Quadratic forms — positive definite, negative definite, indefinite: Exercises

4.5.39 · D4 · Maths › Linear Algebra (Full) › Quadratic forms — positive definite, negative definite, inde

Shuru karne se pehle, ek reference figure hai jo paanch shapes tumhare dimaag mein fix kar deta hai — har classification neeche basically yahi pooch rahi hai "inme se konsi surface hai?"

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

Intuition Reference figure padhna (har colour ek definiteness type hai)

Figure mein paanch surfaces $z = Q (x, y)$ dikhaye gaye hain, har ek apne colour aur type dono ke saath titled hai, origin par ek black dot hai (critical point). Colour → type aise match karo:

Blue bowl — positive definite (PD). Neeche se, har direction upar jaati hai, isliye $Q > 0$ iske around. (Socho "valley".)
Orange dome — negative definite (ND). Upar se, har direction neeche jaati hai, isliye $Q < 0$ . (Socho "hilltop".)
Red saddle — indefinite. Ek direction upar jaati hai, perpendicular wali neeche, isliye $Q$ dono signs leta hai.
Green trough (flat-bottomed) — positive semidefinite (PSD). Ek direction mein bowl jaisa hai lekin doosri direction mein bilkul flat hai — flat line zero-eigenvalue direction hai, jahan $x \neq = 0$ ke liye $Q = 0$ hota hai.
Gray inverted trough (flat-topped) — negative semidefinite (NSD). Ek direction mein dome, doosri mein flat; green trough ka mirror image. Jab bhi koi solution neeche "blue bowl / orange dome / red saddle / green trough / gray trough" kahe, woh is labelled picture ki taraf point kar raha hai.

Level 1 — Recognition

Yahan $A$ pehle se diagonal hai, isliye $Q = λ_{1} x^{2} + λ_{2} y^{2} + \dots$ aur diagonal entries hi eigenvalues hain. Bas signs padh lo.

Recall Solution L1.1

WHAT: matrix hai $A = (5003)$ , diagonal hai, isliye eigenvalues hain $5$ aur $3$ . WHY: ek diagonal matrix ke liye koi cross terms nahi hote, isliye diagonal numbers exactly $λ_{i}$ hain $\sum λ_{i} y_{i}^{2}$ mein. Dono $> 0$ hain ⇒ har input positive output deta hai (jab tak $x = 0$ na ho). Positive definite (blue bowl).

Recall Solution L1.2

$A = (- 2 0 0 - 7)$ , eigenvalues $- 2, - 7$ , dono $< 0$ . Har output $x \neq = 0$ ke liye negative hai. Negative definite (orange dome).

Recall Solution L1.3

$λ = 1, - 4$ : opposite signs hain. $(1, 0)$ daalo → $+ 1$ ; $(0, 1)$ daalo → $- 4$ . Dono signs aate hain. Indefinite (figure mein red saddle).

Recall Solution L1.4

$A = (0006)$ , eigenvalues $0, 6$ . Kabhi negative nahi, lekin $Q (1, 0) = 0$ jabki $x \neq = 0$ . Positive semidefinite (PD nahi) — figure mein yeh green flat-bottomed trough hai. Zero eigenvalue hi asli baat hai: yeh flat direction $x$ -axis hai.

Recall Solution L1.5

$A = (- 5 0 00)$ , eigenvalues $- 5, 0$ . Kabhi positive nahi (dono eigenvalues $\leq 0$ hain), lekin $Q (0, 1) = 0$ jabki $x \neq = 0$ . Negative semidefinite (NSD) — gray flat-topped trough. Yeh L1.4 ka mirror hai: saare eigenvalues $\leq 0$ hain kam se kam ek zero ke saath. Surface ek downward trough hai jo poori $y$ -axis ke along flat hai — $x$ -direction mein dome, $y$ -direction mein level.

Level 2 — Application

Ab matrices mein cross terms hain. $A$ ko sahi build karna pehla kaam hai.

Phir Sylvester's leading principal minors use karo ( $D_{1} = A_{11}$ , $D_{2} = det A$ ) aur eigenvalues se cross-check karo.

