Exercises Chapter II.

Page 64 Exercises Chapter II. 5. Let A = (1, 2) and B = ( 2, 6). Sketch vectors of the form X = c 1 A + c 2 B for various values of c 1 and c 2. Which vectors in R 2 can be written in this manner? B y 2A + B 3A + B x For any pair of numbers x, y, set c 1 = 3x + y and c 2 = 2x + y. Then (x, y) = c 1 A + c 2 B. 2 So all vectors in R 2 can be written this way. 7. Let A = (1, 1, 2), B = (0, 1, 0). a. Find all vectors x in R 3 such that x = t 1 A + t 2 B for some constants t 1 and t 2. If x = (x 1, x 2, x 3 ) = t 1 (1, 1, 2) + t 2 (0, 1, 0) = (t 1, t 1 + t 2, 2t 1 ), then we must have x 3 = 2x 1. b. Is the vector (0,0,1) one of the vectors you found in part a? This vector is not a linear combination of the vectors A and B since 1 2 0. 9. Let x = (1, 2), y = ( 1, 1). Describe the following sets of vectors. a. {tx + (1 t)y : 0 t 1}. The tips of this set of vectors make up the straight line joining the two points (1, 2) y and ( 1, 1). ( 1,1) (1,2) x A

b. {t 1 x + t 2 y : 0 t 1 1, 0 t 2 1} The tips of this set of vectors fill up the parallelogram whose vertices are: (1, 2), (3, 0), ( 1, 1), and (0, 0). y ( 1, 1) (1, 2) x c. {tx + (1 t)y : t any real number} x 2 y = ( 1, 1) x = (1, 2) This set is the straight line through the two points x and y. d. {t 1 x + t 2 y : 0 t 1, 1 t 2 } The set is the infinite triangular region bounded on one side by the line y + tx, for 0 t, and on the other side by the line ty for 1 t. x 1 y + tx ty y x e. {t 1 x + t 2 y : t 1 and t 2 arbitrary} If either of these vectors is a multiple of the other, then the set is the straight line passing through these points and the origin. If they are not multiples of each other, which is the case for our x and y, then the set is all of R 2. 2

Page 69 9. Let V 0 be all 2 2 matrices for which the 1,1 entry is zero. Let V 1 be all 2 2 matrices for which the 1,1 entry is nonzero. Determine if either of these two sets is a vector space. Define addition and multiplication as we did in M mn. The set V 0 is a vector space. Adding two such matrices gives us a matrix with a zero in the 1,1 position, and a scalar multiple of such a matrix also gives us a matrix with a zero in the 1,1 position. It is an easy matter to check that each of the axioms for a vector space is satisfied by the vectors in V 0. The set V 1 is not a vector space. It is not closed under vector addition or scalar multiplication. If A V 1, then so is A, but A + ( A A) has a zero in the 1,1 position so this vector is not in V 1. Similarly 0A is not in V 1. 10 Let V = {A, B,..., Z}; that is, V is the set of capital letters. Define A + B = C, B + C = D, A + E = F ; i.e., the sum of two letters is the first letter larger than either of the summands, unless Z is one of the letters to be added. In that case the sum will always be A. Can you define a way to multiply the elements in V by real numbers in such a way that V becomes a vector space? The answer is no. One reason is that there is no zero vector. 11. Let V = {(x, 0, y): x and y are arbitrary real numbers}. Define addition and scalar multiplication as follows: Is V a vector space? (x 1, 0, y 1 ) + (x 2, 0, y 2 ) = (x 1 + x 2, y 1 + y 2 ) c(x, 0, y) = (cx, cy) V is not a vector space. It is not closed under vector addition or scalar multiplication. 12. Let V c = {(x 1, x 2 ): x 1 + 2x 2 = c}. a. For each value of c sketch V c. y x c = 1 c = 0 c = 1 3

b. For what values of c, if any, is V c a vector space? The only value of c for which V c is a vector space is c = 0. If c 0, then V c is not closed under either of the algebraic operations. For example the vector (c, 0) is in V c for any c, but (c, 0) + (c, 0) = (2c, 0) is not back in V c unless c = 0. 14. Let V c = {p: p is in P 4 and p(c) = 0}. For what values of c is V c a vector space? V c is a vector space for every value of c. 16. Let V 1 = {(x 1, x 2 ): x 2 1 + x 2 2 = 1}, V 2 = {(x 1, x 2 ): x 2 1 + x 2 2} 1, V 3 = {(x 1, x 2 ): x 1 0, x 2 0}, and V 4 = {(x 1, x 2 ): x 1 + x 2 0}. Which of these subsets of R 2 is a vector space? None of these subsets of R 2 is a vector space, for they are not closed under either of the vector space operations. 4

