Symmetric matrices — spectral theorem (real eigenvalues, orthogonal eigenvectors): Worked examples

4.5.38 · D3 · Maths › Linear Algebra (Full) › Symmetric matrices — spectral theorem (real eigenvalues, ort

Yeh page parent topic ki "worked-out drill" child hai. Parent ne prove kiya kyun real symmetric matrices aisa behave karti hain. Yahan hum har tarah ka case cover karenge jo ek problem mein aa sakta hai, aur har ek ko numeric answer tak grind karenge.

Shuru karne se pehle, ek plain-words reminder taaki koi cheez use hone se pehle build ho jaye.

The scenario matrix

Har symmetric-matrix problem inhi cells mein se ek mein aati hai. Neeche ke examples labeled hain us cell se jo woh cover karte hain, taaki saath milke koi gap na rahe.

#	Case class	Kya unusual hai	Example
C1	Dono eigenvalues positive	ordinary "stretch/stretch"	Ex 1
C2	Mixed signs (ek $+$ , ek $-$ )	andar ek reflection chhupa hai	Ex 2
C3	Ek zero eigenvalue (singular $A$ )	$A$ ek poori direction ko $0$ par squash karta hai	Ex 3
C4	Repeated eigenvalue (multiplicity 2)	eigenvectors forced nahi — tum choose karte ho	Ex 4
C5	Diagonal $A$ (degenerate/limiting)	eigenvectors khud axes hain	Ex 5
C6	$3 \times 3$ ek clean block ke saath	3D mein poora spectral decomposition	Ex 6
C7	Word problem (real-world)	pehle symmetric matrix banana padega	Ex 7
C8	Exam twist — $A$ ko $Q, Λ$ se rebuild karo	theorem ko ulta chalao	Ex 8

Example 1 — C1: both eigenvalues positive

Worked example Statement

$A = (4114)$ ko orthogonally diagonalize karo.

Forecast: trace $= 8$ , det $= 16 - 1 = 15$ . Kaunse do numbers $8$ mein add hote hain aur $15$ mein multiply? Aage padhne se pehle guess karo.

Steps:

$λ^{2} - 8 λ + 15 = 0 = (λ - 5) (λ - 3)$ solve karo, toh $λ = 5, 3$ . Yeh step kyun? Characteristic quadratic directly trace aur det se banti hai — haath se $det (A - λ I)$ expand karne ki zaroorat nahi.
$λ = 5$ : $(A - 5 I) x = (- 1 1 1 - 1) x = 0$ solve karo. Dono rows kehti hain $- x_{1} + x_{2} = 0$ , toh $x = (1, 1)$ . Yeh step kyun? Ek eigenvector koi bhi nonzero vector hai jise shifted matrix $0$ par bhej de; dono rows agree karte hain, jo confirm karta hai eigenvalue sahi hai.
$λ = 3$ : $(A - 3 I) x = (1111) x = 0 \Rightarrow x = (1, - 1)$ .
Normalize karo (har ek ko uski length $2$ se divide karo): $q_{1} = \frac{1}{2} (1, 1)$ , $q_{2} = \frac{1}{2} (1, - 1)$ . Yeh step kyun? Sirf unit-length columns hi $Q^{⊤} = Q^{- 1}$ banate hain, jo $P Λ P^{- 1}$ ko tidy $Q Λ Q^{⊤}$ mein turn karta hai. Dekho Diagonalization aur Orthogonal Matrices. $Q = \frac{1}{2} (11 1 - 1), Λ = (5003) .$

Verify: dot product $(1, 1) \cdot (1, - 1) = 0$ ✓ (orthogonal, jaise distinct eigenvalues ke liye promise tha). Rebuild: $Q Λ Q^{⊤} = (4114) = A$ ✓. Dono $λ > 0$ , toh $A$ stretch karta hai (kabhi flip nahi karta) — yeh positive definite hai, dekho Positive Definite Matrices.

Dono eigenvalues positive matlab har direction outward push hoti hai. Neeche ki picture eigen-axes dikhati hai (woh sirf do directions jinhein $A$ same pointing chhod deta hai) aur kaise ek circle balloon hokar ellipse ban jaati hai.

Figure — Symmetric matrices — spectral theorem (real eigenvalues, orthogonal eigenvectors)

Example 2 — C2: mixed signs (a reflection inside)

Neeche ki magenta arrow reflect hoti eigen-direction hai: same line, reversed arrow.

Example 3 — C3: a zero eigenvalue (singular matrix)

Worked example Statement

$A = (1111)$ ko diagonalize karo.

Forecast: trace $= 2$ , det $= 1 - 1 = 0$ . Ek zero determinant matlab ek eigenvalue $0$ hai (eigenvalues ka product $0$ hai). Doosra poora trace hona chahiye. Unhe guess karo.

