Ma 309: Matrix Algebra Model Solutions to Homework Assignment 6 Prof. Wickerhauser Exercise 4 of Section 4.1, p.196. Write (7,5) = 4*(1,2)+3*(1,-1). Then L(7,5) = L(4*(1,2)+3*(1,-1)) = 4*L(1,2)+3*L(1,-1) by linearity, = 4*(-2,3)+3*(5,2) as given, = (7,18). Exercise 7 of Section 4.1, p.196. a. Linear, by Theorem 1.3.2, parts 1,6,9. b. Linear, by the Algebraic Rules for transposes, p.56. c. Not linear, since L(0)=I != 0. d. Linear, by the rules on p.56 and Theorem 1.3.2. Exercise 8 of Section 4.1, p.196. a. Linear, since for any p,q in P2 and a,b in R, L(a*p(x)+b*q(x)) = x*[a*p(x)+b*q(x)] = a*x*p(x)+b*x*q(x) = a*L(p(x)) + b*L(q(x)), and the result is a polynomials of degree at most 3. b. Not linear, since L(0)=x^2 != 0. c. Linear, since it is the sum of linear transformations p(x)-->p(x), p(x)-->x*p(x), and p(x)-->x^2*p(x), all of which produce polynomials of degree 3 or less from a polynomial p(x) of degree 2 or less. Exercise 9 of Section 4.1, p.196. L is a linear transformation because the definite integral of a*f(t)+b*g(t) is a linear combination of their definite integrals a*F(x)+b*G(x). Riemann's theorem on integrability states that the defeinite integral of a continuous function exists on any closed and bounded interval [0,x], so F(x) is defined at each x in [0,1] if f is continuous on [0,1], so L is defined on C([0,1]). The Fundamental Theorem of the Calculus asserts that F(x) is continuous in x, so L takes values in C([0,1]). We may use the Fundamental Theorem to evaluate: L(e^x) = e^x-1; L(x^2) = (1/3)x^3. Exercise 15 of Section 4.1, p.197. Note that for any scalars a,b and vectors u,v, L2(L1(a*u+b*v)) = L2(a*L1(u)+b*L1(v)), by lin. of L1, = a*L2(L1(u))+b*L2(L1(v)), by lin. of L2, so u --> L2(L1(u)) is a linear transformation. Exercise 19 of Section 4.1, p.197. Check the requirements: (i) for any scalar a and vector v, v in L^{-1}(T) ==> L(v) in T definition ==> L(a*v)=a*L(v) in T lin. of L ==> a*v in L^{-1}(T) definition. (ii) for any vectors u,v, u,v in L^{-1}(T) ==> L(u),L(v) in T definition ==> L(u)+L(v)=L(u+v) in T lin. of L ==> u+v in L^{-1}(T) definition. Hence L^{-1}(T) is a subspace. Exercise 2abc of Section 4.2, p.208. a. 1 1 0 b. 1 0 0 c. -1 1 0 0 0 0 0 1 0 0 -1 1 Exercise 6 of Section 4.2, p.209. A = 1 0 0 1 1 1 Exercise 10abcd of Section 4.2, p.210. All angles are in degrees. a. cos(120) -sin(120) 0 b. 1 0 -3 sin(120) cos(120 0 0 1 5 0 0 1 0 0 1 c. 1/3 0 0 d. 1 0 0 0 1/3 0 0 -1 2 0 0 1 0 0 1 Exercise 11 of Section 4.2, p.211. Evaluate L(a+b*x) = ( a + b/2, a )' = / 1 1/2 \ /a\ = A/a\ \ 1 0 / \b/ \b/ Exercise 13 of Section 4.2, p.211. Evaluate D on an element of Span{e^x, x*e^x, x^2*e^x}: D[a*e^x + b*x*e^x + c*x^2*e^x] = = a*e^x + b*e^x + b*x*e^x + 2*c*x*e^x + c*x^2*e^x = (a+b)*e^x + (b+2*c)*x*e^x + c*x^2*e^x. The expansion coefficients of the output in the given ordered basis may be written as 1*a+1*b+0*c, 0*a+1*b+2*c, and 0*a+0*b+1*c, respectively. These expansions determine the rows of the matrix of D: 1 1 0 0 1 2 0 0 1 Exercise 2abc of Section 4.3, p.217. First solve 1(a): the matrix representing L(x1,x2)=(-x1,x2) with respect to {e1,e2} is A = -1 0 0 1 Write U=[1 1; -1 1]' for the matrix whose columns are the ordered basis u1=[1 1]' and u2=[-1 1]'. Then the matrix B representing L with respect to {u1,u2} is given by B = U^{-1}AU = 0 1 1 0 This follows from Theorem 4.3.1, p.214, and the result may be computed using MatLab: B=inv(U)*A*U. a. The transition matrix from [u1,u2] to [v1,v2] is S=V^{-1}U, as shown in Examples 4 and 5, pp.167-168. With v1=[2 1]', v2=[1 0]' and V=[2 1;1 0]', MatLab gives S = inv(V)*U = 1 1 -1 3 b. Since S^{-1} is the transition matrix from [v1,v2] to [u1,u2], we may compute the matrix of L in the V basis as C=SBS^{-1}, using Theorem 4.3.1 on p.214. Using MatLab, it is C = S*B*inv(S) = 1 0 -4 -1 c. Note: our matrix "C" is called "A" in the text. We compute L(v1) = L(2,1) = (-2,1)' = 1*v1+(-4)*v2 = c11*v1+c21*v2 L(v2) = L(1,0) = (-1,0)' = 0*v1+(-1)*v2 = c21*v1+c22*v2 These are indeed the coefficients of C. Exercise 4 of Section 4.3, p.218. With V=[v1,v2,v3]=[1 1 1; 1 2 0; 0 -2 1]', the transition matrix for the change of basis {v1,v2,v3} --> {e1,e2,e3} is V = -2 1 2 3 -1 -2 2 -1 -1 We follow Example 3, p.167, and use MatLab for the computation. If A=[3 -1 -2; 2 0 -2; 2 -1 -1] is the matrix representing L with respect to {e1,e2,e3}, then by Theorem 4.3.1 the matrix representing L with respect to {v1,v2,v3} is B=VAV^{-1}, which we compute with MatLab: B = V*A*inv(V) = -21 10 16 -11 6 8 -22 10 17 Exercise 5 of Section 4.3, p.219. a. L(1) = 0 = 0*1 + 0*x + 0*x^2, L(x) = x = 0*1 + 1*x + 0*x^2, L(x^2) = 2x^2+2 = 2*1 + 0*x + 2*x^2. Hence the matrix of L with respect to {1,x,x^2} is A = 0 0 2 0 1 0 0 0 2 b. L(1) = 0 = 0*1 + 0*x + 0*(1+x^2), L(x) = x = 0*1 + 1*x + 0*(1+x^2), L(1+x^2) = 2x^2+2 = 0*1 + 0*x + 2*(1+x^2). Hence the matrix of L with respect to {1,x,1+x^2} is B = 0 0 0 0 1 0 0 0 2 c. By Th.4.3.1, S is given by the transition matrix {1,x,x^2} --> {1,x,1+x^2}. By Example 7, p.172, this transition matrix is the inverse of T = 1 0 0 0 1 0 1 0 1 Using MatLab we get S = inv(T) = 1 0 0 0 1 0 -1 0 1 d. Following Example 2 on p.216, we compute L^n(p(x)) in the basis {1,x,1+x^2}, for which L has a diagonal matrix representation B, as L^n(a0 + a1*x + a2*(1+x^2)) = B^n[a0,a1,a2]' = a1*x + 2^n*a2*(1+x^2). Exercise 7 of Section 4.3, p.219. A similar to B ==> B = S^{-1}AS for some nonsingular S. B similar to C ==> C = T^{-1}BT for some nonsingular T. Combining these equalities gives C = T^{-1}S^{-1}AST = (ST)^{-1} A ST, where we use Th.1.4.1 to obtain the last equality with the nonsingular matrix ST. Thus A is similar to C by definition, using ST. Exercise 10 of Section 4.3, p.219. Since A and B are similar, there is a nonsingular matrix Q such that B = Q^{-1}AQ = TS, where we exhibit T=Q^{-1}A and nonsingular S=Q to obtain the last equality. But then we also obtain A = (QQ^{-1})A = Q(Q^{-1}A) = ST. Exercise 11 of Section 4.3, p.219. By definition, A similar to B <=> B=S^{-1}AS for some nonsingular matrix S. But then det(B) = [1/det(S)]det(A)det(S) = det(A) by Th.2.2.3 and Exercise 6, p.113. Exercise 12 of Section 4.3, p.219. a. By definition, A similar to B <=> B=S^{-1}AS for some nonsingular matrix S ==> B' = S' A' (S^{-1})' By Rule 4 on p.56. But (S^{-1})'=(S')^{-1} by Exercise 17, p.65, so B' = S' A' (S')^{-1} and thus A' is similar to B' by definition, using the nonsingular matrix T = (S')^{-1}. b. We use induction to prove the stronger theorem that if B=S^{-1}AS for some nonsingular matrix S, then B^k=S^{-1} A^k S for positive integer k. The result is evidently true if k=1. Now suppose that the equation holds for k. We compute B^{k+1} = B^k B = S^{-1} A^k S S^{-1} A S = S^{-1} A^{k+1} S so the equation holds for k+1 as well. Exercise 13 of Section 4.3, p.219. By definition, A similar to B <=> B=S^{-1}AS for some nonsingular matrix S ==> det(B) = det(A) by Exercise 11, p.219. But then det(B) != 0 since det(A) != 0 by Th.2.2.2. Hence B is nonsingular by Th.2.2.2. We may thus find the inverse of both sides of the equality B=S^{-1}AS with Th.1.4.1: B^{-1} = S^{-1} A^{-1} S But then by definition, B^{-1} and A^{-1} are similar.