Eigenvalues and eigenvectors — characteristic polynomial: Question bank

4.5.29 · D5 · Maths › Linear Algebra (Full) › Eigenvalues and eigenvectors — characteristic polynomial

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Figure — Eigenvalues and eigenvectors — characteristic polynomial

Eigenvector ek aisi direction hai jo matrix se guzarne ke baad bina mudi rahti hai — $A x$ wapas usi dashed line par aata hai jahan $x$ tha, sirf $λ$ se rescale hoke. Ek generic vector (grey) apni line se hat jaata hai. Har claim ke saamne yahi image rakho.

True or false — justify karo

True or false: Agar

det (A) = 0

hai to

0

A

ka ek eigenvalue hai.

True.

det (A) = det (A - 0 \cdot I) = p (0)

, isliye

det (A) = 0

ka matlab hai

λ = 0

characteristic polynomial ka root hai — equivalently null space nontrivial hai, jo ek nonzero

x

deta hai jahan

A x = 0

True or false: Har

2 \times 2

real matrix ke do real eigenvalues hote hain.

False. Polynomial

λ^{2} - tr (A) λ + det (A)

ka discriminant negative ho sakta hai; rotation

(01 - 1 0)

λ = \pm i

milta hai, wahan real eigenvalues exist hi nahi karte (dekho Complex Numbers).

True or false: Ek eigenvalue ke kaafi saare eigenvectors ho sakte hain, lekin ek eigenvector sirf ek hi eigenvalue ka hota hai.

True. Agar

A x = λ x

aur

A x = μ x

ho to

(λ - μ) x = 0

; kyunki

x \neq = 0

hai, isliye

λ = μ

milta hai. Lekin ek

λ

ke paas eigenvectors ka poora subspace ho sakta hai.

True or false: Agar

x

ek eigenvector hai to

5 x

bhi usi eigenvalue ke saath eigenvector hai.

True.

A (5 x) = 5 A x = 5 λ x = λ (5 x)

. Koi bhi nonzero scalar multiple usi magic line par rahta hai — eigenvectors directions hain, fixed-length arrows nahi.

True or false: Ek

n \times n

matrix ka characteristic polynomial hamesha exactly degree

n

ka hota hai.

True.

λ

ke terms sirf diagonal factors

(a_{ii} - λ)

se aate hain; unka product ek leading

(- λ)^{n}

contribute karta hai, isliye degree exactly

n

hoti hai aur kabhi kam nahi hoti.

True or false: Ek matrix aur uske transpose ke eigenvalues same hote hain.

True.

det (A^{⊤} - λ I) = det ((A - λ I)^{⊤}) = det (A - λ I)

, kyunki transpose karne se determinant nahi badlta. Same polynomial ⇒ same eigenvalues (halanki eigenvectors alag ho sakte hain).

True or false: Agar

A

invertible hai aur uska eigenvalue

λ

hai, to

A^{- 1}

ka eigenvalue

1/ λ

hoga.

True.

A x = λ x

se,

A^{- 1}

se multiply karo:

x = λ A^{- 1} x

, to

A^{- 1} x = (1/ λ) x

. Invertibility guarantee karti hai ki

λ \neq = 0

, isliye divide karna legal hai.

True or false: Jo do matrices same characteristic polynomial rakhte hain, woh same matrix hote hain.

False.

(3013)

aur

3 I

dono

(3 - λ)^{2}

dete hain, phir bhi pehla defective hai aur doosra nahi. Polynomial eigenvalues fix karta hai, poori matrix nahi.

True or false:

A^{2}

ke eigenvalues,

A

ke eigenvalues ke squares hote hain.

True.

A x = λ x \Rightarrow A^{2} x = A (λ x) = λ^{2} x

. Wahi eigenvector bana rehta hai eigenvalue

λ^{2}

ke saath (ulta —

A^{2}

ka har eigenvalue kisi

λ^{2}

ke form mein hoga — yeh bhi

C

par hold karta hai).

Error dhundho

Flaw nikalo: "

A - λ

singular hai, to

det (A - λ) = 0

set karo."

Tum scalar

λ

ko matrix

A

se subtract nahi kar sakte — woh alag worlds mein rehte hain. Sahi object

A - λ I

hai; identity

I

scalar ko matrix mein badal deta hai taaki subtraction define ho sake.

Flaw nikalo: "Maine

λ = 3

nikala, phir eigenvector padhne ke liye

3

wapas

det (A - λ I)

mein daala."

Determinant ek single number hai, usme koi direction stored nahi hoti. Eigenvectors system

(A - 3 I) x = 0

solve karne se milte hain — yaani

A - 3 I

ka null space.

Flaw nikalo: "

A 0 = λ 0

true hai, isliye

0

har

λ

ke liye eigenvector hai."

Equation true hai lekin vacuous hai. Eigenvectors define hote hain nonzero hone ke liye;

0

ko allow karne se har scalar "kaam karta", concept ko meaningless bana deta.

Flaw nikalo: "

λ = 2

double root hai, isliye zaroor do independent eigenvectors honge."

Yeh algebraic aur geometric multiplicity ko confuse kar raha hai. Double root sirf algebraic multiplicity

2

guarantee karta hai, lekin geometric multiplicity sirf

1

bhi ho sakti hai (ek defective matrix jaise

(2012)

), jo Diagonalization rok deta hai.

