Ma 309: Matrix Algebra Model Solutions to Homework Assignment 4 Prof. Wickerhauser Exercise 2 of Section 3.1, p.131. (a) ||x1|| = sqrt(2^2+1^2) = sqrt(5); ||x2|| = sqrt(6^2+3^2) = sqrt(45) = 3*sqrt(5). (b) x3=[8 4]'; ||x3|| = sqrt(8^2+4^2) = sqrt(80) = 4*sqrt(5) = ||x1||+||x2||. (c) x1 and x2 are parallel, so the length of thei sum is the sum of their lengths. Exercise 4 of Section 3.1, p.132. Either cite Theorem 1.3.2, p.46, or check the axioms directly from the definitions as follows: Write A=[aij], B=[bij], and C=[cij] to signify the coefficients of arbitrary mxn matrices A,B,C, respectively. Let s,t be any scalars. A1: A+B = [aij+bij] = [bij+aij] = B+A, since the coefficients are scalars and scalar addition is commutative. A2: (A+B)+C = [(aij+bij)+cij] = [aij+(bij+cij)] = A+(B+C), since the coefficients are sclars and scalar addition is associative. A3: Exhibit the mxn zero matrix. A4: Given A, we have -A = [-aij], the additive inverse of A. A5: s(A+B)=s[aij+bij]=[s*aij+s*bij]=[s*aij]+[s*bij]=sA+sB, by the definition of scalar multiplication for matrices, since scalar multiplication is distributive over scalar addition. A6: (s+t)A=[(s+t)*aij]=[s*aij+t*aij]=[s*aij]+[t*aij]=sA+tA, by the definition of scalar multiplication for matrices, since scalar multiplication is distributive over scalar addition. A7: (s*t)A=[(s*t)*aij]=[s*(t*aij)]=s[t*aij]=s(tA), by the definition of scalar multiplication for matrices, since scalar multiplication is associative. A8: 1A=[1*aij]=[aij]=A, by the definition of scalar multiplication for matrices, since 1 is the multiplicative identity for scalars. Exercise 7 of Section 3.1, p.132. Let V be a vector space containing vectors 0 and 0'. Suppose that 0 and 0' both satisfy x+O=x and x+0' =x for every vector x in V. We may substitute x <-- 0' in the first equation and x <-- 0 in the second equation to conclude that 0' = 0'+0 = 0+0' = 0, since vector addition is commutative by Axiom 1. Exercise 9 of Section 3.1, p.132. (a) Let b be any scalar and let 0 be the identity for vector addition. Then 0=0+0 by Axiom 3, so b0=b(0+0)=b0+b0, We may add the inverse -b0 to both sides and use the associativity of vector addition: 0 = -b0+b0 = -b0+(b0+b0) = (-b0+b0)+b0 = 0+b0 = b0 Thus b0=0. Note how this differs from the proof of Theorem 3.1.1(i). (b) Suppose that a is a scalar, x is a vector, 0 is the vector additive identity, and ax=0. If the equation holds by theorem 3.1.1(i) because a=0 (the scalar zero), then the conclusion "either a=0 or x=0" is true because of its first clause. Otherwise, if a!=0, then we may multiply both sides by 1/a to get x=(1/a)0=0 by part (a), and the conclusion "either a=0 or x=0" is satified because of its second clause. Exercise 10 of Section 3.1, p.132. Axiom 3 fails to hold for this addition. Suppose there is an additive identity 0=[x,y]. But then [1,1](+)0=[1,1]. However, from the definition, [1,1](+)[x,y] = [1+x,0], and we arrive at the contradition 1=0 from the second coordinate. Exercise 3 of Section 3.2, p.142. (a) Subspace. (b) Subspace. (c) Not a subspace: if A=(aij) and a12=1, then 2A=(2*aij) does not have 1 as its 1,2 coefficient. Hence the set is not closed under scalar multiplication. Also, there is no additive identity, since 0 is not an element of the set. Also, the set cannot contain both a matrix and its additive inverse. (d) Subspace. (e) Subspace. (f) Not a subspace, as it is not closed under addition. Counterexample: the adding the following singular matrices A,B gives the nonsingular matrix A+B=I: A = 1 0 B = 0 0 0 0 0 1 Exercise 4 of Section 3.2, p.143. Use Gauss-Jordan reduction, look for free parameters: (a) 2 1 : 0 --> 2 1 : 0 3 2 : 0 0 1/2 : 0 ==> no free parameters; Nullspace is {0} (b) 1 2 -3 -1 : 0 --> 1 2 -3 0 : 0 -2 -4 6 4 : 0 0 0 0 2 : 0 ==> solution is x4=0, x1= -2*x2+3*x3, with two free parameters x2,x3 ==> Nullspace is {(-2*s+3*t,s,t,0): s,t in R}. (c) 1 3 -4 : 0 1 0 -1 : 0 2 -1 -1 : 0 --> 0 1 -1 : 0 -1 -3 4 : 0 0 0 0 : 0 ==> solution is x1=x3 