Inner product spaces — dot product generalization: Question bank

4.5.33 · D5 · Maths › Linear Algebra (Full) › Inner product spaces — dot product generalization

Teen axioms jinpe hum baar baar aate hain: symmetry, linearity in the first slot, aur positive-definiteness ( $⟨ v, v ⟩ \geq 0$ with equality only for $0$ ). Yahan ka almost har trap asliyat mein yahi hai — "tumne kaun sa axiom bhool gaye?"

True or false — justify

True or false: kisi real space par koi bhi symmetric bilinear form ek inner product hota hai.

False — usse positive-definite bhi hona chahiye.

⟨ u, v ⟩ = u_{1} v_{1} - u_{2} v_{2}

symmetric aur bilinear hai lekin

⟨(0, 1), (0, 1)⟩ = - 1

, toh lengths imaginary ho jaayengi.

True or false: agar

⟨ v, v ⟩ = 0

toh

v

ka zero vector hona zaroori nahi.

False for a genuine inner product — positive-definiteness force karta hai ki

⟨ v, v ⟩ = 0 ⟺ v = 0

. Agar koi nonzero

v

0

mile, toh woh form sirf "positive-semidefinite" hai, inner product nahi.

True or false: do vectors ek inner product ke under orthogonal ho sakte hain lekin doosre ke under nahi.

True — orthogonality ka matlab hai

⟨ u, v ⟩ = 0

, aur woh number chosen inner product par depend karta hai. Jaise ki

x

aur

1 - x

⟨ p, q ⟩ = p (0) q (0) + p (1) q (1)

ke under orthogonal hain lekin

\int_{0}^{1} pq d x

ke under nahi.

True or false: zero vector har vector ke saath orthogonal hota hai.

True — linearity se

⟨ 0, v ⟩ = ⟨ 0 \cdot 0, v ⟩ = 0 \cdot ⟨ 0, v ⟩ = 0

for all

v

. Zero vector woh ek hi vector hai jo sab cheez se, apne aap se bhi, orthogonal hai.

True or false:

∥ u + v ∥ = ∥ u ∥ + ∥ v ∥

sirf tab hi ho sakta hai jab

u = v

False — triangle inequality mein equality tab hoti hai jab

u, v

ek hi direction mein ho (

v = c u

with

c \geq 0

), zaroor equal nahi. Cross term

⟨ u, v ⟩

ko apni ceiling

∥ u ∥∥ v ∥

pe pahunchna hoga.

True or false:

⟨ u, v ⟩ = 3 u_{1} v_{1} - 2 u_{2} v_{2}

mein ek weight ko negate karne se bhi inner product milta hai.

False — negative weight positive-definiteness khatam kar deta hai:

⟨(0, 1), (0, 1)⟩ = - 2 < 0

. Ek weighted inner product mein har weight strictly positive honi chahiye.

True or false: Cauchy–Schwarz ek theorem hai jo prove karna padta hai, na ki ek axiom jo assume karte hain.

True — yeh positive-definiteness se

⟨ u - t v, u - t v ⟩ \geq 0

ke zariye derive hota hai. Yeh woh keemat hai jo axioms ada karte hain taaki

cos θ

[- 1, 1]

mein rahe. Dekho Cauchy-Schwarz Inequality.

True or false: agar

⟨ u, v ⟩ = 0

for all

v

, toh

u = 0

True — khaas taur par

v = u

lo, jisse

⟨ u, u ⟩ = 0

milta hai, toh positive-definiteness force karta hai

u = 0

. Sirf zero vector hi poore space se orthogonal hota hai.

Spot the error

Spot the error: "Maine

cos θ = 1.3

calculate kiya, toh angle thoda unusual hi hai."

Cosine kabhi

1

se exceed nahi kar sakta; Cauchy–Schwarz guarantee karta hai

∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥∥ v ∥

> 1

ki value matlab ya toh arithmetic mein galti hai ya woh form real inner product nahi hai.

Spot the error: "Symmetry se main

⟨ c u, v ⟩ = ⟨ u, c v ⟩

likh sakta hoon, toh

c

ko kisi bhi slot se freely nikaal sakta hoon."

Conclusion sahi hai (real inner products ke liye) lekin reason linearity plus symmetry hai, sirf symmetry nahi. Symmetry sirf slots swap karti hai; scaling linearity karta hai.

Spot the error: "

⟨ p, q ⟩ = p^{'} (0) q^{'} (0)

degree-

\leq 1

polynomials par ek valid inner product hai."

Yeh positive-definiteness fail karta hai: kisi bhi constant nonzero

p

ke liye

p^{'} = 0

hai, toh

⟨ p, p ⟩ = 0

jabki

p \neq = 0

. Nonzero vectors ki ek poori line ki "zero length" ho jaati hai.

Spot the error: "Kyunki

\int_{0}^{1} sin (2 π x) cos (2 π x) d x = 0

, yeh functions graph par literally

9 0^{\circ}

par hain."

Yeh integral inner product mein orthogonal hain, jo ki "

9 0^{\circ}

" ka abstract meaning hai — yeh curves ke beech koi visible right angle nahi hai. Orthogonality inner product mein rehti hai, page par nahi.

Spot the error: "Distance

d (u, v) = ∥ u ∥ - ∥ v ∥

hai kyunki subtraction measure karta hai ki lengths kitni door hain."

