Eigenvalues and eigenvectors — characteristic polynomial: Exercises

4.5.29 · D4 · Maths › Linear Algebra (Full) › Eigenvalues and eigenvectors — characteristic polynomial

Shuru karne se pehle, ek shared picture. Ek eigenvector woh direction hoti hai jise matrix turn nahi karta — woh sirf usse stretch karta hai. Figure mein ek generic vector ko rotate aur stretch hote dikhaaya gaya hai (orange), jabki ek eigenvector apni hi line par rehti hai (green).

Figure — Eigenvalues and eigenvectors — characteristic polynomial

Level 1 — Recognition

Exercise L1.1

$A = (4007)$ ke liye, kya $x = (10)$ ek eigenvector hai? Agar haan, toh kis eigenvalue ke saath?

Recall Solution

$A x$ compute karo aur $x$ ke kisi scalar multiple se compare karo. $A x = (4007) (10) = (40) = 4 (10) .$ WHAT humne kiya: matrix apply ki. WHY: definition $A x = λ x$ ek test hai — vector plug in karo aur dekho ki output input ka multiple hai ya nahi. RESULT: output input ka $4$ guna hai, same direction mein, toh haan — $== λ = 4 ==$ ke saath eigenvector hai.

Exercise L1.2

Ek general $2 \times 2$ matrix $A = (a c b d)$ ke liye characteristic polynomial $p (λ) = det (A - λ I)$ symbolically likhein. Uski degree batayein.

Recall Solution

Har diagonal entry se $λ$ subtract karo (yahi matlab hai $A - λ I$ ka: $I$ mein diagonal par $1$ 's hote hain, toh $λ I$ mein wahan $λ$ 's hote hain): $A - λ I = (a - λ c b d - λ) .$ Determinant lo (main-diagonal product minus off-diagonal product): $p (λ) = (a - λ) (d - λ) - b c = λ^{2} - (a + d) λ + (a d - b c) .$ Degree $= 2$ , jo matrix ki size se match karta hai (ek $n \times n$ matrix degree $n$ deta hai).

Level 2 — Application

Exercise L2.1

$A = (2123)$ ke dono eigenvalues nikalo aur unhe trace/determinant check se verify karo.

Recall Solution

$p (λ) = (2 - λ) (3 - λ) - (2) (1) = λ^{2} - 5 λ + 6 - 2 = λ^{2} - 5 λ + 4.$ WHY trace/det shortcut kaam karta hai: $p (λ) = λ^{2} - tr (A) λ + det (A)$ , toh hum expect karte hain coefficient $- tr (A) = - 5$ aur constant $det (A) = 4$ . Factor karo: $λ^{2} - 5 λ + 4 = (λ - 1) (λ - 4) \Rightarrow λ_{1} = 1, λ_{2} = 4.$ Check (dekho Trace of a Matrix aur Determinants): $λ_{1} + λ_{2} = 5 = tr (A)$ ✓, $λ_{1} λ_{2} = 4 = det (A)$ ✓.

Exercise L2.2

$A = (2123)$ ka $λ = 4$ ke liye ek eigenvector nikalo.

Recall Solution

$(A - 4 I) x = 0$ solve karo — eigenvector $A - λ I$ ke null space mein rehta hai. $A - 4 I = (- 2 1 2 - 1) .$ Row 1 kehta hai $- 2 x_{1} + 2 x_{2} = 0 \Rightarrow x_{1} = x_{2}$ . (Row 2 bhi yehi kehta hai: $x_{1} - x_{2} = 0$ .) Ek free direction: $x = (11) .$ Verify: $A (1 1) = (1 + 3 2 + 2) = (4 4) = 4 (1 1)$ ✓.

Exercise L2.3

Triangular matrix $A = 500 9 - 1 0 - 3 42$ ke eigenvalues nikalo.

Recall Solution

WHY koi kaam nahi karna padta: ek triangular matrix ke liye $det (A - λ I) = \prod_{i} (a_{ii} - λ)$ , kyunki ek triangular matrix ka determinant uski diagonal entries ka product hota hai. Toh $p (λ) = (5 - λ) (- 1 - λ) (2 - λ) = 0 \Rightarrow λ = 5, - 1, 2.$ Eigenvalues exactly diagonal entries $== 5, - 1, 2 ==$ hain.

