Whats the answer for these eigenvalues and eigenvectors questions?

Question

Q1:verify that λ_i is an eigenvalue of A and x_i is a corrcsponding eigenvector.

1-A=⌈ 1 0 ⌉

⌊ 0 -1⌋

λ₁ = 1, x₁ = (1,0)

λ₂ =-1,x₂ = (0,1)

2-A=⌈ 4 -5 ⌉

⌊ 2 -3 ⌋

λ₁ = -1, x₁ = (1,1)

λ₂ = 2,x₂ = (5,2)

Q2:determin whether x is an eigenvector of A.

1-A=⌈ 7 2 ⌉

⌊ 2 4 ⌋

(a) x =(1,2)

(b) x =(2,1)

(c) x =(1,-2)

(d) x =(-1,0)

Q3:find (a) the characteristic equation and (b) the eigenvalues (and corresponding eigenvectors) of the matrix.

1-⌈ 6 -3 ⌉

⌊ -2 1 ⌋

2-⌈ 2 -2 3 ⌉

0 3 -2

⌊ 0 -1 2 ⌋

Answer 1

The second part of Q3 is the interesting part.

The secular equation is obtained by replacing the first diagonal element with 2 - λ , the second diagonal element with 3 - λ , the third diagonal element with 2 - λ , leaving the off-diagonal elements as before and then setting the determinant to zero. This results in the equation:

(2 - λ)² (3 - λ) - 2 ( 2 - λ) = 0 ( the secular equation)

by inspection it is seen that one solution is λ = 2 . The other two solutions can be obtained by dividing through by ( 2 - λ) to produce : (2- λ) (3 - λ) -2 = 0. This quadratic equation is easily solved giving

the other two solutions λ = 1 and λ = 4

So the three eigen values are 1, 2 , 4

The eigen vector corresponding to λ =1 is found by writing M V₁ = V₁ (where M is the given matrix) and then requiring that V₁ ⋅ V₁ = 1 . This is a bit messy, but straightforward. The result is

V₁ = sqrt(1/3) ( 1, -1, -1 )^T

Similarly, the equation for V₂ is M V₂ = 2 V₂ (and also V₂ ⋅ V₂ =1 )

V₂ = (1, 0 , 0)^T

For V₃ M V₃ = 4 V₃ (and V₃ ⋅ V₃ = 1 )

V₃ = sqrt(1/69) (7 , -4, 2)^T