Ma 309: Matrix Algebra Model Solutions to Homework Assignment 5 Prof. Wickerhauser Exercise 3 of Section 3.4, p.161. a. By Theorem 3.3.1, {x1,x2} is a linearly independent set in R^2 since det[x1;x2]=2 is nonzero. Since it has 2 vectors and R^2 is 2-dimensional, {x1,x2} is therefore a spanning set for R^2 by Theorem 3.4.3. Hence {x1,x2} is a basis for R^2. b. {x1,x2,x3} must be linearly dependent by Theorem 3.4.1. c. Note that Span{x1,x2,x3}=Span{x1,x2}, since x3=(33/2)*x1-(13/2)*x2 is a linear combination of x1 and x2. Since {x1,x2} are linearly independent, it follows that Span{x1,x2} and thus Span{x1,x2,x3} are both 2-dimensional by the definition on p.158. Exercise 4 of Section 3.4, p.161. Note that x2= -x1 and x3= -2*x1. Hence x2 and x3 are linear combinations of x1, so Span{x1,x2,x3}=Span{x1}. Since x1 is nonzero, it is linearly independent, so the dimension is one. Exercise 7 of Section 3.4, p.161. We separate the parameters: S = {a(1,1,0,0)+b(1,-1,1,0)+c(0,2,0,1): a,b,c in R} = Span{(1,1,0,0), (1,-1,1,0), (0,2,0,1)} The three vectors {(1,1,0,0), (1,-1,1,0), (0,2,0,1)} are linearly independent, since (0,0,0,0)=a(1,1,0,0)+b(1,-1,1,0)+c(0,2,0,1) implies b=c=0 from the last two coefficients and thus a=0 from the first coefficient. Hence {(1,1,0,0), (1,-1,1,0), (0,2,0,1)} is a basis for Span{(1,1,0,0), (1,-1,1,0), (0,2,0,1)} by the definition on p.156, so Span{(1,1,0,0), (1,-1,1,0), (0,2,0,1)} is three-dimensional by the definition on p.158. Exercise 11 of Section 3.4, p.162. Separate the parameters: ax^2+bx+2a+3b = a(x^2+2)+b(x+3). Hence S=Span{x^2+2, x+3}. But {x^2+2, x+3} is a linearly independent set in P3 by Theorem 3.3.3, since the Wronskian is det/ x^2+2 x+3 \ = -2x^2-6x+2 != 0 at x=0, \ 2x 1 / so S has dimension 2 by the definition on p.158. Exercise 17 of Section 3.4, p.162. Follow the hint. Suppose {u1,u2} and {v1,v2} are bases for two-dimensional subspaces U,V of R^3, respectively. Since R^3 is 3-dimensional, the set {u1,u2,v1,v2} must be linearly dependent. Hence there are scalars a,b,c,d, not all 0, such that a*u1 + b*u2 + c*v1 + d*v2 = 0. Now define x = a*u1 + b*u2. Then x belongs to U = Span{u1,u2}. But since a*u1 + b*u2 = -c*v1 - d*v2, we see that x belongs to V = Span{v1,v2} as well, so x is in the intersection of U and V. We claim that x is not 0, for then since {u1,u2} is linearly independent we would have a=b=0, and since {v1,v2} is linearly independent we would have c=d=0, contradicting the hypothesis that not all of a,b,c,d are zero. Exercise 1b of Section 3.5, p.173. b. Following Example 3 on p.167, the transition matrix is (u1,u2) = / 1 2 \ \ 2 5 / Exercise 2b of Section 3.5, p.173. b. Following Example 2 on p.167, the transition matrix is the inverse (u1,u2)^{-1} = / 5 -2 \ \ -2 1 / Exercise 3b of Section 3.5, p.173. b. Write U=(u1,u2) and V=(v1,v2). Following Example 5 on p.169, the transition matrix from [v1,v2] to [u1,u2] is U^{-1}V, which we compute with MatLab: <>U=[ [1 2]' [2 5]' ] <>V = [ [3 2]' [4 3]' ] <>inv(U)*V 1. 2. 3. 4. 11. 14. 2. 5. 2. 3. -4. -5. Exercise 8 of Section 3.5, p.174. Following Example 5, if U=(u1,u2) and V=(v1,v2) and S is the transition matrix from [v1,v2] to [u1,u2], then S= U^{-1}V. Hence U=VS^{-1}, which we compute using MatLab: <>v1=[2 6]'; v2=[1 4]'; V=[v1 v2] 2. 1. 6. 4. <>S=[4 1; 2 1] <>V*inv(S) 4. 1. 0. 1. 2. 1. -1. 5. Exercise 10 of Section 3.5, p.174. Following Example 7 on p.172, first write the second ordered basis in terms of the first: 1 = 1*1 + 0*x + 0* x^2 1+x = 1*1 + 1*x + 0* x^2 1+x+x^2 = 1*1 + 1*x + 1* x^2 The coefficients at right give the transition matrix S from [1,1+x,1+x+x^2] to [1,x,x^2]: S = 1 0 0 1 1 0 1 1 1 Hence, the transition matrix from [1,x,x^2] to [1,1+x,1+x+x^2] is S^{-1}, which we compute with MatLab: <>S=[1 0 0; 1 1 0; 1 1 1]; inv(S) 1. 0. 0. -1. 1. 0. 0. -1. 