Finding eigenspaces: Exercises — Hinglish deep dive

4.5.30 · D4 · Maths › Linear Algebra (Full) › Finding eigenspaces

Poora game, ek sentence mein jo aap parent se jaante ho:

Yahan $I$ identity matrix hai (diagonal pe 1's, baaki 0's — woh matrix jo kuch nahi karta), aur $Null (M)$ null space hai: saare vectors $v$ jaise $M v = 0$ . Agar koi bhi phrase shaky lagta hai, pehle Null space and solving homogeneous systems revisit karo.

LEVEL 1 — Recognition

Problem 1.1

Aapko bataya gaya hai ki ek matrix $A$ ke liye, vector $v = (2 - 3)$ satisfy karta hai $A v = (8 - 12)$ . Kya $v$ ek eigenvector hai? Agar haan, toh uska eigenvalue kya hai?

Recall Solution 1.1

Kya check karna hai: ek eigenvector $A v = λ v$ satisfy karta hai — output wahi arrow scaled hona chahiye, turned nahi. Output ko input se componentwise compare karo: $\frac{8}{2} = 4, \frac{- 12}{- 3} = 4.$ Dono components same factor $4$ se scale ho rahe hain, isliye $A v = 4 v$ . Answer: Haan, $v$ ek eigenvector hai eigenvalue $== λ = 4 ==$ ke saath.

Problem 1.2

$A = (70 0 - 2)$ ke liye, eigenvalues inspection se likho aur har ek ke liye ek eigenvector bhi likho.

Recall Solution 1.2

Kyun inspection kaam karta hai: ek diagonal matrix har coordinate axis ko independently scale karta hai — woh unhe kabhi mix nahi karta. Isliye standard axis arrows already eigenvectors hain.

$A (10) = (70) = 7 (10)$ → $λ = 7$ , eigenvector $(10)$ .
$A (01) = (0 - 2) = - 2 (01)$ → $λ = - 2$ , eigenvector $(01)$ . Answer: eigenvalues $== 7 ==$ aur $== - 2 ==$ dono axes ke saath.

LEVEL 2 — Application

Problem 2.1

Saare eigenvalues aur eigenspaces nikalo $A = (4213) .$

Recall Solution 2.1

Step 1 — characteristic equation. $A - λ I = (4 - λ 2 1 3 - λ)$ banao (sirf diagonal se $λ$ ghatao). Iska determinant hai $det (A - λ I) = (4 - λ) (3 - λ) - (1) (2) = λ^{2} - 7 λ + 10.$ $0$ ke barabar karo: $λ^{2} - 7 λ + 10 = (λ - 2) (λ - 5) = 0$ , toh $λ = 2, 5$ .

Step 2 — $E_{5}$ . $A - 5 I = (- 1 2 1 - 2)$ . Row 2, row 1 ka $- 2 \times$ hai, toh yeh reduce hota hai $(10 - 1 0)$ mein, matlab $v_{1} = v_{2}$ . Maano $v_{2} = t$ : $E_{5} = span {(11)} .$ Step 3 — $E_{2}$ . $A - 2 I = (2211) \to (2010)$ , toh $2 v_{1} + v_{2} = 0 \Rightarrow v_{2} = - 2 v_{1}$ : $E_{2} = span {(1 - 2)} .$ Check: $A (11) = (55) = 5 (11)$ ✓.

Neeche wali picture dikhati hai kyun woh do lines special hain: lines se baahar har arrow turn ho jaata hai, do eigen-lines sirf stretch hoti hain.

Problem 2.2

Eigenvalues aur eigenspaces nikalo $A = (01 - 1 0) (ek 90^{\circ} rotation).$

Recall Solution 2.2

Step 1. $det (- λ 1 - 1 - λ) = λ^{2} + 1 = 0$ . Iska matlab: $λ^{2} = - 1$ , toh $λ = \pm i$ — koi real eigenvalues nahi. Kyun, geometrically: yeh matrix har real arrow ko $9 0^{\circ}$ rotate karta hai. Koi real arrow apni line pe nahi rehta, toh koi real eigenvector exist nahi kar sakta. Eigenvalues complex hain. Answer: eigenvalues $== \pm i ==$ ; koi real eigenspaces nahi hain. ( $C$ ke upar har eigenspace ek complex line hai, lekin $R$ ke upar eigenspace sirf ${0}$ hai.)

