MA 305 Spring 2K Homework 6 Solutions > read("/afs/eos.ncsu.edu/users/k/kaltofen/www/courses/LinAlgebra/Maple/initlib.mpl"): > with(refpkg): > with(lsqpkg): > with(linalg): Warning, new definition for norm Warning, new definition for trace Problem 1 (a.) T(a*(x1,y1,z1) + b*(x2,y2,z2)) = T((a*x1+b*x2,a*y1+b*y2,a*z1+b*z2) = ((a*z1+b*z2)-(a*x1+b*x2), (a*z1+b*z2)-(a*y1+b*y2)) = (a*(z1-x1) + b*(z2-x2), a*(z1-y1) + b*(z2-y2)) = a*(z1-x1,z2-y2) + b*(z2-x2,z2-y2) = a*T(x1,y1,z1) + b*T(x2,y2,z2) Therefore, this is a linear transormation. The matrix is > A:=matrix(2,3,[-1,0,1,0,-1,1]); [-1 0 1] A := [ ] [ 0 -1 1] which is the matrix that satisfies T(x,y,z) = A*(x,y,z). (b.) T(a*(x1,y1) + b*(x2,y2)) = T((a*x1+b*x2,a*y1+b*y2,a*z1+b*z2) = ((a*x1+b*x2)-(a*y1+b*y2), (a*x1+b*x2)+(a*y1+b*y2), 2*(a*x1+b*x2)-3*(a*y1+b*y2)) = (a*(x1-y1) + b*(x2-y2), a*(x1+y1) + b*(x2+y2), a*(2*x1-3*y1) + b*(2*x2-3*y2)) = a*(x1-y1, x1+y1, 2*x1-3*y1) + b*(x2-y2, x2+y2, 2*x2-3*y2) = a*T(x1,y1) + T(x2,y2) Therefore, this is a linear transormation. The matrix is > A:=matrix(3,2,[1,-1,1,1,2,-3]); [1 -1] [ ] A := [1 1] [ ] [2 -3] which is the matrix that satisfies T(x,y) = A*(x,y). (c.) Consider the matrix > A:=matrix(2,2,[1,0,1,0]); [1 0] A := [ ] [1 0] Then, T(A+A) = T(2A) which is > matrix(2,2,[16,0,0,0]); [16 0] [ ] [ 0 0] but T(A)+T(A) = 2T(A) which is > matrix(2,2,[4,0,0,0]); [4 0] [ ] [0 0] Thus this is not a linear transformation. Problem 2 > A := matrix(4,4,[5,1,2,0,-1,0,2,0,3,0,-2,0,2,1,2,1]); [ 5 1 2 0] [ ] [-1 0 2 0] A := [ ] [ 3 0 -2 0] [ ] [ 2 1 2 1] (a.) > p:=charpoly(A,'lambda'); 4 3 2 p := lambda - 4 lambda - 12 lambda + 11 lambda + 4 > factor(p); 3 2 (lambda - 1) (lambda - 3 lambda - 15 lambda - 4) (b.) > lambda := eigenvals(A); (1/3) 1 (1/3) lambda := 1, 1/2 %1 + 12 ------- + 1, - 1/4 %1 (1/3) %1 1 / (1/3) 1 \ - 6 ------- + 1 + 1/2 I sqrt(3) |1/2 %1 - 12 -------|, (1/3) | (1/3)| %1 \ %1 / (1/3) 1 - 1/4 %1 - 6 ------- + 1 (1/3) %1 / (1/3) 1 \ - 1/2 I sqrt(3) |1/2 %1 - 12 -------| | (1/3)| \ %1 / %1 := 84 + 12 I sqrt(47) > simplify(lambda[1]);combine(lambda[2]);simplify(lambda[3]);simplify(lambda[4]); 1 (1/3) 1 1/2 (84 + 12 I sqrt(47)) + 12 ------------------------- + 1 (1/3) (84 + 12 I sqrt(47)) (2/3) (1/3) (2/3) 1/4 (-%1 - 