Steps:

$λ^{2} - 2 λ + 0 = 0 = λ (λ - 2)$ , toh $λ = 2, 0$ . Yeh step kyun? Ek zero eigenvalue matlab $A$ ek poori direction ko origin par flatten kar deta hai — woh direction null space hai.
$λ = 2$ : $(- 1 1 1 - 1) x = 0 \Rightarrow x = (1, 1)$ .
$λ = 0$ : seedha $A x = 0$ $\Rightarrow (1, 1) \cdot (x_{1}, x_{2})$ ... $x_{1} + x_{2} = 0$ solve karo $\Rightarrow x = (1, - 1)$ . Yeh step kyun? $λ = 0$ ka eigenvector null space hi hai: line $x = (t, - t)$ par sab kuch $0$ par bhej diya jaata hai.
$Q = \frac{1}{2} (11 1 - 1)$ , $Λ = (2000)$ .

Verify: $A (1, - 1) = (0, 0)$ ✓ (squash hua). $A (1, 1) = (2, 2) = 2 (1, 1)$ ✓. Spectral sum: $A = 2 q_{1} q_{1}^{⊤} = 2 \cdot \frac{1}{2} (1111) = A$ ✓ ( $λ = 0$ wala piece kuch contribute nahi karta — exactly yahi reason hai kyun singular matrices rank lose karti hain).

Neeche ki orange line squashed direction hai — us par har point origin par land karta hai.

Example 4 — C4: a repeated eigenvalue

Worked example Statement

$A = (3003)$ ko diagonalize karo... ruko, yeh toh bahut easy hai. Disguised version $A = (3003)$ lo lekin prove karo ki eigenvalue repeat hota hai aur koi bhi orthonormal pair kaam karta hai.

Forecast: trace $= 6$ , det $= 9$ . Do numbers jo $6$ mein sum aur $9$ mein multiply hote hain: dono $3$ hone chahiye. Toh eigenvectors kya hain?

Steps:

$λ^{2} - 6 λ + 9 = (λ - 3)^{2}$ , toh $λ = 3$ twice. Yeh step kyun? Ek repeated root warn karta hai ki eigenvector ek single line par force nahi hai.
$(A - 3 I) x = (0000) x = 0$ solve karo. Har vector yeh solve karta hai. Yeh step kyun? Shifted matrix sab zeros hai, toh eigenspace poora plane hai — ek 2D eigenspace, kabhi defective nahi (parent ka promise symmetric matrices ke liye).
Kyunki koi bhi do perpendicular unit vectors valid $Q$ banate hain, axes $q_{1} = (1, 0)$ , $q_{2} = (0, 1)$ choose karo, YA rotated pair $\frac{1}{2} (1, 1), \frac{1}{2} (1, - 1)$ . Dono dete hain $Λ = (3003)$ . Yeh step kyun? Jab eigen-directions distinguish nahi kar sakte, tum ek orthonormal basis choose karte ho (higher dimensions mein Gram-Schmidt Process use karo).

Verify: Rotated pair ke liye, $Q Λ Q^{⊤} = 3 Q Q^{⊤} = 3 I = A$ ✓ kyunki kisi bhi orthogonal $Q$ ke liye $Q Q^{⊤} = I$ . Scalar-multiple-of-identity tumhari axis choice ke liye invariant hoti hai — yahi "repeated eigenvalue" ka geometric meaning hai.

Example 5 — C5: diagonal matrix (limiting / degenerate case)

Example 6 — C6: a $3 \times 3$ block matrix (full 3D decomposition)

Worked example Statement

$A = 500021012$ ko diagonalize karo.

Forecast: Top-left $2 \times 2$ bottom block se decoupled hai. Teen eigenvalues kya hain?

Steps:

Isolated $5$ deta hai $λ = 5$ eigenvector $(1, 0, 0)$ ke saath. Yeh step kyun? Row/column 1 mein koi coupling nahi, toh $e_{1}$ seedha pass hota hai: $A e_{1} = 5 e_{1}$ .
Bottom block $(2112)$ : trace $4$ , det $3$ , toh $λ = 3, 1$ ( $y, z$ slots mein eigenvectors $(1, 1)$ aur $(1, - 1)$ ). Yeh step kyun? Sub-block par Ex 1 ka exact $2 \times 2$ trick reuse karo; block-diagonal problems independent chhote problems mein split hoti hain.
Full eigenvectors: $λ = 5 : (1, 0, 0)$ ; $λ = 3 : (0, 1, 1)$ ; $λ = 1 : (0, 1, - 1)$ . Last do ko $2$ se normalize karo. $Q = 100 0 \frac{1}{2} \frac{1}{2} 0 \frac{1}{2} - \frac{1}{2}, Λ = 500030001 .$

Verify: Teeno eigenvectors pairwise orthogonal: $(1, 0, 0) \cdot (0, 1, 1) = 0$ , $(1, 0, 0) \cdot (0, 1, - 1) = 0$ , $(0, 1, 1) \cdot (0, 1, - 1) = 1 - 1 = 0$ ✓. Rebuild $Q Λ Q^{⊤} = A$ ✓. Yeh Singular Value Decomposition se link karta hai: symmetric positive-definite $A$ ke liye, SVD equals spectral decomposition.

