Question

(2) Consider the matrix (a) Find a value of x such that A is diagonalizable. (b) Find a value of x such that A is not diagonalizable.

Answer 1

A = [1 x 0 1 1 x 0 1 1] A - lambda I = [1 - lambda x 0 1 (1 - lambda x 0 1 (1 - lambda)] det (A - lambda I) = 0 implies (1 - lambda) [(1 - lambda)^2 - x] - 1 [x(1 - lambda)] = 0 implies (1 - lambda)^3 - (1 - lambda) x - (1 - lambda)x = 0 implies (1 - lambda) [(1 - lambda)^2 - 2x] = 0 This implies (1 - lambda) 0 = or (1 - lambda)^2 - 2x = 0 implies lambda = 1 or 1 + lambda^2 - 2h - 2x = 0 implies lambda = 1 or lambda^2 - 2 lambda + (1 - 2x) = 0 Now we solve lambda^2 - 2 lambda + (1 - 2x) = 0 D = b^2 - 4ac = (- 2)^2 - 4 times 1 time (1 - 2x) = 4 - (4 - 8x) = 4 - 4 + 8x = 8x \r\n Hence d = -b plusminus squareroot D/2a = 2 plusminus squareroot 8x/2 times 1 = 2 plusminus 2 squareroot 2 times squareroot x/2