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

4.5.39 · D3 · Maths › Linear Algebra (Full) › Quadratic forms — positive definite, negative definite, inde

Hum poori note mein do tools use karte hain — eigenvalue test (via Spectral Theorem for Symmetric Matrices) aur Sylvester's determinant test. Hum Hessian Matrix / Second Derivative Test payoff ke saath close karte hain.

The scenario matrix

Har $2 \times 2$ symmetric quadratic form neeche diye ek hi row mein aata hai. Columns tumhe batate hain ki do tests ko kya kehna chahiye, taaki tum khud check kar sako.

#	Cell (kya khaas hai isme)	Eigenvalue signs	Sylvester $D_{1}, D_{2}$	Type
A	Dono dials upar	$+, +$	$+, +$	PD
B	Dono dials neeche	$-, -$	$-, +$	ND
C	Ek upar, ek neeche	$+, -$	any, $-$	Indefinite
D	Ek dial flat (up branch)	$+, 0$	$+, 0$	PSD (PD nahi)
E	Ek dial flat (down branch)	$-, 0$	$-, 0$	NSD (ND nahi)
F	Dono dials flat — degenerate zero machine	$0, 0$	$0, 0$	PSD bhi & NSD bhi (sirf zero output)
G	Limiting family — ek parameter PD → PSD → indefinite tip karta hai	$0$ cross karta hai	$D_{2}$ crosses $0$	transitions
H	Trap: " $D_{1} > 0$ toh PD hai" (Sylvester misread)	—	$+, -$	Indefinite (PD nahi)
I	Word problem — real Hessian, ek cost minimize karo	$+, +$	$+, +$	PD → minimum
J	Exam twist — $3 \times 3$ , mixed	mixed	—	Indefinite

Neeche ke examples A–J sab ko cover karte hain.

Example 1 — Cell A (Positive definite)

Worked example Cell A: dono dials upar

$Q (x, y) = 5 x^{2} + 4 x y + 2 y^{2}$ ko classify karo.

Forecast: Saare coefficients positive hain aur cross term chhota hai. Guess: PD (ek bowl). Aage padhne se pehle apna guess likh lo.

$A$ banao. Diagonals square coefficients $5$ aur $2$ hain; off-diagonal cross coefficient ka aadha hai, $4/2 = 2$ . $A = (5222) .$ Yeh step kyun? Tests symmetric matrix pe kaam karte hain, raw formula pe nahi.
Sylvester. $D_{1} = 5 > 0$ , aur $D_{2} = det A = 5 \cdot 2 - 2 \cdot 2 = 6 > 0$ . Yeh step kyun? PD $⟺$ har leading minor positive ho — sabse fast haath se check.
Dono positive ⇒ PD.

Verify (eigenvalues, aur sum of squares): $λ$ solve karo $(5 - λ) (2 - λ) - 4 = 0 \Rightarrow λ^{2} - 7 λ + 6 = 0 \Rightarrow λ = 1, 6$ — dono $> 0$ ✓. Square bhi complete karo: $5 x^{2} + 4 x y + 2 y^{2} = 5 (x + \frac{2}{5} y)^{2} + \frac{6}{5} y^{2} > 0$ jab tak $x = y = 0$ na ho. Ek genuine sum of positive squares. ✓

Example 2 — Cell B (Negative definite)

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

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

Saddle dekho: pale-yellow ridge (upar jaata hua) $λ = 3$ eigendirection hai, chalk-blue valley (neeche jaata hua) $λ = - 1$ direction hai. Center pe khade ho, ek taraf upar jaate ho aur doosri taraf neeche — yahi "indefinite" dikhta hai.

Example 4 — Cell D (PSD, PD nahi)

Example 5 — Cell E (NSD, ND nahi)

Example 6 — Cell F (Degenerate zero machine)

Example 7 — Cell G (Limiting family: ek parameter type tip karta hai) — with figure

Worked example Cell G: dekho definiteness continuously kaise change hoti hai

Ek parameterized machine: $Q_{t} (x, y) = x^{2} + 2 t x y + y^{2}$ ek real dial $t$ ke liye. Woh values of $t$ dhundho jahan type change hoti hai.

Forecast: $t = 0$ pe yeh $x^{2} + y^{2}$ hai (PD). Bade $t$ pe cross term dominate karta hai (indefinite, jaise Example 3). Guess: $t = 0$ ke paas PD, aur kahin tip karta hai.

$A$ banao: $A = (1 t t 1)$ .
$D_{2}$ track karo. $D_{1} = 1 > 0$ (fixed), $D_{2} = 1 - t^{2}$ . Yahi switch hai. Yeh step kyun? $D_{1}$ kabhi sign nahi badalta, toh poori kahani $D_{2} = 1 - t^{2}$ ke sign mein hai.
Teen regimes padho:
- $∣ t ∣ < 1$ : $D_{2} > 0$ with $D_{1} > 0$ ⇒ PD.
- $∣ t ∣ = 1$ : $D_{2} = 0$ ⇒ borderline PSD (yeh Example 4 ki matrix hai jab $t = - 1$ , aur $(x + y)^{2}$ jab $t = 1$ ).
- $∣ t ∣ > 1$ : $D_{2} < 0$ ⇒ indefinite.
Eigenvalue confirmation. $λ = 1 \pm t$ . Dono positive $⟺ - 1 < t < 1$ ; ek $t = \pm 1$ pe zero hit karta hai; $∣ t ∣ > 1$ pe opposite signs. Same story.