Page 79 5. Let A = {(1, 1, 0), ( 1, 0, 1)}. Determine which if any of the following vectors is in S[A]: (1, 1, 1), (0, 1, 1), (2, 3, 1). We are looking for constants c 1 and c 2 such that c 1 (1, 1, 0) + c 2 ( 1, 0, 1) = (x 1, x 2, x 3 ), where (x 1, x 2, x 3 ) is one of the given vectors. This vector equation leads to the following system of equations: c 1 c 2 = x 1, c 1 = x 2, c 2 = x 3. This system will have a solution if and only if x 1 = c 1 c 2 = ( x 2 ) x 3 = (x 2 + x 3 ). For the three vectors we have (1, 1, 1) : x 1 = 1, (x 2 + x 3 ) = 0 this vector is not in S[A] (0, 1, 1) : x 1 = 0, (x 2 + x 3 ) = 0 (0, 1, 1) = (1, 1, 0) + ( 1, 0, 1) (2, 3, 1) : x 1 = 2, (x 2 + x 3 ) = 2 this vector is not in S[A]. 6. Let A = {(1, 2), (6, 3)}. Show that S[A] = R 2. Let (x 1, x 2 ) be any vector in R 2. We want to find constants c 1 and c 2 such that c 1 (1, 2) + c 2 (6, 3) = (x 1, x 2 ). This leads to the system of equations: c 1 + 6c 2 = x 1 2c 1 + 3c 2 = x 2 The solution to this system is c 1 = x 1 + 2x 2 and c 2 = 2x 1 x 2. 3 9 [ ] 1 0 3 13. Let A =. Let K be the solution set of Ax = 0, for x in R 4 4 6 3. From Example 9 we know that K is a subspace of R 3. Find a vector x 0 such that S[x 0 ] = K. The matrix A is row equivalent to the matrix Thus, K = S[( 3, 3/2, 1)]. x = ( 3x 3, 3x 3 /2, x 3 ) = x 3 ( 3, 3/2, 1). [ ] 1 0 3. Thus, if x is in K we have 0 1 3/2 5

16. Find spanning sets for each of the following vector spaces: a. V = {(x 1, x 2, x 3 ): x 1 + x 2 6x 3 = 0}) If x is in V, then we must have: x = ( x 2 + 6x 3, x 2, x 3 ) = ( x 2, x 2, 0) + (6x 3, 0, x 3 ) = x 2 ( 1, 1, 0) + x 3 (6, 0, 1) Thus, the vectors {( 1, 1, 0), (6, 0, 1)} form a basis for V. b. V = {p: p is in P 2 and p(1) = 0} If p is in P 2, then p(t) = a 2 t 2 + a 1 t + a 0, and if this polynomial is zero when t = 1, then we have: a 2 + a 1 + a 0 = 0 So if p is also in V, then p(t) = ( a 1 a 0 )t 2 + a 1 t + a 0 = a 1 ( t 2 + t) + a 0 ( t 2 + 1) So a spanning set for V is { t 2 + t, t 2 + 1} c. V = {A = [a ij ]: A is in M 23 and 3 j=1 a ij = 0 for i = 1, 2} If A is in V, then A = [ a1,2 a 1,3 a 12 ] a 1,3 a 2,2 a 2,3 a 2,2 a 2,3 = [ ] [ ] [ ] [ ] 1 1 0 1 0 1 0 0 0 0 0 0 a 1,2 + a 0 0 0 1,3 + a 0 0 0 2,2 + a 1 1 0 2,3 1 0 1 Thus, the set {[ ] 1 1 0, 0 0 0 [ ] 1 0 1, 0 0 0 [ ] 0 0 0, 1 1 0 [ 0 0 ]} 0 1 0 1 is a basis for V. 27. Let W = {(x 1, x 2 ): x 2 1 + x 2 2 c}. For which values of c is W a subspace of R 2? The only value of c for which W is a subspace is c = 0. 6

