EigenValues & EigenVectors

Question

Find the eigenvalues λ1<λ2<λ3 and associated unit vectors v1,v2,v3 of the symmetric matrix

A =

Answer 1

To find eigenvalues use the characteristic equation given by

det(A-λ1) = 0, where 1 is the identity matrix.

This gives

λ³ - 3 λ² - 54 λ = 0

λ (λ + 6) (λ - 9) = 0

λ = 0, -6, 9

For λ = 0

You need to find the null space for

[[2,2,5],[2,2,5],[5,5,-1]] as A x - 0 1 x = (A x - 0 1) x = A x = 0.

REFF(A) = [[1,1,0],[0,0,1],[0,0,0]]

which gives

x₁ = -x₂

x₂ = x₂

x₃ = 0

[x₁, x₂, x₃] = x₂ [1, -1, 0]

With eigenvector [1, -1, 0].

To normalize the vector you divide by the length

u = v/√(v•v)

u = [1/√2, -1/√2, 0]

For λ = -6

You need to find the null space for

[[8,2,5],[2,8,5],[5,5,5]] as A x + 6 1 x = (A x + 6 1) x = 0.

REFF(A + 6 1) = [[1,0,1/2],[0,1,1/2],[0,0,0]]

which gives

x₁ = -1/2 x₃

x₂ = -1/2 x₃

x₃ = x₃

[x₁, x₂, x₃] = x3 [-1/2, -1/2, 1]

With eigenvector [-1, -1, 2].

To normalize the vector you divide by the length

u = v/√(v•v)

u = [-1/√6, -1/√6, 2/√6]

For λ = 9

You need to find the null space for

[[-7,2,5],[2,-7,5],[5,5,-8]] as A x + 6 1 x = (A x - 9 1) x = 0.

REFF(A - 9 1) = [[1,0,-1],[0,1,-1],[0,0,0]]

which gives

x₁ = x₃

x₂ = x₃

x₃ = x₃

[x₁, x₂, x₃] = x3 [1, 1, 1]

With eigenvector [1, 1, 1].

To normalize the vector you divide by the length

u = v/√(v•v)

u = [1/√3, 1/√3, 1/√3]