Distance hai

∥ u - v ∥

, yani difference vector ka norm, norms ka difference nahi. Equal length ke do vectors kaafi door ho sakte hain agar woh alag directions mein point kar rahe hon. Dekho Norms and Distance.

Spot the error: "Gram–Schmidt mein main ek vector ko normalize kar sakta hoon chahe

⟨ v, v ⟩ = 0

ho."

Nahi kar sakte —

∥ v ∥ = ⟨ v, v ⟩ = 0

se divide karna undefined hai. Yeh situation ek nonzero

v

ke liye real inner product ke under kabhi nahi aati, precisely positive-definiteness ki wajah se. Dekho Orthogonality and Gram-Schmidt.

Why questions

Why must the discriminant in the Cauchy–Schwarz proof be

\leq 0

, not

< 0

Quadratic

f (t) = ∥ v ∥^{2} t^{2} - 2 ⟨ u, v ⟩ t + ∥ u ∥^{2}

hai

\geq 0

(strictly neeche kabhi nahi), toh woh axis ko touch kar sakta hai — isse

= 0

milta hai aur equality case milti hai jahan

u, v

parallel hain.

Why does positive-definiteness, not symmetry, guarantee real lengths?

Length hai

⟨ v, v ⟩

; sirf positivity (

⟨ v, v ⟩ \geq 0

) hi ensure karti hai ki root ke andar ka number non-negative ho taaki square root real rahe.

Why is the integral

\int_{0}^{1} f g d x

a natural inner product for functions?

Yeh Dot Product

\sum u_{i} v_{i}

ko mimic karta hai, "coordinates"

x

ke ek continuum par

f (x) g (x)

ko sum karke, aur yeh teeno axioms satisfy karta hai, jisse functions ko genuine length, angle, aur orthogonality milti hai — yahi Fourier Series ki neenv hai.

Why does the triangle inequality need Cauchy–Schwarz?

∥ u + v ∥^{2}

expand karne par ek cross term

2 ⟨ u, v ⟩

aata hai; sirf Cauchy–Schwarz hi isse

2∥ u ∥∥ v ∥

se upar bound karta hai, jo ki

(∥ u ∥ + ∥ v ∥)^{2}

ka square close karne ke liye exactly chahiye.

Why do we insist linearity holds in the first slot specifically?

Real inner product ke liye, symmetry phir dusre slot mein linearity free mein de deti hai, toh ek slot axiom ke roop mein kaafi hai. Dono state karna redundant hoga.

Why is orthogonal projection only well-defined once we have an inner product?

Projection ke liye "component of

u

along

v

" ki notion chahiye, jo hai

\frac{⟨ u , v ⟩}{⟨ v , v ⟩} v

— dono dots ke liye inner product chahiye taaki woh numbers bhi ban sakein. Dekho Orthogonal Projections.

Edge cases

Edge case: kya

⟨ u, v ⟩ = 0

hi ek aisa tarika hai jis se do nonzero vectors Cauchy–Schwarz mein equality dete hain?

Nahi — equality

∣ ⟨ u, v ⟩ ∣ = ∥ u ∥∥ v ∥

tab hoti hai jab woh parallel hoon (linearly dependent), orthogonal ke bilkul opposite extreme jahan value

0

hoti hai.

Edge case: kisi vector aur apne aap ke beech angle kya hota hai?

cos θ = \frac{⟨ v , v ⟩}{∥ v ∥ ^{2}} = 1

, toh

θ = 0^{\circ}

— ek nonzero vector apne aap ke saath perfectly aligned hota hai, bilkul waise jaisi intuition demand karti hai.

Edge case: kya

cos θ

exactly

- 1

ho sakta hai?

Haan, jab

v = c u

with

c < 0

(anti-parallel). Cauchy–Schwarz dono ends

\pm 1

par tight hota hai, jo same-direction aur opposite-direction vectors correspond karte hain.

Edge case: zero vector space

{0}

par, kya koi inner product exist karta hai?

Trivially haan — ek hi pair hai

(0, 0) \mapsto 0

, jo sare axioms vacuously satisfy karta hai, positive-definiteness bhi, kyunki

0

hi zero vector hai.

Edge case: agar ek vector

0

ho toh

cos θ

ka kya hoga?

Yeh undefined hai — formula

∥ 0 ∥ = 0

se divide karta hai. Zero vector ki koi direction nahi hoti, toh "uske saath angle" meaningless hai, chahe

⟨ 0, v ⟩ = 0

isse sab se orthogonal banata ho.

Edge case: kya ek weighted inner product change karta hai ki kaun se vectors orthogonal hain?

Haan —

⟨ u, v ⟩ = 3 u_{1} v_{1} + 2 u_{2} v_{2}

ke under, pair

(2, - 3), (1, 1)

deta hai

6 - 6 = 0

(orthogonal), jabki standard dot product deta hai

- 1 \neq = 0

. Weights geometry ko reshape kar dete hain.

Recall Har trap ki ek-line summary

Yahan ka almost har "concept trap" ya toh positive-definiteness ki chupi hui failure hai, ya Cauchy–Schwarz ka galat use (angles $[- 1, 1]$ mein rehne chahiye), ya phir yeh bhool jaana ki orthogonality chosen inner product ke relative hoti hai. Jab stuck ho, pehle yeh teen check karo.