LEVEL 3 — Analysis

Problem 3.1

Consider $A = 300030007 .$ Har eigenvalue nikalo, uski algebraic multiplicity (characteristic polynomial mein repeat count) aur uski geometric multiplicity ( $dim E_{λ}$ ). Diagonalizability pe comment karo.

Recall Solution 3.1

Characteristic polynomial (diagonal ⇒ $(diag - λ)$ ka product): $det (A - λ I) = (3 - λ)^{2} (7 - λ) .$ Toh $λ = 3$ (algebraic multiplicity $2$ ) aur $λ = 7$ (algebraic multiplicity $1$ ).

$E_{3}$ : $A - 3 I = 000000004$ , rank $1$ , toh $dim E_{3} = 3 - 1 = 2$ . Basis: $100, 010$ . $E_{7}$ : $A - 7 I$ ka rank $2$ hai, toh $dim E_{7} = 3 - 2 = 1$ , basis $001$ .

Comment: geometric = algebraic dono eigenvalues ke liye ( $2 = 2$ aur $1 = 1$ ), toh $A$ diagonalizable hai — actually yeh already diagonal hai.

Problem 3.2

Ke liye $A = (6026),$ eigenvalue nikalo, uski algebraic aur geometric multiplicities nikalo, aur decide karo ki $A$ diagonalizable hai ya nahi.

Recall Solution 3.2

Characteristic polynomial: upper-triangular hai, toh $det (A - λ I) = (6 - λ)^{2}$ . Ek eigenvalue $λ = 6$ , algebraic multiplicity $2$ . $E_{6}$ : $A - 6 I = (0020)$ , rank $1$ . Equation $2 v_{2} = 0 \Rightarrow v_{2} = 0$ , $v_{1}$ free: $E_{6} = span {(10)}, dim E_{6} = 2 - 1 = 1.$ Analysis: geometric multiplicity $1 <$ algebraic $2$ . Rank-Nullity theorem se null space sirf 1-dimensional hai. Toh $A$ diagonalizable nahi hai — off-diagonal $2$ doosri independent direction cheen leta hai.

LEVEL 4 — Synthesis

Problem 4.1

Symmetric matrix ke eigenvalues aur eigenspaces nikalo $A = 2 - 1 0 - 1 2 - 1 0 - 1 2 .$

Recall Solution 4.1

Step 1 — characteristic polynomial. $det (A - λ I)$ ko first row ke along expand karke: $det (A - λ I) = (2 - λ) [(2 - λ)^{2} - 1] - (- 1) [- (2 - λ)] .$ $= (2 - λ) [(2 - λ)^{2} - 1] - (2 - λ) = (2 - λ) [(2 - λ)^{2} - 2] .$ $= 0$ set karo: ya toh $2 - λ = 0 \Rightarrow λ = 2$ , ya $(2 - λ)^{2} = 2 \Rightarrow λ = 2 \pm 2$ . Eigenvalues: $λ = 2, 2 + 2, 2 - 2$ . Teen distinct real values — expected, kyunki symmetric matrices ke eigenvalues hamesha real hote hain.

$E_{2}$ : $A - 2 I = 0 - 1 0 - 1 0 - 1 0 - 1 0$ . Rows se $v_{2} = 0$ aur $v_{1} + v_{3} = 0$ milta hai. Free $v_{3} = t$ : $E_{2} = span ⎩ ⎨ ⎧ 10 - 1 ⎭ ⎬ ⎫ .$ $E_{2 + 2}$ : $(A - (2 + 2) I) v = 0$ solve karo. Row 1: $- 2 v_{1} - v_{2} = 0 \Rightarrow v_{2} = - 2 v_{1}$ . Row 3: $- v_{2} - 2 v_{3} = 0 \Rightarrow v_{3} = - v_{2} / 2 = v_{1}$ . $v_{1} = 1$ lo: $E_{2 + 2} = span ⎩ ⎨ ⎧ 1 - 2 1 ⎭ ⎬ ⎫ .$ $E_{2 - 2}$ : same algebra $+ 2$ ke saath: $v_{2} = 2 v_{1}$ , $v_{3} = v_{1}$ : $E_{2 - 2} = span ⎩ ⎨ ⎧ 121 ⎭ ⎬ ⎫ .$ Spectral bonus: teen eigenvectors mutually orthogonal hain (check karo $10 - 1 \cdot 1 - 2 1 = 1 + 0 - 1 = 0$ ✓), bilkul waise jaisa spectral theorem symmetric $A$ ke liye promise karta hai.