1. Exercise 1abc of Section 3.6, p.180. a. Put the matrix into reduced row echelon form: 1 3 2 1 3 2 1 0 2 2 1 4 --> 0 -5 0 --> 0 1 0 4 7 8 0 -5 0 0 0 0 Only 2 of these rows are nonzero and thus linearly independent. They give a basis for the (2-dimensional) row space: (1,0,2), (0,1,0) The leading ones are in columns 1,2, so those columns of the original matrix form a basis for the (2-dimensional) column space: (1,2,4)', (3,1,7)', (2,4,8)' There is one free parameter corresponding to column 3 of the reduced row echelon form, which has no leading one, so the nullspace is 1-dimensional and has the basis (-2, 0, 1). b. Put the matrix into reduced row echelon form: -3 1 3 4 -3 1 3 4 0 7 0 -2 1 2 -1 -2 --> 1 2 -1 -2 --> 1 2 -1 -2 -3 8 4 2 0 7 1 -2 0 7 1 -2 0 7 0 -2 0 1 0 -2/7 --> 1 2 -1 -2 --> 1 0 0 -10/7 0 0 1 0 0 0 1 0 These 3 rows are all nonzero and thus linearly independent. They give a basis for the (3-dimensional) row space: (0,1,0,-2/7), (1,0,0,-10/7), (0,0,1,0) The leading ones are in columns 1,2,3, so those columns of the original matrix form a basis for the (3-dimensional) column space: (-3,1,3)', (1,2,8)', (3,-1,4)' There is one free parameter corresponding to column 4 of the reduced row echelon form, which has no leading one, so the nullspace is 1-dimensional and has the basis (10/7, 2/7, 0, 1) c. Put the matrix into reduced row echelon form: 1 3 -2 1 1 3 -2 1 1 3 -2 1 2 1 3 2 --> 0 -5 7 0 --> 0 -5 7 0 --> 3 4 5 6 0 -5 11 3 0 0 4 3 1 3 -2 1 1 3 0 5/2 1 0 0 -13/20 --> 0 1 -7/5 0 --> 0 1 0 21/20 --> 0 1 0 21/20 0 0 1 3/4 0 0 1 3/4 0 0 1 3/4 These 3 rows are all nonzero and thus linearly independent. They give a basis for the (3-dimensional) row space: (20,0,0,-13), (0,20,0,21), (0,0,4,3) The leading ones are in columns 1,2,3, so those columns of the original matrix form a basis for the (3-dimensional) column space: (1,2,3)', (3,1,4)', (-2,3,5) There is one free parameter corresponding to column 4 of the reduced row echelon form, which has no leading one, so the nullspace is 1-dimensional and has the basis (-13, 21, 15, 20) Exercise 4bef of Section 3.6, p.181. b. NO. Vector b is not in the column space of A, so Ax=b is not consistent. Show this by row reduction of (A|b): 3 6 | 1 0 0 | -2 <-- zero row of A with nonzero RHS. 1 2 | 1 --> 1 2 | 1 e. YES. Vector b is in the column space of A, so Ax=b is consistent. Show this by row reduction of (A|b): 0 1 | 2 0 1 | 2 1 0 | 5 --> 1 0 | 5 0 1 | 2 0 0 | 0 <-- zero row with zero RHS. f. YES. Vector b is in the column space of A, so Ax=b is consistent. Show this by row reduction of (A|b): 1 2 | 5 1 2 | 5 2 4 | 10 --> 0 0 | 0 <- 1 2 | 5 0 0 | 0 <-- zero rows with zero RHS. Exercise 5bef of Section 3.6, p.181. b. No solution: system is inconsistent. e. Unique solution: each column has a leading 1. f. Infinitely many solutions: column 2 has no leading 1 so there is one free parameter. Exercise 13ab of Section 3.6, p.182. a. If x is in N(A), then (BA)x= B(Ax)=B(0)=0, so x is in N(BA). Thus N(BA) contains N(A). Conversely, if x is in N(BA), then 0=(BA)x=B(Ax), so Ax is in N(B). But N(B)={0} since B is nonsingular, so Ax=0, so x is in N(A). Thus N(A) contains N(BA). Combining both containments, we conclude that N(A)=N(BA) and thus Rank(A)=Rank(BA). b. Apply part a to (AC)' = C'A', noting that C' is nonsingular since C is nonsingular. Thus Rank((AC)')=Rank(C'A')=Rank(A'). By Theorem 3.6.6, Rank(A')=Rank(A) and Rank((AC)')=Rank(AC). Combining these equalities yields Rank(A)=Rank(AC). Exercise 18ab of Section 3.6, p.183. a. Write x=(x1,...,xm) and y=(y1,...,yn). Then xy' is the matrix whose columns are all multiples of x: xy' = [y1*x ... yn*x], so the column space of xy' is Span{x}. But also, xy' is the matrix whose rows are all multiples of y: xy' = [x1*y; x2*y; ... xm*y], so the row space of xy' is Span{y'}. Since one nonzero vector forms a linearly independent set, and x and y are both nonzero, these spans are bases. b. By Theorem 3.6.5, since A=xy' is an mxn matrix, dim N(A) + Rank(A) = n, so since Rank(A) = dim RowSpace(A) = 1 by part a, we have Nullity(A)=dim N(A)=n-1. Exercise 19abc of Section 3.6, p.183. Write C(A) for the column space of matrix A and R(A) for the row space of A. a. x in C(AB) ==> x=(AB)y for some y in R^r ==> x=A(By)=Az for some z=By in R^n ==> x is in C(A) Thus C(AB) is a subset of C(A). But C(AB) is a linear span and therefore a subspace of C(A). b. Apply part a to the matrix C'=(AB)'=B'A' to conclude that R(AB)=C((AB)')=C(B'A') is a subset of R(B)=C(B'). Again, R(AB) is a linear span and therefore a subspace of R(B). c. By Theorem 3.6.6, Rank(A)=dim C(A)=dim R(A). By part a, Rank(AB)=dim C(AB)=