Quadratic forms — positive definite, negative definite, indefinite: Question bank

4.5.39 · D5 · Maths › Linear Algebra (Full) › Quadratic forms — positive definite, negative definite, inde

Prerequisites jo open rakhne worth hain: Eigenvalues and Eigenvectors, Symmetric Matrices, Spectral Theorem, Determinants, Hessian Matrix, Second Derivative Test.

Error: kisi nonzero vector par zero hit karna bina kabhi negative hue matlab positive semidefinite hai, indefinite nahi. Indefinite ke liye strictly positive aur strictly negative dono outputs chahiye.

A

symmetric hai

λ = 2, 0, 3

ke saath, isliye yeh positive definite hai."

Error: zero eigenvalue strictness khatam kar deta hai —

Q = 2 y_{1}^{2} + 0 \cdot y_{2}^{2} + 3 y_{3}^{2}

doosre eigenvector ke direction mein

0

ke equal hota hai. Yeh positive semidefinite hai.

"Sylvester's criterion kehta hai

D_{1} \geq 0, \dots, D_{n} \geq 0

PSD prove karta hai."

Error: leading-minor test cleanly sirf strict PD (

> 0

) certify karta hai. Semidefiniteness ke liye saare principal minors chahiye (har same-index diagonal sub-block), sirf leading (top-left) wale nahi —

(00 0 - 1)

D_{1} = D_{2} = 0

hai phir bhi negative semidefinite (NSD) hai.

Q = x^{2} - y^{2}

D_{1} = 1 > 0

hai, isliye yeh positive definite lagta hai shuru mein."

Error:

D_{2} = - 1 < 0

turant PD ko rule out karta hai, aur negative

D_{2}

(bina zeros ke) matlab indefinite hai; kuch bhi conclude karne se pehle saare minors padhne padenge.

"Cross term

4 x y

ka matlab hai

A_{12} = 4

Error: symmetry cross-coefficient ko aadha kar deta hai, isliye

A_{12} = A_{21} = 4/2 = 2

. Sirf sum

A_{12} + A_{21}

x y

ke coefficient ke equal hota hai.

"Kyunki Hessian ka positive trace hai, isliye critical point ek minimum hai."

Error: trace eigenvalues ka sum hai; yeh positive ho sakta hai jab ek eigenvalue negative ho (jaise

λ = 3, - 1

), ek saddle deta hai. Tumhe saare eigenvalues positive chahiye.

Why questions

A

ke liye symmetric kyun hona zaroori hai eigenvalue test ke kaam karne ke liye?

Spectral Theorem real eigenvalues aur orthogonal (rotation) diagonalization

A = Q Λ Q^{⊤}

sirf symmetric

A

ke liye guarantee karta hai; symmetry ke bina eigenvalues complex ho sakte hain aur sign classification toot jaata hai.

Change of variable

y = Q^{⊤} x

se hum definiteness "read off" kyun kar lete hain?

Yeh axes ko rotate karta hai taaki

Q (x) = \sum λ_{i} y_{i}^{2}

pure sum of squares ban jaaye; har

y_{i}^{2} \geq 0

, isliye poora sign purely

λ_{i}

ke signs se tay hota hai.

Rotation

Q

yeh kyun nahi badalta ki

x

zero hai ya nahi?

Q

orthogonal hai, isliye invertible hai, toh

x = 0 ⟺ y = 0

; "for all

x \neq = 0

" wali condition coordinate change ke baad bhi intact rehti hai.

Mixed-sign eigenvalues kyun guarantee karte hain ki dono output signs actually hote hain?

Positive

λ

ke eigenvector ko plug karo toh positive value milegi aur negative

λ

ke eigenvector ko plug karo toh negative value milegi — sum-of-squares form har direction ko reachable banata hai.

Second Derivative Test kyun Hessian Matrix ki definiteness tak reduce hota hai?

Critical point ke paas

f (x) \approx f_{0} + \frac{1}{2} x^{⊤} H x

; quadratic term dominate karta hai, isliye PD

H

har jagah upar curve karta hai (min), ND neeche (max), indefinite saddle deta hai.