Recall Solution L2.1

$A$ build karo: diagonals $3, 3$ ; $x y$ -coefficient $2$ hai, aur (jaisa upar dikhaya) off-diagonal uska half hota hai kyunki cross term do baar count hota hai, isliye $A_{12} = A_{21} = \frac{2}{2} = 1$ . Toh $A = (3113)$ . Sylvester (WHY yeh PD certify karta hai): upar wale pivot fact se, $D_{1} > 0$ aur $D_{2} > 0$ milkar matlab hai dono pivots $d_{1} = D_{1} = 3$ aur $d_{2} = D_{2} / D_{1} = 8/3$ positive hain, isliye $Q$ positive squares ka sum hai — PD ki definition yahi hai. Compute karo: $D_{1} = 3 > 0$ ; $D_{2} = 9 - 1 = 8 > 0$ . Sab positive ⇒ PD. Eigenvalues se cross-check: $(3 - λ)^{2} - 1 = 0 \Rightarrow λ = 2, 4$ , dono $> 0$ . ✓

Recall Solution L2.2

$A$ build karo: diagonals $- 4, - 4$ ; $x y$ -coefficient $4$ hai, halved (do baar count hota hai) ⇒ off-diagonal $\frac{4}{2} = 2$ . Toh $A = (- 4 2 2 - 4)$ . ND ke liye Sylvester — WHY pattern $D_{1} < 0, D_{2} > 0$ hota hai: $A$ ND hai exactly jab $- A$ PD ho. PD maangta hai ki $- A$ ke saare leading minors positive hon. Lekin ek $k \times k$ block ko $- 1$ se scale karne par uska determinant $(- 1)^{k}$ se multiply hota hai, isliye $- A$ ka $k$ -th minor $(- 1)^{k} D_{k}$ ke barabar hai. Har $(- 1)^{k} D_{k} > 0$ demand karne par milta hai $D_{1} < 0$ ( $k = 1$ se: $- D_{1} > 0$ ) aur $D_{2} > 0$ ( $k = 2$ se: $D_{2} > 0$ ) — yeh alternating pattern hai. Check karo: $D_{1} = - 4 < 0$ ✓; $D_{2} = 16 - 4 = 12 > 0$ ✓. ND. Eigenvalues: $(- 4 - λ)^{2} - 4 = 0 \Rightarrow - 4 - λ = \pm 2 \Rightarrow λ = - 2, - 6$ , dono $< 0$ . ✓

Recall Solution L2.3

$A$ build karo: diagonals $1, 1$ ; $x y$ -coefficient $6$ , halved ⇒ off-diagonal $3$ . Toh $A = (1331)$ . $D_{1} = 1 > 0$ lekin $D_{2} = 1 - 9 = - 8 < 0$ . PD nahi hai (iske liye $D_{2} > 0$ chahiye), aur $D_{1} < 0$ bhi fail hota hai ND pattern ke liye. Ek negative $D_{2} = det A = λ_{1} λ_{2}$ ka matlab hai eigenvalues ke signs opposite hain. Indefinite. Check: $λ = 4, - 2$ . ✓

Recall Solution L2.4

$A$ build karo: diagonals $2, 2, 2$ ; sirf cross term $2 x y$ hai, halved ⇒ $A_{12} = A_{21} = 1$ ; koi $x z$ ya $y z$ nahi, isliye woh off-diagonals $0$ hain. Toh $A = 210120002$ . $D_{1} = 2 > 0$ ; $D_{2} = det (2112) = 3 > 0$ ; $D_{3} = det A$ . WHY $D_{3} = 2 \cdot D_{2}$ (last column ke along cofactor expansion): ek determinant ko kisi bhi row ya column ke along expand kiya ja sakta hai — tum us line par chalte ho, har entry ko uski row aur column delete karne ke baad bache chhote determinant (uske "minor") se multiply karte ho, aur checkerboard sign $(- 1)^{i + j}$ lagate ho. Teesra column yahan $(0, 0, 2)^{⊤}$ hai, isliye sirf ek term bachta hai yaani position $(3, 3)$ mein $2$ : sign $(- 1)^{3 + 3} = + 1$ , aur row 3 & column 3 delete karne par exactly top-left $2 \times 2$ block bachta hai jiska determinant $D_{2}$ hai. Isliye $D_{3} = 2 \cdot D_{2} = 2 \cdot 3 = 6 > 0$ . Teeno $> 0$ hain ⇒ PD.