and x2=x3, with one free parameter x3. ==> Nullspace is {(s,s,0): s in R} (d) 1 1 -1 2 : 0 1 2 0 9 : 0 2 2 -3 1 : 0 --> 0 0 1 3 : 0 -1 -1 0 -5 : 0 0 0 0 0 : 0 ==> solution is x1=-2*x2-9*x4 and x3=-3*x4, with two free parameters x2,x4. ==> Nullspace is {(-2*s-3*t,s,-3*t,t): s,t in R} Exercise 10 of Section 3.2, p.143. (a) Spanning set: Gauss-Jordan reducible to 3x3 identity. (b) Spanning set: set from (a) with additional vector. (c) Not a spanning set: determinant is zero. (d) Not a spanning set: three parallel vectors. (e) Not a spanning set: only two vectors. Exercise 12 of Section 3.2, p.144. (a) Yes. If v is a linear combination of x1,...,xk, then it is a linear combination of x1,...,xk,x{k+1} too, since we may take 0 for the coefficient of x{k+1}. (b) Not in all cases. For example, x1=[1,0]' and x2=[0,1]' is a spanning set for V=R^2, but deleting either x1 or x2 results in a set that does not span V. Exercise 18 of Section 3.2, p.144. Check conditions i,ii on page 134: (i) Suppose x is in U intersect V and a is any scalar. x in U ==> ax in U, since U is a subspace; x in V ==> ax in V, since V is a subspace; Therefore, ax is in U intersect V. (ii) Suppose x,y are in U intersect V. x,y in U ==> x+y in U, since U is a subspace; x,y in V ==> x+y in V, since V is a subspace; Therefore, x+y is in U intersect V. Since 0 is in both U and V, it is in their intersection. Hence U intersect V is a nonempty set and satisfies the conditions to be a subspace. Exercise 2 of Section 3.3, p.154. (a) Linearly independent: can be Gauss-Jordan reduced to I. (b) Linearly dependent: 4 vectors in R^3. (c) Linearly dependent: 3x3 matrix is singular. (d) Linearly dependent: second and third vector are scalar multiples of the first vector. (e) Linearly independent: a(1,1,3)+b(0,2,1)=(0,0,0) ==> a=0 from first coordinate ==> b=0 from third cordinate. Exercise 4 of Section 3.3, p.155. (a) Linearly independent. (b) Linearly independent. (c) Linearly dependent. Exercise 5 of Section 3.3, p.155. (a) Not necessarily. x1=[1,0] and x2=[0,1] are linearly independent in V=R^2, but since they are also a spanning set for V, any other vector x3 can be written as a linear combination of x1 and x2, so (x1,x2,x3) cannot be linearly independent. (b) Yes. If, then there must be a nontrivial linear combination for 0 using a proper subset of the vectors x1,...,xk. We may add in the missing vectors with zeros for coefficients to get a nontrivial linear combination for 0 using all the vectors x1,...,xk. This contradicts the assumption that the vectors x1,...,xk are linearly independent. Exercise 8 of Section 3.3, p.155. Check the 3x3 Wronskian W = det(A), where A is the matrix cos x 1 sin^2(x/2) -sin x 0 sin(x/2)cos(x/2)=(sin x)/2 -cos x 0 (cos x)/2 The second and third rows are linearly dependent for each x in R, since they are multiples (by sin x and cos x, respectively) of the vector (-1,0,1/2). Hence W(x)=0 for all x in R, and in particular for all x in [-pi,pi]. Hence Theorem 3.3.3 gives no information. Note that cos x = cos(2*x/2) = 1-2*sin^2(x/2) for any x in R, by the cosine double-angle formula. Hence, for all x in R, 1*cos x + (-1)*1 + 2*sin^2(x/2) = 0 Hence the vectors cos x,1, and sin^2(x/2) are linearly dependent in C[-pi,pi]. Exercise 9 of Section 3.3, p.155. Check the 2x2 Wronskian W = det(A), where A is the matrix cos(x+a) sin x -sin(x+a) cos x Thus W = [cos x][cos(x+a)] - [sin x][-sin(x+a)] = [cos x][cos(x+a)] + [sin x][sin(x+a)] = cos(x-[x+a]) = cos(-a) = cos a, where we have used the angle subtraction formula for cosine: cos(A-B)=cos(A)cos(B)+sin(A)sin(B). Then W=0 for a in [-pi,pi] if and only if cos a = 0 if and only if a=pi/2 or a= -pi/2. By Theorem 3.3.3, sin x and cos(x+a) are linearly independent in C[-pi,pi] for all a except possibly a=pi/2 or a= -pi/2. For a=pi/2 or a= -pi/2, we note that for all x in R, cos(x+pi/2)= -sin x and cos(x-pi/2)=sin x. Hence the functions cos(x+a) and sin x are linearly dependent in C[-pi,pi] if a=pi/2 or a= -pi/2.