LEVEL 5 — Mastery

Problem 5.1

Ek matrix design karo. Ek $2 \times 2$ real matrix $A$ (diagonal nahi) banao jiske eigenvalues $λ = 1$ eigenvector $(11)$ ke saath aur $λ = 4$ eigenvector $(1 - 1)$ ke saath hon.

Recall Solution 5.1

Tool kya hai — diagonalization ulta chalaana. Agar eigenvectors $P$ ke columns hain aur eigenvalues diagonal $D$ pe baithe hain, toh $A = P D P^{- 1}$ $A$ ko reconstruct karta hai. Yeh kyun kaam karta hai: construction se $A P = P D$ column-by-column kehta hai $A (col i) = λ_{i} (col i)$ — exactly eigen-equation. $P = (11 1 - 1), D = (1004), P^{- 1} = \frac{1}{- 2} (- 1 - 1 - 1 1) = \frac{1}{2} (11 1 - 1) .$ Multiply karo: $P D = (11 4 - 4), A = P D P^{- 1} = \frac{1}{2} (11 4 - 4) (11 1 - 1) = \frac{1}{2} (5 - 3 - 3 5) .$ $A = (2.5 - 1.5 - 1.5 2.5) .$ Verify karo: $A (11) = (11)$ (= $1 \cdot$ ) ✓, aur $A (1 - 1) = (4 - 4)$ (= $4 \cdot$ ) ✓.

Problem 5.2

Free parameter ko constrain karo. Real number $c$ ki kaunsi value(s) ke liye $A = (20 c 2)$ ka eigenspace 2-dimensional hoga? Characteristic polynomial use karke explain karo.

Recall Solution 5.2

Characteristic polynomial har $c$ ke liye $(2 - λ)^{2}$ hai (triangular), toh $λ = 2$ algebraic multiplicity $2$ ke saath hamesha hoga. Eigenspace hai $Null (A - 2 I) = Null (00 c 0)$ .

Agar $c \neq = 0$ : rank $= 1$ , toh $dim E_{2} = 2 - 1 = 1$ — sirf ek line.
Agar $c = 0$ : matrix zero matrix hai, rank $= 0$ , toh $dim E_{2} = 2 - 0 = 2 = R^{2}$ . Answer: sirf $== c = 0 ==$ 2-dimensional eigenspace deta hai (aur tabhi $A$ diagonalizable hai). Off-diagonal entry exactly wahi hai jo eigenspace ko collapse karti hai.

Wrap-up recall

Recall Poori ladder ka ek-line summary

Eigenvalues $det (A - λ I) = 0$ se aate hain; eigenspaces $Null (A - λ I)$ hain; dimension $n - rank$ hai; repeat count sirf woh dimension bound karta hai; aur aap poori cheez $A = P D P^{- 1}$ ke saath ulta bhi chala sakte ho.

Connections

Finding eigenspaces — woh parent recipe jise yeh exercises drill karti hain.
Characteristic polynomial — jahan in problems ka har $λ$ paida hota hai.
Null space and solving homogeneous systems — yahan har eigenspace ke peeche yahi machinery hai.
Determinants — har characteristic equation mein use hota hai.
Rank-Nullity theorem — $dim E_{λ} = n - rank$ formula.
Diagonalization — L3 aur L5 test karte hain kab yeh succeed karta hai aur ise kaise reverse karein.
Symmetric matrices and spectral theorem — L4 ke orthogonal eigenvectors.