Question

Define T: P2 ? P2 by Tao + ax + a2x2) = (-3a1 + 5a2) + (-4a0 + 4a1-10a2)x + 4a2x2 Find the eigenvalues. (Enter your answers from smallest to largest.) a1, A2, A3) = | |-2.4.6 Find the corresponding coordinate eigenvectors of T relative to the standard basis f1, x, x2 ?,0,1) X2 =?-5,1

Answer 1

File Home sert Page Layout References Mailng Review View Find Calibri (Body) -п! AA, Aa . a Copy e Replace Select Editing Blu-k x, x a-里_a, E. 仨· Normal , No Spaci Heading 1 Heading 2 Title Subtitle Subtle Em. , ㄧ Paragraph we know that {1,x,xy is the standard basis for P2. Also, T(1)--4x, T(x)--344x and T(x2)-5-10x+4x2. Hence the standard matrix of T is A = 0 -4 0 -10 4 It may be observed that the entries in the columns of A are the scalar multiples of 1 and the coefficients of x, x2 in T(1), T(x) and T(x2) respectively The eigenvalues of T are same as the eigenvalues of A which are solutions to its characteristic equation det(A-Als)-0 or, (A-6)(A-4)(A+2) = 0. Thus,人1--2,人2-4 andん-6 are the 3 eigenvalues of T. Further, the eigenvector of A corresponding to the eigenvalue-2 is solution to the equation (A+2l3)X-0. To solve this equation, we will reduce A+213 to its RREF which is 3/2 0 0 0 0 Now, if Χ-(x,y,z)", then the equation (A+21JX-0 is equivalent tox-3y/2-0 or, x-3y/2 and z-0 so that X (3y/2,y,0) y/2(3,2,0).Hence,v1 (3,2,0) is the eigenvector of A corresponding to the eigenvalue -2 Similarly, the eigenvector of A corresponding to the eigenvalue 4 is solution to the equation (A-4l3)X-0. 10:08 AM 5/3201 3

File Home sert Page Layout References Mailng Review View Cut copy Find e Replace Calibri (Body) -п! AA, Aa AaBbCcD AaBbccD AaBbC AaBbCc AaB AaBbCc. AoBbccD Heading 2 . Blu-kx,x' a-里_a, E. 仨· , No Spaci ChangeSelect Normal Heading 1 Title Subtitle Subtle Em. , ㄧ Paragraph Editing The RREF of A-4l, is 5/2 0 0 0 Now, if X- (x,y,z)T, then the equation (A-4ls)X- 0 is equivalent to x +5z/2 0 or, zx--5z/2 and y-5z 0 or, y = 5z so that X =(-Sz/2,5z,z)" = z/2(-5,10,2)".Hence,v,-(-5, 10, 2)" is the eigenvector of A corresponding to the eigenvalue 4.Similarly, the eigenvector of A corresponding to the eigenvalue 6 is solution to the equation (A-6l3)X- 0. The RREF of A-6l3 is 0 0 0 Now, if X- (x,y,z)T, then the equation (A-61s)X0 is equivalent to x ty/2-0 or, x -y/2 and z -0 so thatX = (-y/2,y,0)"-y/2(-1,2,0)".Hence,Vs=(-1,2,0)" is the eigenvector of A corresponding to the eigenvalue 6 The corresponding eigenvectors of T are x1 3+2x, x2--5+10x +2x2 and x3 --1+2x. 10:09 AM 5/3/2018