Flaw nikalo: "Trace

7

hai, isliye koi ek eigenvalue zaroor

7

hoga."

Trace saare eigenvalues ka sum hai, kisi ek ka nahi.

(5002)

ke liye trace

7

is tarah split hota hai:

5 + 2

; koi bhi eigenvalue

7

nahi hai (dekho Trace of a Matrix).

Flaw nikalo: "Maine

det (λ I - A)

se compute kiya aur mere dost ne

det (A - λ I)

use kiya; hamare polynomials alag aaye to koi ek galat hai."

Dono theek hain.

det (λ I - A) = (- 1)^{n} det (A - λ I)

, isliye odd

n

ke liye polynomials ek global sign se alag hain — lekin sign flip roots kabhi nahi badlta, to eigenvalues match karte hain.

Why questions

Hum

(A - λ I) x = 0

ka nonzero solution kyun maangte hain, koi bhi solution kyun nahi?

x = 0

hamesha ek homogeneous system solve karta hai, isliye woh koi information nahi deta. Sirf ek nonzero solution ek genuine invariant direction reveal karta hai — aur use require karna

A - λ I

ko singular hone par majboor karta hai.

"Singular" exactly "

det = 0

" mein kyun translate hota hai?

Ek square matrix invertible hai tab aur sirf tab jab uska determinant nonzero ho. Isliye "not invertible" (singular) aur "

det = 0

" literally same statement hain, jo eigenvalue hunt ko ek solvable polynomial equation mein badal deta hai.

Ek triangular matrix ke eigenvalues sirf uske diagonal entries kyun hote hain?

Ek triangular

A

ke liye,

A - λ I

triangular rehta hai, aur ek triangular determinant uske diagonals ka product hota hai:

\prod_{i} (a_{ii} - λ)

. Product ko zero set karne par

λ = a_{ii}

milta hai har

i

ke liye.

Ek real matrix ko complex eigenvalues allow kyun karne chahiye?

Kyunki

det (A - λ I)

ek real polynomial hai jo shayad koi real root na rakhe (jaise

λ^{2} + 1

). Fundamental Theorem of Algebra sirf

C

par

n

roots guarantee karta hai; unhe refuse karna rotations ko "eigenvalue-less" chhod deta.

Eigenvalues ka sum trace ke barabar kyun hota hai?

2 \times 2

ke liye

p (λ) = λ^{2} - tr (A) λ + det (A)

likhne aur Vieta's formulas apply karne par,

λ

ka coefficient (sign tak) root sum hota hai. Generally

λ^{n - 1}

ka coefficient diagonal collect karta hai, trace deta hai.

Cayley–Hamilton theorem hamein characteristic polynomial se

A^{- 1}

compute kyun karne deta hai?

Cayley–Hamilton kehta hai ki

A

apna

p (λ)

satisfy karta hai.

p (A) = 0

rearrange karne se

I

A

ke polynomial ke roop mein isolate kar sakte hain; constant term

det (A)

(invertible hone par nonzero) se divide karne par

A^{- 1}

sirf

A

ki powers use karke express ho jaata hai.

Edge cases

Zero matrix

0_{n \times n}

ke eigenvalues kya hain?

Sirf

λ = 0

, algebraic multiplicity

n

ke saath;

p (λ) = (- λ)^{n}

. Har nonzero vector ek eigenvector hai, isliye geometric multiplicity bhi

n

hai — jitna possible ho sake utna un-defective.

Identity matrix

I

ke eigenvalues kya hain?

Sirf

λ = 1

repeated

n

times, kyunki

I x = 1 \cdot x

sab

x

ke liye. Har direction fixed hai, isliye poora space eigenspace hai — geometric multiplicity

n

Agar

A

ek projection hai (

A^{2} = A

), to kaunse eigenvalues possible hain?

Sirf

0

aur

1

A^{2} x = A x

se hume

λ^{2} = λ

milta hai, isliye

λ (λ - 1) = 0

— vectors ya to kill ho jaate hain (

λ = 0

) ya rakhe jaate hain (

λ = 1

Kya ek

3 \times 3

real matrix mein exactly do real eigenvalues ho sakte hain aur koi teesra real nahi?

"Exactly do isolated" nahi chhod sakte ek missing — complex roots conjugate pairs mein aate hain, isliye ek real cubic mein ya to 3 real roots hote hain ya 1 real + ek conjugate pair. Ek akela unpaired complex root impossible hai.

Jab ek eigenvalue repeated ho lekin matrix diagonalizable ho, to eigenvectors ka kya hota hai?

Repeated

λ

ek full-dimensional eigenspace own karta hai (geometric = algebraic multiplicity), jaise

2 I

mein

λ = 2

do baar aata hai aur poora plane eigenspace hai. Yahi equality exactly Diagonalization condition hai.

0

ka eigenvalue matrix ki rank ke baare mein kya batata hai?

Iska matlab hai null space nontrivial hai, isliye matrix singular hai aur rank-deficient hai.

λ = 0

ke eigenspace ki dimension nullity ke barabar hai, yaani

n - rank (A)

Recall One-line self-check

Q: Same characteristic polynomial ⇒ same matrix? → Nahi; defective vs diagonalizable isko share kar sakte hain.
Q: Zero determinant ⇒ kaunsa eigenvalue? → $λ = 0$ .
Q: Projection eigenvalues? → Sirf $0$ aur $1$ .