Finding eigenspaces — Hinglish notes

4.5.30 · Maths › Linear Algebra (Full)

HUM KYA FIND KAR RAHE HAIN

$E_{λ}$ ek subspace kyun hai? Kyunki yeh matrix $(A - λ I)$ ka null space hai, aur har null space ek subspace hota hai (addition aur scalar multiplication ke under closed, $0$ contain karta hai). Yahi key trick hai — eigenspace find karna = ek homogeneous system solve karna.

RECIPE KO HUM SCRATCH SE KAISE DERIVE KARTE HAIN

Hum nonzero $v$ chahte hain jiske liye $A v = λ v$ ho.

Step 1 — Sab kuch ek taraf le jao. $A v - λ v = 0 .$ Kyun? $v$ ko collect karne ke liye aur isko ek single matrix times $v$ mein convert karne ke liye.

Step 2 — $v$ ko factor out karo. Hum $(A - λ) v$ nahi likh sakte kyunki $A$ ek matrix hai aur $λ$ ek scalar hai — woh alag worlds mein rehte hain. Identity insert karo: $λ v = λ I v$ . $(A - λ I) v = 0 .$ Kyun? Ab $A - λ I$ ek genuine matrix hai, toh yeh ek clean homogeneous system ban jata hai.

Step 3 — Nonzero solution ki demand karo. Ek homogeneous system $M v = 0$ ka nonzero solution tab hoga iff $M$ not invertible ho, yaani $det (A - λ I) = 0.$ Kyun? Agar $A - λ I$ invertible hota, toh ek hi solution hota $v = 0$ — jo kisi kaam ka nahi. Toh singularity exactly woh condition hai jab eigenvectors exist karte hain. $λ$ mein yeh equation characteristic equation kehlati hai.

Step 4 — Har eigenvalue $λ$ ke liye, $(A - λ I) v = 0$ solve karo. Solution set (null space) hi eigenspace hai. Uska basis = independent eigenvectors ka set.

WORKED EXAMPLE 1 — ek clean $2 \times 2$

$A = (2112) .$

Characteristic equation. $det (2 - λ 1 1 2 - λ) = (2 - λ)^{2} - 1 = 0.$ Yeh step kyun? Eigenvectors sirf wahan exist karte hain jahan matrix singular ho, toh hum determinant ko $0$ set karte hain.

Expand karo: $(2 - λ)^{2} = 1 \Rightarrow 2 - λ = \pm 1 \Rightarrow λ = 1, 3.$

$λ = 3$ ke liye eigenspace. $A - 3 I = (- 1 1 1 - 1)$ banao. Row reduce karo → $(10 - 1 0)$ , jisse $v_{1} = v_{2}$ milta hai. Kyun? Free variable $v_{2} = t$ ⇒ $v = t (1 1)$ . $E_{3} = span {(11)} .$

$λ = 1$ ke liye eigenspace. $A - I = (1111)$ → $v_{1} = - v_{2}$ . $E_{1} = span {(1 - 1)} .$

Check: $A (1 1) = (3 3) = 3 (1 1)$ ✓.

WORKED EXAMPLE 2 — ek repeated eigenvalue jisme 2D eigenspace hai

$A = (5005) .$ $det (A - λ I) = (5 - λ)^{2} = 0 \Rightarrow λ = 5$ (algebraic multiplicity 2).

$A - 5 I = (0000)$ . Har vector $(A - 5 I) v = 0$ satisfy karta hai, toh $E_{5} = span {(10), (01)} = R^{2} .$ Yeh kyun matter karta hai: eigenspace dimension (geometric multiplicity) yahan $2$ hai — repeat count ke barabar.

WORKED EXAMPLE 3 — repeated eigenvalue lekin ek "deficient" eigenspace

$A = (5015) .$ Phir se $det (A - λ I) = (5 - λ)^{2}$ , toh $λ = 5$ (multiplicity 2).

$A - 5 I = (0010)$ → equation $v_{2} = 0$ , $v_{1}$ free. $E_{5} = span {(10)}, dim E_{5} = 1.$ Yahi lesson hai: algebraic multiplicity $2$ lekin geometric multiplicity $1$ . Geometric ≤ algebraic hamesha, aur jab strictly less ho toh matrix not diagonalizable hoti hai.

Recall Pehle predict karo phir verify karo

Example 3 padhne se pehle predict karo: " $λ = 5$ do baar repeat hota hai, toh eigenspace poori $R^{2}$ hogi, hai na?" Verify karo: Nahi! Example 3 dikhata hai ki yeh sirf ek line bhi ho sakti hai. Off-diagonal $1$ doosri direction ko bigad deta hai. Repeat count sirf eigenspace dimension ki ek upper bound hai.

Common mistakes

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho ek trampoline hai jo, jab tum push karo, sab kuch bahar ki taraf stretch karta hai aur twist bhi karta hai. Tumne jo bhi arrows draw kiye hain woh twist hokar naye direction mein point karne lagte hain. Lekin kuch special arrows sirf bade ya chhote hote hain bina mude — jaise ek wheel ki spokes jo sirf grow karti hain. Woh jadu wale non-turning arrows eigenvectors hain, aur jitna woh grow karte hain woh eigenvalue hai. Agar do alag direction wale arrows dono ek hi growth amount ke liye magic hain, toh unke beech ki poori flat region bhi magic hai — woh poori region eigenspace hai. Use find karne ke liye tum poochho, "kis stretch amount ke liye machine kuch arrows ko zero kar deti hai jab tum woh stretch subtract karo?" — aur tum woh flattening equation solve karo.

Active recall

Eigenspace of

λ

ko kaun si equation define karti hai?

(A - λ I) v = 0

, yaani

E_{λ} = Null (A - λ I)

Eigenvalue ke liye

det (A - λ I) = 0

kyun hona chahiye?

Ek homogeneous system ka nonzero solution tab exist karta hai iff matrix singular ho, yaani determinant zero ho.

Eigenspace hamesha subspace kyun hota hai?

Yeh ek null space hai, aur har null space ek subspace hota hai.

Kya

λ = 0

eigenvalue ho sakta hai?

Haan; iska matlab hai

A

singular hai aur

E_{0} = Null (A)

. Sirf vector nonzero hona chahiye.

Eigenspace ki dimension ka formula?

dim E_{λ} = n - rank (A - λ I)

(geometric multiplicity).

Geometric aur algebraic multiplicity mein kya relation hai?

1 \leq

geometric

\leq

algebraic; matrix diagonalizable hai iff dono har

λ

ke liye equal hon.

Kya

0

eigenvector hai?

Nahi, lekin yeh har eigenspace mein hota hai.

A = (2112)

ke liye eigenvalues kya hain?

λ = 1

aur

λ = 3

Connections

Characteristic polynomial — woh eigenvalues produce karta hai jo tum yahan use karte ho.
Null space and solving homogeneous systems — step 4 ka engine.
Diagonalization — tab possible hai jab eigenspaces ki dimensions ka sum $n$ ho.
Determinants — step 3 mein singularity test.
Rank-Nullity theorem — $dim E_{λ} = n - rank (A - λ I)$ deta hai.
Symmetric matrices and spectral theorem — full, orthogonal eigenspaces guarantee karta hai.