Example 7 — C7: word problem (build the matrix first)

Worked example Statement

Ek stretchy sheet deform hoti hai taaki position $(x, y)$ par point move ho $(x + 0.5 y, 0.5 x + y)$ par. Strain symmetric hai. Do perpendicular directions of pure stretch aur unke stretch factors dhundho.

Forecast: Transformation matrix hai $A = (1 0.5 0.5 1)$ . Trace $= 2$ , det $= 1 - 0.25 = 0.75$ . Stretch factors guess karo.

Steps:

Confirm karo $A$ symmetric hai: off-diagonals dono $0.5$ ✓, toh Spectral Theorem apply hota hai aur pure-stretch axes exist karti hain. Yeh step kyun? Physically, ek symmetric strain mein principal axes hote hain bina shear ke — theorem unhe guarantee karta hai.
$λ^{2} - 2 λ + 0.75 = 0$ . $λ = \frac{2 \pm 4 - 3}{2} = \frac{2 \pm 1}{2}$ use karke, toh $λ = 1.5, 0.5$ . Yeh step kyun? Yeh stretch factors hain: $1.5$ (50% stretch) aur $0.5$ (half tak compress).
$λ = 1.5$ : direction $(1, 1)$ ( $4 5^{\circ}$ diagonal). $λ = 0.5$ : direction $(1, - 1)$ (anti-diagonal). Yeh step kyun? $(1, 1)$ ke along sheet sabse zyada stretch hoti hai; perpendicular $(1, - 1)$ ke along compress hoti hai. Yeh exactly ek ellipse ki axes ki Quadratic Forms geometry hai.

Verify: $A (1, 1) = (1.5, 1.5) = 1.5 (1, 1)$ ✓; $A (1, - 1) = (0.5, - 0.5) = 0.5 (1, - 1)$ ✓. Directions perpendicular: $(1, 1) \cdot (1, - 1) = 0$ ✓. Units: stretch factors dimensionless ratios hain (new length / old length) ✓.

Example 8 — C8: exam twist (run the theorem backwards)

Worked example Statement

Tumhe diya gaya hai $Q = \frac{1}{2} (11 - 1 1)$ aur $Λ = (6002)$ . Symmetric matrix $A = Q Λ Q^{⊤}$ reconstruct karo aur confirm karo ki yeh symmetric hai.

Forecast: Eigenvalues $6$ aur $2$ (dono positive) aur $4 5^{\circ}$ -rotated eigen-axes ke saath, guess karo ki kya $A$ ke equal diagonal entries hain aur nonzero off-diagonal.

Steps:

Pehle $Λ Q^{⊤}$ : $Λ Q^{⊤} = (6002) \cdot \frac{1}{2} (1 - 1 11) = \frac{1}{2} (6 - 2 62)$ . Yeh step kyun? Right-to-left multiply karne se har matrix chota rehta hai; $Q^{⊤}$ woh $Q$ hai jiske rows aur columns swap hain.
Phir $Q (Λ Q^{⊤}) = \frac{1}{2} (11 - 1 1) \cdot \frac{1}{2} (6 - 2 62) = \frac{1}{2} (8448) = (4224)$ . Yeh step kyun? Yeh spectral theorem constructively use hai — tum chosen orthonormal axes aur chosen eigenvalues se koi bhi symmetric matrix manufacture kar sakte ho.

Verify: $A = (4224)$ mein $A = A^{⊤}$ ✓ (off-diagonals dono $2$ ). Trace $= 8 = 6 + 2$ ✓, det $= 16 - 4 = 12 = 6 \times 2$ ✓. Recovered eigenvalues $Λ$ se match karte hain.

Recall Quick self-test

Ek symmetric $2 \times 2$ matrix ke liye $det A < 0$ diya hai, eigenvalues ke baare mein kya keh sakte ho? ::: Unki opposite signs hain (product negative hai), toh $A$ ek eigen-axis ke along reflect karta hai (Case C2). Ek symmetric matrix ka $det = 0$ hai. Ek eigenvalue kya equal hai, aur geometrically kya hota hai? ::: Ek eigenvalue $0$ hai; woh eigen-direction origin par squash ho jaati hai — $A$ singular hai (Case C3). Ek symmetric matrix par repeated eigenvalue ke liye, kya eigenvectors uniquely determined hain? ::: Nahi — eigenspace full-dimensional hai, toh tum ek orthonormal basis choose karte ho (Gram–Schmidt) (Case C4).