24 + 4 %1 + I sqrt(3) %1 - 24 I sqrt(3)) / (1/3) / %1 / %1 := 84 + 12 I sqrt(47) (2/3) (1/3) (2/3) %1 + 24 - 4 %1 + I sqrt(3) %1 - 24 I sqrt(3) - 1/4 ----------------------------------------------------------- (1/3) %1 %1 := 84 + 12 I sqrt(47) These are the four eigenvalues of the matrix. Notice three of them are complex. (c.) > eigenvectors(A); (1/3) 1 [ [1, 1, {[0, 0, 0, 1]}], [1/2 %1 + 12 ------- + 1, 1, {[ (1/3) [ %1 [ (1/3) 1 (1/3) 1 1/6 %1 + 4 ------- + 1, - 1/2 %1 - 12 ------- (1/3) (1/3) %1 %1 / (1/3) 1 \2 11 (1/3) - 19/3 + 1/3 |1/2 %1 + 12 ------- + 1| , 1, -- %1 | (1/3) | 42 \ %1 / 1 / (1/3) 1 \2 + 44/7 ------- + 15/7 - 2/21 |1/2 %1 + 12 ------- + 1| (1/3) | (1/3) | %1 \ %1 / ] [ (1/3) 1 ]}], [%4, 1, {[- 1/12 %1 - 2 ------- + 1 + 1/6 %2, ] [ (1/3) ] [ %1 (1/3) 1 2 1/4 %1 + 6 ------- - 19/3 - 1/2 %2 + 1/3 %4 , 1, (1/3) %1 11 (1/3) 1 11 2] - -- %1 - 22/7 ------- + 15/7 + -- %2 - 2/21 %4 ]}], [ 84 (1/3) 42 ] %1 ] [ (1/3) 1 %3, 1, {[- 1/12 %1 - 2 ------- + 1 - 1/6 %2, [ (1/3) [ %1 (1/3) 1 2 1/4 %1 + 6 ------- - 19/3 + 1/2 %2 + 1/3 %3 , 1, (1/3) %1 11 (1/3) 1 11 2] - -- %1 - 22/7 ------- + 15/7 - -- %2 - 2/21 %3 ]}] 84 (1/3) 42 ] %1 ] %1 := 84 + 12 I sqrt(47) / (1/3) 1 \ %2 := I sqrt(3) |1/2 %1 - 12 -------| | (1/3)| \ %1 / (1/3) 1 %3 := - 1/4 %1 - 6 ------- + 1 - 1/2 %2 (1/3) %1 (1/3) 1 %4 := - 1/4 %1 - 6 ------- + 1 + 1/2 %2 (1/3) %1 Problem 3 (a.) > A := matrix(3,3,[2,0,-1,2,-1,0,-12,0,1]); [ 2 0 -1] [ ] A := [ 2 -1 0] [ ] [-12 0 1] > sol := eigenvectors(A); sol := [-1, 1, {[0, 1, 0]}], [-2, 1, {[1, -2, 4]}], [5, 1, {[3, 1, -9]}] > u := C[1]*exp(-2*t) + 3*C[3]*exp(5*t); u := C[1] exp(-2 t) + 3 C[3] exp(5 t) > v := (-2)*C[1]*exp(-2*t) + C[2]*exp(-t) + C[3]*exp(5*t); v := -2 C[1] exp(-2 t) + C[2] exp(-t) + C[3] exp(5 t) > w := 4*C[1]*exp(-2*t) + (-9)*C[3]*exp(5*t); w := 4 C[1] exp(-2 t) - 9 C[3] exp(5 t) These are the three general solutions to the system. (b.) > sols:=solve(simplify(subs(t=0,{u=7,v=4,w=3})),{C[1],C[2],C[3]}); 25 sols := {C[3] = --, C[1] = 24/7, C[2] = 29/3} 21 > assign(sols); > u; 24/7 exp(-2 t) + 25/7 exp(5 t) > v; 25 - 48/7 exp(-2 t) + 29/3 exp(-t) + -- exp(5 t) 21 > w; 96/7 exp(-2 t) - 75/7 exp(5 t) These are the three particular solutions for the given initial conditions.