Question

Problem 2. In each part below, either diagonalize the given linear transformation, if possible, or else explain why this is impossible. (That is, find a basis B such that the coordinate matrix [T\B or explain why no such basis exists.) (а) Т: Р2 —> Р2 given by T(p) — ар'. (b) Т:P, — P2 given by T(р) — р(2л — 1). (c) T R2x2 R2x2 given by T(A) = A+ AT. (d) T: С +С given by T(a + bi) is diagonal = a - bi -1 Ax, where A = (e) T : R2 ->R2 given by T(ī) -1 3 -2 2 2 (f) T R3 -R3 given by T(F) Ax, where A = 1 -2 5 2 1 TT

Answer 1

T: p_2 rightarrow p_2 given by T(p) = xp' {1, x, x^2} in the standard basis of p_2 so, T(1) = x.0 = 0 = 0.1 + 0.x + 0.x^2 T(x) = x 1 = x = 0 1 + 1 x + 0 x^2 (x^2) = x.2 = x.2 x = 2 x^2 = 0.1 + 0.x + 2x x^2 Therefore the matrix is (0 0 0 0 1 0 0 0 2) The matrix itself a diagonal matrix. so it has three distinct eigen values 0, 1, 2 Hence T is diagonalizable \r\n T: p_2 rightarrow p_2 given by T(p) = p (2x - 1) Therefore (1, x, x^2) is the standard basis of p_2 So, T(1) = 1 (2x - 1) = 2x - 1 = (-1) 1 + 2 x + 0 x^2 T(x) = x(2x - 1) = -x + 2X^2 = 0 1 + (-1) x + 2 x^2