$ . $A = (1331) .$
Compute BOTH minors. $D_{1} = 1 > 0$ , but $D_{2} = 1 - 9 = - 8 < 0$ . Why this step? PD needs every $D_{k} > 0$ . One positive minor proves nothing on its own.
A negative $D_{2}$ ⇒ indefinite (it's not PD, and it's not the alternating ND pattern either).

Verify: eigenvalues $λ = 1 \pm 3 = 4, - 2$ — mixed signs, indefinite ✓. Witness: $Q (1, 1) = 8 > 0$ , $Q (1, - 1) = 1 - 6 + 1 = - 4 < 0$ ✓.

Example 9 — Cell I (Word problem: minimizing a real cost)

Worked example Cell I: is this cost surface a genuine minimum?

A factory's daily cost (in thousands) as a function of two throughput dials $x, y$ is $f (x, y) = 3 x^{2} - 2 x y + 3 y^{2} - 8 x - 4 y + 20.$ You solved $\nabla f = 0$ and found a critical point. Is it a minimum, a maximum, or a saddle?

Forecast: The purely quadratic part is $3 x^{2} - 2 x y + 3 y^{2}$ , big positive squares, tiny cross term. Guess: a real minimum (bowl-shaped cost).

The Hessian is constant here. $\frac{\partial ^{2} f}{\partial x ^{2}} = 6, \frac{\partial ^{2} f}{\partial y ^{2}} = 6, \frac{\partial ^{2} f}{\partial x \partial y} = - 2$ , so $H = (6 - 2 - 2 6) .$ Why this step? The Second Derivative Test classifies a critical point by the definiteness of the Hessian quadratic form.
Test $H$ with Sylvester. $D_{1} = 6 > 0$ , $D_{2} = 36 - 4 = 32 > 0$ ⇒ PD.
PD Hessian ⇒ local minimum (bowl). So yes — cost really bottoms out there.

Verify (locate it, units check): solving $\nabla f = 0$ : $6 x - 2 y = 8$ and $- 2 x + 6 y = 4$ ⇒ $x = \frac{7}{4}, y = \frac{5}{4}$ . Eigenvalues of $H$ : $λ = 6 \pm 2 = 8, 4$ , both $> 0$ ✓ PD. Cost has units of thousands and $H$ 's entries are (thousands)/(dial $^{2}$ ), consistent — a positive curvature in every direction means every deviation raises cost ✓.

Example 10 — Cell J (Exam twist: a $3 \times 3$ )

Worked example Cell J: three variables, don't panic

Classify $Q (x, y, z) = x^{2} + y^{2} - z^{2} + 2 x y$ .

Forecast: The $- z^{2}$ smells like a downward direction while $x, y$ look upward. Guess: indefinite — but check whether the $2 x y$ ruins the $x, y$ block.

Build the $3 \times 3$ $A$ : diagonals $1, 1, - 1$ ; the only cross term is $2 x y$ so $A_{12} = A_{21} = 1$ , all other off-diagonals $0$ . $A = 110110 00 - 1 .$
Block structure ⇒ eigenvalues are easy. The top-left $2 \times 2$ block $(1111)$ has eigenvalues $0$ and $2$ ; the isolated $- 1$ gives $λ = - 1$ . So the spectrum is ${2, 0, - 1}$ . Why this step? A block-diagonal matrix's eigenvalues are just the union of the blocks' eigenvalues — no messy cubic needed.
Mixed signs (a $+ 2$ and a $- 1$ ) ⇒ indefinite. The extra zero eigenvalue also means it is not strictly definite in any case.

Verify: witnesses — $Q (1, 1, 0) = 1 + 1 + 2 = 4 > 0$ ; $Q (0, 0, 1) = - 1 < 0$ . Both signs, plus a null direction $(1, - 1, 0)$ giving $Q = 1 + 1 - 2 = 0$ . Indefinite confirmed ✓.

Recall

Recall Quick self-test across the matrix

Which cell has $D_{1} > 0, D_{2} = 0$ and eigenvalues $0, +$ ? ::: Cell D — PSD, not PD. Which single cell is simultaneously PSD and NSD? ::: Cell F — the zero machine (only zero output). In the family $x^{2} + 2 t x y + y^{2}$ , at what $t$ does PD end? ::: $∣ t ∣ = 1$ (there $D_{2} = 1 - t^{2} = 0$ ). Why is $D_{1} > 0$ alone NOT enough for PD? ::: PD needs the whole chain $D_{1} > 0, D_{2} > 0, \dots$ ; one minor can be positive while $D_{2} < 0$ (indefinite) — the Cell H trap. A PD Hessian at a critical point signals what? ::: A local minimum (bowl).