DET([-1 - k, 2; 2, 1 - k]) = k^2 - 5 = 0
k = - √5 ∨ k = √5
[-1 - √5, 2; 2, 1 - √5] * [a; b] = [0; 0]
a = b·(√5/2 - 1/2)
Eigenvektor zum Eigenwert √5 ist [b·(√5/2 - 1/2); b]
Den anderen kannst du genau so machen
DET([1 - k, 2, 0; 0, 2 - k, 0; 1, 2, 0 - k]) = k·(1 - k)·(k - 2) = 0
k = 2 ∨ k = 1 ∨ k = 0
[1 - 0, 2, 0; 0, 2 - 0, 0; 1, 2, 0 - 0]·[a; b; c] = [0; 0; 0] --> [0; 0; c]
[1 - 1, 2, 0; 0, 2 - 1, 0; 1, 2, 0 - 1]·[a; b; c] = [0; 0; 0] --> [c; 0; c]
[1 - 2, 2, 0; 0, 2 - 2, 0; 1, 2, 0 - 2]·[a; b; c] = [0; 0; 0] --> [c, c/2, c]