Level 3 — Analysis

Ab $A$ mein ek parameter hai. Parameter ki wo range dhundho jis par required type mile. Yahan Sylvester apni value dikhata hai.

Recall Solution L3.1

Sylvester se PD: chahiye $D_{1} = 2 > 0$ (hamesha true) aur $D_{2} = 16 - k^{2} > 0$ . $16 - k^{2} > 0 ⟺ k^{2} < 16 ⟺ - 4 < k < 4$ . Answer: $- 4 < k < 4$ . $k = \pm 4$ par, $D_{2} = 0$ ⇒ PSD boundary; iske baad, indefinite.

Recall Solution L3.2

$A$ build karo: $x y$ -coefficient $2 k$ hai, halved ⇒ off-diagonal $k$ . Toh $A = (1 k k 4)$ , $D_{2} = 4 - k^{2} = λ_{1} λ_{2}$ . Indefinite ⇔ eigenvalues opposite signs ⇔ $D_{2} < 0$ : $4 - k^{2} < 0 ⟺ k^{2} > 4 ⟺ k < - 2 or k > 2$ . Answer: $∣ k ∣ > 2$ .

Recall Solution L3.3

ND ke liye chahiye $D_{1} = k < 0$ aur $D_{2} = k^{2} - 1 > 0$ (L2.2 wala alternating pattern). $k^{2} - 1 > 0 ⟺ k < - 1$ ya $k > 1$ . $k < 0$ se intersect karo: $k < - 1$ . Answer: $k < - 1$ . (Eigenvalues $k \pm 1$ , dono negative exactly jab $k < - 1$ . ✓)

Level 4 — Synthesis

Form ko uske origin (ek formula, ek Hessian, ek completed square) ke saath combine karo. Ek problem mein multiple tools.

Recall Solution L4.1

Step 1 — critical point check. $f_{x} = 3 x^{2} - 3 = 0 \Rightarrow x = \pm 1$ ; $f_{y} = 2 y = 0 \Rightarrow y = 0$ . Toh $(1, 0)$ critical hai. ✓ Step 2 — Hessian. $f_{xx} = 6 x$ , $f_{y y} = 2$ , $f_{x y} = 0$ . $(1, 0)$ par: $H = (6002)$ . Step 3 — $H$ classify karo. Eigenvalues $6, 2 > 0$ ⇒ PD ⇒ local minimum (blue bowl). Bonus — doosra critical point $(- 1, 0)$ : $H = (- 6 0 02)$ , $λ = - 6, 2$ mixed ⇒ indefinite ⇒ saddle (red surface). Yeh poora Second Derivative Test hai.

Recall Solution L4.2

WHAT ( $x$ mein complete the square karo): $x^{2} + 4 x y + 5 y^{2} = (x + 2 y)^{2} - 4 y^{2} + 5 y^{2} = (x + 2 y)^{2} + y^{2} .$ WHY yeh PD prove karta hai: do squares ka sum $\geq 0$ hota hai, aur $0$ sirf tab hota hai jab $x + 2 y = 0$ aur $y = 0$ , yaani $x = y = 0$ . Toh $Q > 0$ sabhi $x \neq = 0$ ke liye ⇒ PD. Implied eigenvalue signs: PD ⇒ dono eigenvalues $> 0$ (indeed $D_{1} = 1 > 0$ , $D_{2} = 5 - 4 = 1 > 0$ , toh $λ_{1} λ_{2} = 1 > 0$ aur $λ_{1} + λ_{2} = 6 > 0$ : dono positive hain). Yeh completing-the-square factorisation exactly woh idea hai jo Cholesky Decomposition ke peechhe hai — PD precisely woh condition hai jo tumhe $A = L L^{⊤}$ likhne deti hai.

Recall Solution L4.3

$D_{1} = 9 > 0$ hamesha. Switch tab hota hai jab $D_{2} = 9 k - 36$ sign change kare: $9 k - 36 = 0 \Rightarrow k = 4$ .