Sylvester's criterion negative definiteness detect kyun nahi kar paata agar tum sirf "all minors negative" check karo?

Kyunki ND,

- A

ka PD hai, aur har

k \times k

minor

(- 1)^{k}

se scale hota hai; even-sized minors wapas positive ho jaate hain,

-, +, -

alternating pattern banate hain instead of all-negative.

x^{2} + (x + y)^{2} + y^{2}

jaisi sum of squares automatically positive definite kyun hai?

Har squared term

\geq 0

hai, aur unhe simultaneously vanish karne ka ek hi tarika hai

x = y = 0

; nonzero input se kam se kam ek positive term force hota hai, isliye

Q > 0

Cholesky Decomposition positive definiteness ki evidence kyun hai?

A = L L^{⊤}

with real invertible

L

deta hai

Q = ∥ L^{⊤} x ∥^{2} > 0

for

x \neq = 0

; aisa factorization exactly PD matrices ke liye exist karta hai, isliye Cholesky ka successful hona khud ek PD certificate hai.

Edge cases

Kya zero matrix positive semidefinite hai, negative semidefinite hai, ya dono?

Dono —

Q (x) = 0

har

x

ke liye

Q \geq 0

aur

Q \leq 0

dono satisfy karta hai simultaneously; yeh degenerate case hai jo har boundary par ek saath baitha hai.

Kya ek

1 \times 1

matrix indefinite ho sakti hai?

Nahi — ek single eigenvalue ek saath positive aur negative nahi ho sakta, isliye scalar

[a]

sirf PD (

a > 0

), ND (

a < 0

), ya degenerate zero case ho sakta hai; indefiniteness ke liye kam se kam do eigenvalues chahiye.

A = (0110)

kya hai?

Indefinite — eigenvalues

1

aur

- 1

hain; note karo

Q = 2 x y

(1, - 1)

par negative aur

(1, 1)

par positive hai, haalaanki iska zero diagonal hai.

Agar

det A = 0

hai, toh kaun se definiteness types abhi bhi possible hain?

PSD, NSD, ya indefinite — zero determinant matlab ek zero eigenvalue hai, jo strict PD aur ND ko rule out karta hai lekin har "semi" ya mixed case ko open rakhta hai.

Kya ek PSD matrix mein tiny

ε I

add karne se woh positive definite ban jaati hai?

Haan — yeh har eigenvalue ko

ε > 0

se upar shift karta hai, isliye koi bhi zero eigenvalue strictly positive ban jaata hai; yeh "regularization" trick exactly wahi hai jo Least Squares singular normal-equations matrix ko stabilize karne ke liye use karta hai.

Agar

A

aur

B

dono positive definite hain, kya

A + B

positive definite hai?

Haan —

Q_{A + B} (x) = x^{⊤} A x + x^{⊤} B x

do strictly positive numbers ka sum hai

x \neq = 0

ke liye; definiteness addition ke under preserve hoti hai (haalaanki general products ke under nahi).

Agar

A

PD hai aur

c < 0

ek scalar hai, toh

c A

kya hai?

Negative definite — scaling har eigenvalue ko negative

c

se multiply karta hai, all-positive ko all-negative mein flip karta hai.

Kya ek positive definite matrix mein negative off-diagonal entry ho sakti hai?

Haan — off-diagonals koi bhi sign ke ho sakte hain; jaise

(2 - 1 - 1 2)

ke eigenvalues

1, 3

hain aur negative

- 1

entries ke bawajood PD hai.

Recall Ek-line survival guide

PD = all $λ > 0$ = all leading minors $> 0$ = Cholesky exists. ND = signs flip karo, minors alternate karte hain. Semidefinite = ek zero andar ghus jaata hai. Indefinite = dono signs appear hote hain — aur dhyan raho: positive entries, positive trace, ya positive determinant akele inn mein se kuch bhi prove nahi karta.

Question bank — Quadratic forms — positive definite, negative definite, indefinite

True or false — justify karo

Error dhundho

Why questions

Edge cases