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

Figure do eigenvalues $1 + t$ (pale yellow) aur $1 - t$ (chalk blue) ko dial $t$ ke against plot karta hai. PD woh band hai jahan dono curves axis ke upar hain; $t = \pm 1$ pe ek curve zero ko touch karti hai (PSD); uske baahir, ek curve neeche cross karti hai → indefinite. Definiteness literally yeh hai: "kya dono curves line ke upar hain?"

Example 8 — Cell H (Sylvester trap)

Worked example Cell H: "

$D_{1} > 0$ , toh PD hona chahiye" — isse rokho $Q (x, y) = x^{2} + 6 x y + y^{2}$ ko classify karo.

Forecast: Ek tempting-but-wrong reflex: " $D_{1} = 1 > 0$ , positive-looking squares, PD bol do." Decide karne se pehle khud ko $D_{2}$ compute karne par majboor karo.

$A$ banao: diagonals $1, 1$ ; off-diagonal $6/2 = 3$ . $A = (1331) .$
DONO minors compute karo. $D_{1} = 1 > 0$ , lekin $D_{2} = 1 - 9 = - 8 < 0$ . Yeh step kyun? PD ko har $D_{k} > 0$ chahiye. Ek positive minor apne aap kuch prove nahi karta.
Negative $D_{2}$ ⇒ indefinite (yeh PD nahi hai, aur ND alternating pattern bhi nahi hai).

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: ek real cost minimize karna)

Worked example Cell I: kya yeh cost surface genuinely minimum hai?

Ek factory ki daily cost (thousands mein) do throughput dials $x, y$ ke function ke roop mein hai: $f (x, y) = 3 x^{2} - 2 x y + 3 y^{2} - 8 x - 4 y + 20.$ Tumne $\nabla f = 0$ solve kiya aur ek critical point mila. Kya yeh minimum hai, maximum hai, ya saddle?

Forecast: Purely quadratic part $3 x^{2} - 2 x y + 3 y^{2}$ hai, bade positive squares, chhota cross term. Guess: ek real minimum (bowl-shaped cost).

Hessian yahan constant hai. $\frac{\partial ^{2} f}{\partial x ^{2}} = 6, \frac{\partial ^{2} f}{\partial y ^{2}} = 6, \frac{\partial ^{2} f}{\partial x \partial y} = - 2$ , toh $H = (6 - 2 - 2 6) .$ Yeh step kyun? Second Derivative Test ek critical point ko Hessian quadratic form ki definiteness se classify karta hai.
$H$ ko Sylvester se test karo. $D_{1} = 6 > 0$ , $D_{2} = 36 - 4 = 32 > 0$ ⇒ PD.
PD Hessian ⇒ local minimum (bowl). Toh haan — cost wahan sach mein bottom out karti hai.

Verify (locate karo, units check): $\nabla f = 0$ solve karo: $6 x - 2 y = 8$ aur $- 2 x + 6 y = 4$ ⇒ $x = \frac{7}{4}, y = \frac{5}{4}$ . $H$ ke eigenvalues: $λ = 6 \pm 2 = 8, 4$ , dono $> 0$ ✓ PD. Cost ki units thousands hain aur $H$ ki entries (thousands)/(dial $^{2}$ ) hain, consistent — har direction mein positive curvature matlab har deviation cost badhata hai ✓.

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

Worked example Cell J: teen variables, panic mat karo

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

Forecast: $- z^{2}$ ek downward direction ki smell deta hai jabki $x, y$ upar lagte hain. Guess: indefinite — lekin check karo ki $2 x y$ $x, y$ block ko toh nahi bigaad raha.

$3 \times 3$ $A$ banao: diagonals $1, 1, - 1$ ; ek hi cross term $2 x y$ hai toh $A_{12} = A_{21} = 1$ , baaki saare off-diagonals $0$ . $A = 110110 00 - 1 .$
Block structure ⇒ eigenvalues easy hain. Top-left $2 \times 2$ block $(1111)$ ke eigenvalues $0$ aur $2$ hain; isolated $- 1$ se $λ = - 1$ milta hai. Toh spectrum ${2, 0, - 1}$ hai. Yeh step kyun? Block-diagonal matrix ke eigenvalues sirf blocks ke eigenvalues ka union hote hain — koi messy cubic nahi chahiye.
Mixed signs ( $+ 2$ aur $- 1$ ) ⇒ indefinite. Extra zero eigenvalue ka matlab yeh bhi hai ki yeh kisi bhi case mein strictly definite nahi hai.

Verify: witnesses — $Q (1, 1, 0) = 1 + 1 + 2 = 4 > 0$ ; $Q (0, 0, 1) = - 1 < 0$ . Dono signs, plus ek null direction $(1, - 1, 0)$ jisme $Q = 1 + 1 - 2 = 0$ . Indefinite confirm ✓.

Recall

Recall Poore matrix mein quick self-test

Kis cell mein $D_{1} > 0, D_{2} = 0$ aur eigenvalues $0, +$ hain? ::: Cell D — PSD, PD nahi. Kaun sa ek cell PSD aur NSD dono ek saath hai? ::: Cell F — zero machine (sirf zero output). Family $x^{2} + 2 t x y + y^{2}$ mein, kis $t$ pe PD khatam hota hai? ::: $∣ t ∣ = 1$ (wahan $D_{2} = 1 - t^{2} = 0$ ). Akela $D_{1} > 0$ PD ke liye kyun ENOUGH nahi hai? ::: PD ko poori chain $D_{1} > 0, D_{2} > 0, \dots$ chahiye; ek minor positive ho sakta hai jabki $D_{2} < 0$ (indefinite) — Cell H trap. Critical point pe PD Hessian kya signal karta hai? ::: Ek local minimum (bowl).