Question

Let a be a 66 matrix with characteristic polynomial f() = (1+)(1)2(2)3. Prove that it is not possible to nd three linearly independent vectors v1; v2; v3 in r6 such that av1 = v1; av2 = v2 and av3 = v3.

Answer 1

p(x) =4x² - 4x - 3

=> 4x² - 6x + 2x - 3

=> 2x( 2x -3) + 1(2x + 3)

=> (2x + 1) (2x -3)

p(x) = 0

(2x + 1) = 0 or (2x-3) = 0

x=-1/2 or x= 3/2

Hence, -1/2 and 3/2 are the zeroes of p(x).