Level 3 — Analysis

Exercise L3.1

$A = (4014)$ ke liye, eigenvalue, uski algebraic multiplicity, aur uski geometric multiplicity nikalo. Kya $A$ defective hai?

Recall Solution

Triangular hai, toh $p (λ) = (4 - λ)^{2}$ . Root $λ = 4$ do baar aata hai → algebraic multiplicity $= 2$ . Geometric multiplicity = independent eigenvectors ki sankhya = $A - 4 I$ ke null space ki dimension: $A - 4 I = (0010) .$ Yeh $x_{2} = 0$ force karta hai, $x_{1}$ free hai: sirf ek direction $(0 1)$ . Toh geometric multiplicity $= 1$ . Kyunki geometric $(1) <$ algebraic $(2)$ , matrix $== defective ==$ hai (diagonalizable nahi — dekho Diagonalization).

Exercise L3.2

Rotation matrix $R = (01 - 1 0)$ har real vector ko $9 0^{\circ}$ rotate karta hai. Characteristic polynomial ke zariye samjhao ki iske koi real eigenvector kyun nahi hain, aur iske complex eigenvalues nikalo.

Recall Solution

$p (λ) = det (- λ 1 - 1 - λ) = (- λ) (- λ) - (- 1) (1) = λ^{2} + 1.$ WHY koi real roots nahi: $λ^{2} + 1 = 0$ ka matlab hai $λ^{2} = - 1$ , aur koi bhi real number negative par square nahi hota. Geometrically yeh picture se match karta hai — ek $9 0^{\circ}$ rotation har nonzero vector ko uski apni line se hataa deta hai, toh koi bhi real direction preserved nahi hoti. Complex Numbers use karke: $λ = \pm i .$ Check: sum $= i + (- i) = 0 = tr (R)$ ✓, product $= i \cdot (- i) = - i^{2} = 1 = det (R)$ ✓.

Exercise L3.3

$A = (1221)$ ke saare eigenvalues aur uske do eigenvectors nikalo. Note karo ki eigenvectors perpendicular hain — yahan yeh expected kyun hai?

Recall Solution

$p (λ) = (1 - λ)^{2} - 4 = λ^{2} - 2 λ - 3 = (λ - 3) (λ + 1) \Rightarrow λ = 3, - 1.$ $λ = 3$ ke liye: $A - 3 I = (- 2 2 2 - 2)$ , toh $- x_{1} + x_{2} = 0$ , deta hai $x = (1 1)$ . $λ = - 1$ ke liye: $A + I = (2222)$ , toh $x_{1} + x_{2} = 0$ , deta hai $x = (- 1 1)$ . Dot product $(1 1) \cdot (- 1 1) = 1 - 1 = 0$ → perpendicular. WHY expected: $A$ symmetric hai ( $A = A^{⊤}$ ), aur symmetric matrices mein hamesha distinct eigenvalues ke liye perpendicular eigenvectors hote hain.

Level 4 — Synthesis

Exercise L4.1

Ek $2 \times 2$ matrix ka $tr (A) = 6$ aur $det (A) = 8$ hai. $A$ jaane bina uske eigenvalues nikalo.

Recall Solution

WHY yeh $A$ ke bina kaam karta hai: ek $2 \times 2$ matrix ka characteristic polynomial sirf trace aur determinant par depend karta hai: $p (λ) = λ^{2} - tr (A) λ + det (A)$ . $λ^{2} - 6 λ + 8 = 0 \Rightarrow (λ - 2) (λ - 4) = 0 \Rightarrow λ = 2, 4.$ Check: $2 + 4 = 6$ ✓, $2 \cdot 4 = 8$ ✓.

Exercise L4.2

Given ki $λ = 2$ , $A = (31 k 2)$ ka eigenvalue hai, $k$ nikalo.

Recall Solution

$λ = 2$ root hai yani $p (2) = det (A - 2 I) = 0$ : $A - 2 I = (11 k 0), det = (1) (0) - (k) (1) = - k .$ Set karo $- k = 0 \Rightarrow k = 0$ . Check: $k = 0$ ke saath, $A = (3102)$ triangular hai diagonal $3, 2$ ke saath — eigenvalues $3$ aur $2$ , aur sach mein $2$ unme se ek hai ✓.

