B = T^{-1} * A * T
T * B = A * T
mit T = [a, b; c, d] gilt
[a, b; c, d]·[2/3, - 5/3; 11/3, 4/3] = [1, -2; 3, 1]·[a, b; c, d]
[(2·a + 11·b)/3, (4·b - 5·a)/3; (2·c + 11·d)/3, (4·d - 5·c)/3] = [a - 2·c, b - 2·d; 3·a + c, 3·b + d]
Man kann das lineare Gleichungssystem lösen:
a = (b + 6·d)/5 ∧ c = (d - 9·b)/5
a = 1
c = 2
b = -1
d = 1
Also ist
T = [1, -1; 2, 1]
[1, -1; 2, 1]·[2/3, - 5/3; 11/3, 4/3] = [1, -2; 3, 1]·[1, -1; 2, 1]
[-3, -3; 5, -2] = [-3, -3; 5, -2]
Das passt also.