28. Let W 1 = {p: p is in P 2 and p(1) = 0}. Let W 2 = {p: p is in P 2 and p(2) = 0}. a. Find a spanning set for W 1 W 2. If p W 1 W 2, then p(1) = p(2) = 0, and if p(t) = a 2 t 2 + a 1 t + a 0, then we have: a 0 + a 1 + a 2 = 0 and a 0 + 2a 1 + 4a 2 = 0. This imlies that a 1 = 2a 2 and a 1 = 3a 2. So p(t) = a 2 (2 3t+t 2 ). Thus, {2 3t+t 2 } is a spanning set for W 1 W 2. b. Find spanning sets for W 1 and W 2 each of which contains the spanning set you found in part a. An equation, which characterizes those p W 1 is a 0 + a 1 + a 2 = 0. If we set a 2 = 0, then we must have a 1 = a 0, so a second vector in W 1 is p(t) = 1 t, and a basis for W 1 is {1 t, 2 3t + t 2 }. An equation characterizing those vectors in W 2 is a 0 + 2a 1 + 4a 2 = 0. Setting a 2 = 0, we must have a 1 = a 0 /2. Thus a spanning set for W 2 is {2 t, 2 3t + t 2 }. 29. Example 9 shows that the solution sets of homogeneous systems of equations are subspaces. Find spanning sets for each of the solution spaces of the following systems of equations: a. 2x 1 + 6x 2 x 3 = 0 A spanning set for the solution set is: {(1, 0, 2), (0, 1, 6)} b. 2x 1 + 6x 2 x 3 = 0, x 2 + 4x 3 = 0 A spanning set is {(25/2, 4, 1)}. c. 2x 1 + 6x 2 x 3 = 0, x 2 + 4x 3 = 0, x 1 + x 2 = 0 The solution set is the zero vector, or the vector space consisting solely of the zero vector, and this vector space has no basis. 7

Page 85 4. Determine whether the following sets of vectors are linearly independent. a. {(1, 3), (1, 4), (8, 12)} This set is linearly dependent. 44(1, 3) + 12(1, 4) 7(8, 12) = (0, 0). b. {(1, 1, 4), (8, 3, 2), (0, 2, 1)} This set is linearly independent. The only solution to c 1 (1, 1, 4) + c 2 (8, 3, 2) + c 3 (0, 2, 1) = (0, 0, 0) is c 1 = c 2 = c 3 = 0. c. {(1, 1, 2, 3), ( 2, 3, 0, 4), (8, 7, 4, 18)} This set is linearly dependent. 2(1, 1, 2, 3) + 3( 2, 3, 0, 4) + (8, 7, 4, 18) = (0, 0, 0, 0). 10. Let A = {(0, 1, 1), (1, 0, 1), (1, 1, 0)}. Is A linearly dependent? S[A] =? A is linearly independent and its span is all of R 3 13. Let A = {(x 1, x 2 ): x 2 1 + x 2 2 = 1}. Is A a linearly independent subset of R 2? Is A a spanning subset of R 2? A is not linearly independent as it contains the three vectors {(1, 0), (0, 1), (1/ 2, 1/ 2)}, and they are linearly dependent. A is a spanning subset of R 2. 16. Let V = {p(t): p is in P 2 and 1 p(t)dt = 0}. Find a spanning set of V that has exactly 0 two vectors in it. Show that your set is also linearly independent. Integrating a generic vector in P 2, we have 1 0 p(t)dt = 1 0 (a 0 + a 1 t + a 2 t 2 )dt = a 0 + a 1 2 + a 2 3 = 0. Thus, a 0 = ( a 1 2 + a 2 3 ). A basis for V is { 1 + 2t, 3 + 3t 2 }. If a linear combination of these vectors, c 1 ( 1 + 2t) + c 2 ( 3 + 3t 2 ) is the zero polynomial, then we must have c 1 = c 2 = 0. Thus, the vectors are linearly independent. 8

