Orthogonal sets and orthonormal basis: Question bank

4.5.34 · D5 · Maths › Linear Algebra (Full) › Orthogonal sets and orthonormal basis

True ya false — justify karo

Har claim ek aisa statement hai jise tum true ya false label karo — lekin asli score justification mein hai.

Ek orthogonal set hamesha linearly independent hota hai.

False — sirf nonzero orthogonal sets hoti hain. Zero vector

0 \cdot u_{j} = 0

vacuously satisfy karta hai, isliye

{0, u}

"orthogonal" hai phir bhi dependent hai.

Agar

u \cdot v = 0

toh

{u, v}

orthonormal hai.

False — orthogonal hona sirf yeh batata hai ki woh perpendicular hain; orthonormal ke liye additionally

∥ u ∥ = ∥ v ∥ = 1

bhi chahiye, jo dot product ka zero hona nahi batata.

R^{n}

mein

n

nonzero vectors ka har orthogonal set

R^{n}

ka basis hota hai.

True — woh automatically independent hote hain, aur

R^{n}

mein

n

independent vectors hamesha use span karte hain, isliye basis banate hain.

R^{3}

ke kisi basis mein 4 mutually orthogonal nonzero vectors ho sakte hain.

False — 4 nonzero orthogonal vectors independent hote hain, lekin

R^{3}

mein zyada se zyada 3 independent vectors hote hain, isliye aisa koi set exist hi nahi karta.

Agar

U

ke columns orthonormal hain toh

U

invertible hai.

Saamanya roop mein False —

U^{T} U = I

hold karta hai, lekin agar

U

m \times p

hai jahan

m > p

toh woh square nahi hai aur uska koi inverse nahi hai; sirf square orthonormal-column matrices invertible hoti hain.

Kisi bhi matrix

U

ke liye jiske columns orthonormal hain,

U U^{T} = I

False jab tak

U

square na ho.

U U^{T}

asal mein

U

ke column space par projection hai; yeh

I

tab hi hoga jab woh columns poore

R^{m}

ko span karein.

Kisi vector ko orthogonal matrix se multiply karne par uski length badal sakti hai.

False —

∥ U x ∥ = ∥ x ∥

orthonormal columns ke liye, kyunki

∥ U x ∥^{2} = x^{T} U^{T} U x = x^{T} x

Ek orthogonal matrix ka determinant

+ 1

hota hai.

False — orthogonal matrices mein

det (U) = \pm 1

hota hai;

- 1

reflections ke corresponding hota hai (odd number of them),

+ 1

pure rotations ke liye.

Agar koi set orthonormal hai, toh har vector ko uski length se divide karne par woh unchanged rehta hai.

True — orthonormal vectors ki length pehle se 1 hoti hai, isliye normalize karne se kuch nahi badalta; yahi reason hai ki orthonormal "already-finished" state hai.

Orthonormal basis mein coordinates nikalne ke liye ek linear system solve karna padta hai.

False — yahi toh pura fayda hai:

c_{j} = y \cdot u_{j}

ek single dot product per coordinate hai, koi elimination nahi chahiye.

Error dhundho

Har line mein ek galat statement hai; reveal mein mistake batai gayi hai aur usse theek kiya gaya hai.

{u_{1}, u_{2}}

orthogonal hai, isliye uska coordinate formula

c_{j} = y \cdot u_{j}

hai."

Denominator chhod diya gaya — ek sirf orthogonal (normalized nahi) basis ke liye formula hai

c_{j} = \frac{y \cdot u _{j}}{u _{j} \cdot u _{j}}

; clean version ke liye unit vectors chahiye.

U

ke columns orthogonal hain, isliye

U^{T} U = I

Sirf orthogonal hain, orthonormal nahi — phir

U^{T} U

ek diagonal matrix hogi jisme entries

∥ u_{j} ∥^{2}

hongi,

I

tab hi hogi jab normalize karo.

U

orthogonal hai, isliye uski rows orthonormal nahi bhi ho sakti hain."

Ek square orthogonal matrix ke liye

U^{T} U = U U^{T} = I

hota hai, isliye rows bhi orthonormal hoti hain; square

U

ke liye row- aur column-orthonormality equivalent hain.

"Is orthogonal set mein

0

samete 5 vectors hain, aur yeh independent hai kyunki sab perpendicular hain."

Zero vector sab cheez ke perpendicular hota hai lekin

0 \cdot u + 0 = 0

mein ek nonzero coefficient deta hai, independence todta hai — "nonzero" hypothesis zaroori hai.

y

W

par project karne ke liye

W

ka basis orthonormal hona chahiye."

Sirf orthogonal hona kaafi hai; Orthogonal projection formula

\sum \frac{y \cdot u _{j}}{u _{j} \cdot u _{j}} u_{j}

kisi bhi orthogonal basis ke liye kaam karta hai, denominator non-unit lengths ko handle karta hai.

R^{3}

mein 2 vectors ka orthonormal set ek orthonormal basis hai."

Yeh ek orthonormal set hai, lekin

R^{3}

ke basis ke liye 3 vectors chahiye; yeh sirf us 2-dimensional subspace ka basis hai jise woh span karte hain.