Exercise L4.3

Maano $A = (2112)$ (parent note se, eigenvalues $1$ aur $3$ ). Directly dikhao ki $A^{2}$ ke eigenvalues $1$ aur $9$ hain, yeh rule illustrate karte hue: "agar $A x = λ x$ toh $A^{2} x = λ^{2} x$ ."

Recall Solution

WHY rule kaam karta hai: $A$ do baar apply karo — $A^{2} x = A (A x) = A (λ x) = λ (A x) = λ^{2} x$ . Same eigenvector, eigenvalue squared. Direct confirmation: $A^{2} = (2112)^{2} = (5445)$ . $p_{A^{2}} (μ) = (5 - μ)^{2} - 16 = μ^{2} - 10 μ + 9 = (μ - 1) (μ - 9) \Rightarrow μ = 1, 9.$ Aur sach mein $1 = 1^{2}$ , $9 = 3^{2}$ ✓ — $A$ ke eigenvalues ke squares.

Level 5 — Mastery

Exercise L5.1

Prove karo ki ek matrix $A$ singular hai (matlab $det A = 0$ ) tab aur sirf tab jab $0$ uske eigenvalues mein se ek ho.

Recall Solution

WHAT " $0$ ek eigenvalue hai" ka matlab: $p (0) = det (A - 0 \cdot I) = det (A)$ . Toh $0$ characteristic polynomial ka root tab hoga jab $det (A) = 0$ . $0 is an eigenvalue ⟺ p (0) = 0 ⟺ det (A) = 0 ⟺ A singular .$ Geometric meaning: $A x = 0 \cdot x = 0$ kisi nonzero $x$ ke liye ka matlab hai ki $A$ us direction ko zero par collapse kar deta hai — ek nontrivial null space, jo exactly singularity hai. $■$

Exercise L5.2

$A = (2112)$ ke liye, Cayley–Hamilton theorem verify karo: $A$ apne khud ke characteristic polynomial ko satisfy karta hai, yaani $p (A) = A^{2} - 4 A + 3 I = 0$ .

Recall Solution

Parent note se, $p (λ) = λ^{2} - 4 λ + 3$ . Cayley–Hamilton Theorem kehta hai ki $λ$ ki jagah matrix substitute karne par (aur constant term ki jagah $I$ ) zero matrix milta hai. $A^{2} = (5445), 4 A = (8448), 3 I = (3003) .$ $A^{2} - 4 A + 3 I = (5 - 8 + 3 4 - 4 + 0 4 - 4 + 0 5 - 8 + 3) = (0000) ✓ .$

Exercise L5.3

$3 \times 3$ matrix $A = 214035 00 - 1$ ka characteristic polynomial aur eigenvalues nikalo, aur eigenvalues ka sum aur product batao, unhe trace aur $det$ se confirm karo.

Recall Solution

$A$ lower-triangular hai, toh eigenvalues directly diagonal entries hain: $p (λ) = (2 - λ) (3 - λ) (- 1 - λ) \Rightarrow λ = 2, 3, - 1.$ Sum $= 2 + 3 - 1 = 4 = tr (A)$ (general fact "sum of eigenvalues $=$ trace" sabhi $n$ ke liye valid hai). Product $= 2 \cdot 3 \cdot (- 1) = - 6 = det (A)$ (triangular bhi hai = diagonal ka product). Dono structural facts confirmed ✓.

Recall wrap-up

Recall Quick self-quiz

Eigenvalues nikalne ke liye solve karo ::: $det (A - λ I) = 0$ Diye gaye $λ$ ke liye eigenvectors nikalne ke liye solve karo ::: null space $(A - λ I) x = 0$ Ek triangular matrix ke eigenvalues ::: uski diagonal entries hoti hain Eigenvalues ka sum barabar hota hai ::: trace ke; product barabar hota hai determinant ke Agar $A x = λ x$ , toh $A^{k} x =$ ::: $λ^{k} x$ $0$ ek eigenvalue hai tab aur sirf tab ::: jab $A$ singular ho ( $det A = 0$ )