Eigenvalues & Eigenvectors

Given 2×2 matrix A:

[1, 2] [3, 4]

For eigenvalues, we solve det(A - λI) = 0

A - λI = [1-λ, 2]

[3, 4-λ]

det(A - λI) = (1-λ)(4-λ) - (2)(3)

Expanding:

= 1·4 - 1λ - 4λ + λ² - 6

= λ² - 5λ + -2

Characteristic polynomial:

λ² - 5λ + -2 = 0

Where:

• Trace(A) = 1 + 4 = 5

• det(A) = 1·4 - 2·3 = -2

Using quadratic formula:

λ = (trace ± √(trace² - 4·det)) / 2

λ = (5 ± √(5² - 4·-2)) / 2

λ = (5 ± √33) / 2

λ₁ = (5 + 5.744563) / 2 = 5.372281

λ₂ = (5 - 5.744563) / 2 = -0.372281

Two distinct real eigenvalues

Solve (A - λI)v = 0:

[-4.372281, 2][v₁] [0]

[3, -1.372281][v₂] = [0]

From first equation: -4.372281v₁ + 2v₂ = 0

Eigenvector v1 = [-2, -4.372281]ᵀ

(or any scalar multiple)

Solve (A - λI)v = 0:

[1.372281, 2][v₁] [0]

[3, 4.372281][v₂] = [0]

From first equation: 1.372281v₁ + 2v₂ = 0

Eigenvector v2 = [-2, 1.372281]ᵀ

(or any scalar multiple)

Verify Av = λv for each eigenvalue:

For λ1 = 5.372281:

Av = [-10.744563, -23.489125]ᵀ

λv = [-10.744563, -23.489125]ᵀ

Av = λv ✓

For λ2 = -0.372281:

Av = [0.744563, -0.510875]ᵀ

λv = [0.744563, -0.510875]ᵀ

Av = λv ✓

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EIGENVALUE RESULTS

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Eigenvalues:

λ₁ = 5.372281

λ₂ = -0.372281

Eigenvectors:

v₁ = [-2, -4.372281]ᵀ

v₂ = [-2, 1.372281]ᵀ

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