Eigendecomposition of matrices — Hinglish notes

1.1.14 · AI-ML › Linear Algebra Essentials

1. Eigenvector / eigenvalue KYA hai?

" $v \neq = 0$ " rule KYO? Agar hum $v = 0$ allow karte, toh $A \cdot 0 = λ \cdot 0$ har $λ$ ke liye sach hota — koi matlab nahi. Hum zero vector ko forbid karte hain taaki eigenvalue meaningful ho.

2. Eigenvalues KAISE dhundhe — scratch se derive karo

Defining equation se shuru karo aur rearrange karo (char. polynomial ko memorize mat karo — khud build karo):

$A v = λ v ⟹ A v - λ v = 0 ⟹ (A - λ I) v = 0$

Ab $(A - λ I) v = 0$ ek homogeneous system hai. Hume ek nonzero solution $v$ chahiye. Ek square system $M v = 0$ ka nonzero solution tab aata hai jab $M$ singular ho, yaani

$det (A - λ I) = 0$

Jab ek baar har $λ_{i}$ mil jaaye, use wapas plug karo aur eigenvector(s) ke liye $(A - λ_{i} I) v = 0$ solve karo (null space).

3. Decomposition KAISE assemble kare

Maano $A$ (size $n$ ) ke paas $n$ linearly independent eigenvectors $v_{1}, \dots, v_{n}$ hain jinke eigenvalues $λ_{1}, \dots, λ_{n}$ hain. Eigenvectors ko $V$ ke columns ke roop mein stack karo aur eigenvalues ko $Λ$ ke diagonal pe rakho:

$V = [v_{1} v_{2} \dots v_{n}], Λ = diag (λ_{1}, \dots, λ_{n})$

Dekho $A$ ek saath sab columns ke saath kya karta hai: $A V = [A v_{1} \dots A v_{n}] = [λ_{1} v_{1} \dots λ_{n} v_{n}] = V Λ$

Toh $A V = V Λ$ . Agar $V$ invertible hai (independent eigenvectors chahiye), $V^{- 1}$ se multiply karo:

4. Yeh powerful KYO hai — matrix powers

$A^{2} = (V Λ V^{- 1}) (V Λ V^{- 1}) = V Λ I V^{- 1} V Λ V^{- 1} = V Λ^{2} V^{- 1}$

Induction se: $A^{k} = V Λ^{k} V^{- 1}, Λ^{k} = diag (λ_{1}^{k}, \dots, λ_{n}^{k})$

KYO bada deal hai: ek diagonal matrix ko power karna sirf diagonal entries ko power karna hai — baar baar $O (n^{3})$ multiplications ki jagah $O (n)$ . Isi tarah Markov chains steady states dhundhti hain aur hum long-term behavior analyze karte hain.

5. Special case: symmetric matrices

6. Worked examples

7. Common mistakes (Steel-man karo)

8. Recall

Recall Active recall — answers cover karo

Kaunsa equation ek eigenpair define karta hai? → $A v = λ v$ .
Hum $det (A - λ I) = 0$ KYO require karte hain? → taaki $(A - λ I) v = 0$ ka ek nonzero solution ho (matrix singular).
$V$ mein kya hona chahiye aur kya satisfy karna chahiye? → columns = eigenvectors; linearly independent (invertible) hone chahiye.
$A^{k}$ ka formula? → $V Λ^{k} V^{- 1}$ .
$V$ orthogonal kab hota hai? → jab $A$ symmetric ho (spectral theorem).

Recall 12-saal ke bachche ko explain karo (Feynman)

Socho ek bada rubber sheet hai. Jab tum ise khinchte ho, uspe bane zyaatar arrows naye direction mein ghoom jaate hain. Lekin kuch magic arrows same direction mein point karte rehte hain — woh bas lambe ya chhote ho jaate hain. Woh magic arrows eigenvectors hain, aur "kitna lamba" eigenvalue hai. Agar tum saare magic arrows jaante ho, toh tum sab kuch samajh jaate ho jo kheench karta hai: bas har magic arrow ke saath stretch karo. Eigendecomposition un magic arrows ki list aur unki stretch amounts likhna hai.

9. Connections

Determinants — $det = 0$ singularity test jo eigenvalues produce karta hai.
Diagonalization — eigendecomposition exist karne ki condition ( $n$ independent eigenvectors).
Singular Value Decomposition (SVD) — woh fallback jo hamesha exist karta hai, non-square/defective matrices ke liye bhi.
Principal Component Analysis (PCA) — covariance matrix ka eigendecomposition.
Markov Chains — $A^{k} = V Λ^{k} V^{- 1}$ se steady states.
Symmetric Matrices & Spectral Theorem — orthonormal eigenbasis.

Eigenvector defining equation

A v = λ v

(jisme

v \neq = 0

);

A

v

ko rotate kiye bina scale karta hai.

Hum

A - λ I

mein

I

kyun daalta hai?

λ v = λ I v

ko ek matrix–vector product banana ke liye taaki use

A v

se subtract kiya ja sake.

Characteristic equation

det (A - λ I) = 0

; iske roots eigenvalues hain.

det (A - λ I) = 0

KYO?

Ek homogeneous system

(A - λ I) v = 0

ka nonzero solution tab hota hai jab matrix singular ho, yaani determinant zero ho.

Eigendecomposition formula

A = V Λ V^{- 1}

V

= eigenvector columns,

Λ

= eigenvalues ka diag.

A V = V Λ

derive karo

A [v_{1} \dots v_{n}] = [λ_{1} v_{1} \dots λ_{n} v_{n}] = V Λ

; phir

A = V Λ V^{- 1}

A^{k}

ka formula

A^{k} = V Λ^{k} V^{- 1}

jisme

Λ^{k} = diag (λ_{i}^{k})

Eigendecomposition kab impossible hai?

Jab

A

ke paas

n

linearly independent eigenvectors na hon (defective matrix), toh

V

invertible nahi hota.

Spectral theorem

Symmetric

A = A^{⊤}

⇒ real eigenvalues, orthonormal eigenvectors,

A = Q Λ Q^{⊤}

Kya eigenvectors unique hote hain?

Nahi — koi bhi nonzero scalar multiple kaam karta hai; hum

∥ v ∥ = 1

pe normalize karte hain.