EIGEN VALUES OF A MATRIX

Example 1 :

Find the eigenvavlues of

To find the eigen values, solve the following equation.

det|A - λI| = 0

where I is the 2 × 2 identity matrix.

The characteristic polynomial in this case is

det|A - λI| = (5 - λ)(2 - λ) - 18

det|A - λI| = 10 - 5λ - 2λ + λ² - 18

det|A - λI| = λ² - 7λ - 8

Setting this equal to zero gives the characteristic equation.

λ² - 7λ - 8 = 0

Solve the above quadratic equation by factoring.

(λ - 8)(λ + 1) = 0

λ = 8  or  λ = -1

Therefore, the two eigenvalues of the given matrix are

λ₁ = 8

λ₂ = -1

Example 2 :

Find the eigenvavlues of

To find the eigen values, solve the following equation.

det|B - λI| = 0

where I is the 3 × 3 identity matrix.

The characteristic polynomial in this case is

det|A - λI| = (2 - λ)[(4 - λ)(2 - λ) - 0] - 1[1(2 - λ) - 0] + 0

det|A - λI| = (2 - λ)[8 - 4λ - 2λ + λ²] - 1[2 - λ]

det|A - λI| = (2 - λ)(λ² - 6λ + 8) - (2 - λ)

Factor (2 - λ) out.

det|A - λI| = (2 - λ)(λ² - 6λ + 8 - 1)

det|A - λI| = (2 - λ)(λ² - 6λ + 7)

Setting this equal to zero gives the characteristic equation.

(2 - λ)(λ² - 6λ + 7) = 0

2 - λ = 0  or  λ² - 6λ + 7 = 0

Solving 2 - λ = 0, we get

λ = 2

Solve λ² - 6λ + 7 = 0.

The quadratic equation λ² - 6λ + 7 = 0 can not be solved by factoring. So, we have to use quadratic formula.

Comparing λ² - 6λ + 7 = 0 and ax² + bx + c = 0, we get

a = 1, b = -6, c = 7

Quadratic Formula :

λ = 3 ± √2

λ = 3 + √2  or  3 - √2

Therefore, the three eigenvalues of the given matrix are

λ₁ = 2

λ₂ = 3 + √2

λ₃ = 3 - √2

Notice that since B is a 3 × 3 matrix, it has three eigenvalues.

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