$k > 4$ : $D_{2} > 0$ ⇒ PD.
$k < 4$ : $D_{2} < 0$ ⇒ indefinite.
$k = 4$ par: $D_{2} = det A = 0$ ⇒ ek eigenvalue $0$ hai, doosra ( $trace = 13 > 0$ ) positive hai ⇒ PSD. Knife-edge value $k = 4$ hai, wahan type = PSD.

Level 5 — Mastery

Structural facts prove karo. Yeh "tests sahi kyun hain" wali problems hain.

Recall Solution L5.1

Step 1 — diagonal par rotate karo. Spectral Theorem se, $A = Q Λ Q^{⊤}$ . $y = Q^{⊤} x$ set karo. Kyunki $Q$ orthogonal hai toh length preserve hoti hai, isliye $∥ y ∥ = ∥ x ∥ = 1$ , matlab $\sum y_{i}^{2} = 1$ . Step 2 — rewrite karo. $Q (x) = \sum_{i} λ_{i} y_{i}^{2}$ . Yeh eigenvalues ka ek weighted average hai weights $y_{i}^{2} \geq 0$ ke saath jo $1$ mein sum hote hain. Step 3 — average bound karo. Koi bhi weighted average sabse chhote aur sabse bade ingredient ke beech hota hai: $λ_{m i n} = λ_{m i n} \sum y_{i}^{2} \leq \sum λ_{i} y_{i}^{2} \leq λ_{m a x} \sum y_{i}^{2} = λ_{m a x} .$ Step 4 — definiteness follow karta hai. Agar $λ_{m i n} > 0$ toh $Q (x) \geq λ_{m i n} > 0$ sabhi unit $x$ ke liye (hence scaling se sabhi $x \neq = 0$ ke liye) ⇒ PD. Agar $λ_{m a x} < 0$ ⇒ ND. Agar $λ_{m i n} < 0 < λ_{m a x}$ , bounds zero ke dono taraf hain aur eigenvector directions dono signs realize karte hain ⇒ indefinite. $■$

Recall Solution L5.2

WHAT humein dikhana hai: $x^{⊤} B x > 0$ sabhi $x \neq = 0$ ke liye. Step 1 — $B = C^{⊤} A C$ substitute karo aur regroup karo: $x^{⊤} B x = x^{⊤} C^{⊤} A C x = (C x)^{⊤} A (C x) .$ Humne $(C x)^{⊤} = x^{⊤} C^{⊤}$ use kiya do outer factors ko ek vector mein fold karne ke liye. Step 2 — inner vector name karo. $w = C x$ lo. Tab $x^{⊤} B x = w^{⊤} A w .$ Step 3 — WHY invertibility matter karti hai. Humne $x \neq = 0$ se start kiya. Kyunki $C$ invertible hai, $C x = 0$ force karega $x = C^{- 1} 0 = 0$ , jo contradiction hai. Isliye $w = C x \neq = 0$ . Step 4 — $A$ ki PD apply karo. Kyunki $A$ PD hai aur $w \neq = 0$ , hamein milta hai $w^{⊤} A w > 0$ . Isliye $x^{⊤} B x = w^{⊤} A w > 0 for all x \neq = 0 .$ Toh $B$ positive definite hai. $■$ Yahi reason hai ki variables ka change (jaisa Least Squares mein, jahan normal-equation matrix $C^{⊤} C$ hoti hai) ek form ko positive definite rakhta hai.

Recall Solution L5.3

Sylvester: $D_{1} = 2 > 0$ . $D_{2} = det (2 - 1 - 1 2) = 4 - 1 = 3 > 0$ . $D_{3} = det A$ : top row ke along expand karo $= 2 (4 - 1) - (- 1) (- 2 - 0) = 6 - 2 = 4 > 0$ . Sab $> 0$ ⇒ PD. Eigenvalues: is tridiagonal matrix ke liye woh $2 - 2 cos \frac{k π}{4}$ hain $k = 1, 2, 3$ ke liye, yaani $2 - 2, 2, 2 + 2$ — sab positive hain. ✓ PD confirmed.