"Kyunki

U^{T} U = I

hai, isliye

U^{- 1} = U^{T}

Yeh inference tab hi valid hai jab

U

square ho; tall

U

ke liye,

U^{T}

ek left inverse hai, genuine inverse nahi.

Why questions

Yeh har fact ke reason ko probe karte hain, fact ko nahi.

Dependence relation

\sum c_{i} u_{i} = 0

u_{j}

se dot karne par independence kaise prove hoti hai?

Orthogonality har cross term

u_{i} \cdot u_{j}

(

i \neq = j

) ko zero kar deti hai, sirf

c_{j} ∥ u_{j} ∥^{2} = 0

bachta hai; kyunki

∥ u_{j} ∥^{2} > 0

nonzero vectors ke liye, har

j

ke liye

c_{j} = 0

hoga.

"Orthogonal matrix" term ek historical trap kyun hai?

Isme orthonormal columns aur squareness dono chahiye, sirf orthogonal nahi — naam mein "orthogonal" raha gaya jabki unit length aur squareness bhi required hain.

Orthonormal-column maps dot products kyun preserve karti hain, aur numerically yeh kyun matter karta hai?

(U x) \cdot (U y) = x^{T} U^{T} U y = x \cdot y

, isliye angles aur lengths untouched rehte hain — aisi maps kabhi errors amplify nahi karti, yahi reason hai ki QR decomposition numerically stable hai.

Coordinate denominator mein

∥ u_{j} ∥

ki jagah

u_{j} \cdot u_{j}

kyun hai?

Dot karne ke baad isolated term

c_{j} (u_{j} \cdot u_{j}) = c_{j} ∥ u_{j} ∥^{2}

aata hai, isliye

u_{j} \cdot u_{j} = ∥ u_{j} ∥^{2}

(squared length) se divide karna exactly cancel karta hai.

Orthonormal basis Least squares ko easy kyun banata hai?

Normal-equation matrix

U^{T} U

ban jaati hai

I

, isliye least-squares solution collapse hokar

\hat{x} = U^{T} b

ho jaata hai — system solve karne ki jagah sirf dot products.

Gram-Schmidt process exist kyun karta hai agar orthonormal bases itne convenient hain?

Kyunki zyaadatar diye gaye bases orthogonal nahi hote; Gram-Schmidt kisi arbitrary basis se orthonormal basis banata hai taaki phir tum dot-product shortcuts enjoy kar sako.

Ek symmetric matrix ke do eigenvectors orthonormal kyun choose kiye ja sakte hain?

Ek symmetric matrix ke alag eigenvalues se Eigenvalues and eigenvectors automatically orthogonal hote hain, aur unhe normalize karne par orthonormal eigenbasis milta hai — yahi spectral theorem ka fayda hai.

Edge cases

Degenerate, zero, aur limiting inputs jo definitions quietly cover karti hain.

Kya empty set orthogonal hai?

Vacuously haan — koi distinct pairs hi nahi hain jo

u_{i} \cdot u_{j} = 0

violate karein, isliye definition trivially satisfy hoti hai (aur yeh independent bhi hai).

Kya ek single nonzero vector ek orthogonal set hai?

Haan — check karne ke liye koi distinct pair nahi hai, isliye koi bhi akela nonzero vector orthogonal count hota hai, aur apni length se divide karne par orthonormal ban jaata hai.

Kya ek unit vector ek orthogonal-but-not-orthonormal set ka hissa ho sakta hai?

Haan — orthonormality ke liye sab vectors ki length 1 chahiye; ek unit vector aur ek length-3 vector wala set orthogonal hai lekin orthonormal nahi.

Kisi bhi orthogonal basis mein

y = 0

ka coordinate kya hoga?

Har

c_{j} = \frac{0 \cdot u _{j}}{u _{j} \cdot u _{j}} = 0

, isliye zero vector ke sab coordinates zero hote hain, jaisa expected hai.

Agar

y

exactly

u_{2}

ki direction mein ho, toh uske coordinates kaisa dikhenge?

Sirf

c_{2}

nonzero hoga aur baaki sab

c_{j} = 0

, kyunki perpendicular directions ke liye

y \cdot u_{j} = 0

— decoupling ka ek saaf illustration.

Agar koi ek basis vector

0

ho toh coordinate formula ka kya hoga?

Yeh toot jaata hai —

u_{j} \cdot u_{j} = 0

denominator ko zero kar deta hai, yahi exact reason hai ki zero vector ko basis se forbid kiya jaata hai.

Kya

1 \times 1

matrix

[- 1]

ek orthogonal matrix hai?

Haan — iska single column ki length 1 hai aur

(- 1) (- 1) = 1 = I_{1}

; yeh reflection

x \mapsto - x

represent karta hai aur iska determinant

- 1

hai.

R^{n}

mein sabse bada possible orthonormal set kya hai?

Exactly

n

vectors — isse zyada dimension se aage independent hote, jo impossible hai, isliye ek maximal orthonormal set hamesha ek orthonormal basis hota hai.

Recall Traps ki ek-line summary

Orthogonal ≠ orthonormal ≠ orthogonal-matrix; zero vector independence todta hai; tall orthonormal-column matrices mein $U^{T} U = I$ hota hai lekin $U U^{T} = I$ nahi; aur $u_{j} \cdot u_{j}$ denominator kabhi mat chhodna jab tak lengths sach mein 1 na hon.