# EIGEN VALUES OF A MATRIX

A linear operator A is a zero of its characteristic polynomial. In practical calculations, we set the characteristic polynomial equal to zero, giving the characteristic equation

det|A - λI| = 0

The zeros of the characteristic polynomial, which are the solutions to this equation, are the eigenvalues of the matrix A.

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| = λ- 7λ - 8

Setting this equal to zero gives the characteristic equation.

λ- 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

Note :

For a 2 × 2 matrix, we found two eigenvalues. This is because a 2 × 2 matrix leads to a second-order characteristic polynomial.

This is true in general ; an n × n matrix will lead to an nth order characteristic polynomial with n (not necessarily distinct) solutions.

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)(λ- 6λ + 8) - (2 - λ)

Factor (2 - λ) out.

det|A - λI| = (2 - λ)(λ- 6λ + 8 - 1)

det|A - λI| = (2 - λ)(λ- 6λ + 7)

Setting this equal to zero gives the characteristic equation.

(2 - λ)(λ- 6λ + 7) = 0

2 - λ = 0  or  λ- 6λ + 7 = 0

Solving 2 - λ = 0, we get

λ = 2

Solve λ- 6λ + 7 = 0.

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

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

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

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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