17. Let V = {p(t): p is in P 3 and p (0) = 0, p(1) p(0) = 0}. Find a spanning set of V that has exactly two vectors. Is the set also linearly independent? If p(t) = a 0 +a 1 t+a 2 t 2 +a 3 t 3 is a generic vector in P 3, and it is in V, then the coefficients a i satisfy the following equations: a 1 = 0, a 1 + a 2 + a 3 = 0. Thus, a 0 is arbitrary, a 1 = 0, and a 2 = a 3. So an arbitrary vector in V has the form p(t) = a 0 a 3 t 2 + a 3 t 3 = a 0 + a 3 ( t 2 + t 3 ). The two vectors {1, t 2 + t 3 } are linearly independent, hence they are a basis of V. 18. Let V = {A: A is in M 23 and AB = AC = 0 22 }, where 1 1 1 1 B = 1 1 C = 1 1 1 1 1 1 Find a linearly independent spanning set of V. How many vectors are in your set? [ ] a1 a Let A = 2 a 3. Then the conditions AB = AC = 0 a 4 a 5 a 2,2 lead to the equations: 6 a 1 + a 2 + a 3 = 0, a 1 a 2 + a 3 = 0, a 4 + a 5 + a 6 = 0, a 4 a 5 + a 6 = 0 a 1 a 2 + a 3 = 0, a 1 + a 2 a 3 = 0, a 4 a 5 + a 6 = 0, a 4 + a 5 a 6 = 0 The solution to this system is a 1 = a 3, a 2 = 0, a 4 = a 6, a 5 = 0. Thus, [ ] [ ] [ ] a3 0 a A = 3 1 0 1 0 0 0 = a a 6 0 a 3 + a 6 0 0 0 6 1 0 1 {[ ] [ ]} 1 0 1 0 0 0 and a basis for V is,. 0 0 0 1 0 1 9

Page 96 1. Find a basis for S[(2, 6, 0, 1, 3), (1, 1, 0, 1, 4), (0, 0, 0, 1, 1)]. The three given vectors are linearly independent so they form a basis for the vector space they span. 10. Let K = {(x 1, x 2, x 3 ): x 1 x 2 + 3x 3 = 0}. a. Find a basis for K. A vector (x 1, x 2, x 3 ) is in K if and only if (x 1, x 2, x 3 ) = (x 2 3x 3, x 2, x 3 ) = x 2 (1, 1, 0) + x 3 ( 3, 0, 1). Thus, a basis of K is {(1, 1, 0), ( 3, 0, 1)}. b. dim(k) =? The dimension of K is two. c. Extend your basis of K to a basis of R 3. All we need is to find a third vector, which is not in K. (1, 0, 0) is one such vector. So a basis of R 3 which contains the basis we found for K is {(1, 1, 0), ( 3, 0, 1), (1, 0, 0)}. 12. Let K = {(x 1, x 2, x 3 ): x 1 x 2 = x 3, x 2 + 2x 3 = 0)}. a. Find a basis B for K. The equations which determine K imply that x 2 = 2x 3, and x 1 = 2x 3 +x 3 = x 3. Thus, a basis for K is {( 1, 2, 1)}. b. dim(k) =? dim(k) = 1. c. How many vectors have to be added to B to get a basis for R 3? Since the dimension of R 3 is 3, we need to add two vectors to a basis of K to get a basis of R 3. 24. Let V = {p: p is in P 4, 1 p(t)dt = 0 = 2 p(t)dt}. Find a basis for V. 0 1 If p(t) = a 0 + a 1 t + a 2 t 2 + a 3 t 3 + a 4 t 4 is in V, then we have a 0 + a 1 2 + a 2 3 + a 3 4 + a 4 5 = 0, a 0 + 3a 1 2 + 7a 2 3 + 15a 3 4 The solution to this system is a 0 = 2 3 a 2 + 3 2 a 3 + 14 5 a 4, a 1 = 2a 2 7 2 a 3 6a 4 Thus, a basis of V is { 2 3 2t + t2, 3 2 7 2 t + t3, 14 5 6t + t4}. + 31a 4 5 = 0 10

25. Let A = {(1, 0, 0), (0, 1, 1), (0, 2, 3), (0, 3, 4)(1, 1, 3)}. a. Show that A is linearly dependent. Since (0, 1, 1) + (0, 2, 3) (0, 3, 4) = (0, 0, 0) the set A must be linearly dependent. b. What is the largest number of vectors in A that can form a linearly independent set? Since these vectors are in R 3, the maximum number is 3. c. For which vectors x in A is it true that S[A] = S[A\x}]? Turns out that this is true for every vector in A. d. Find a subset B of A such that B is linearly independent and for which S[B] = S[A]. One such subset is {(1, 0, 0), (0, 1, 1), (0, 2, 3)}. Note that the last two vectors form a basis of the subspace of R 3 whose vectors have a zero in the first slot, and the first vector is a basis for that subspace of R 3 whose last two slots are zero. Thus, between these three vectors, we have a basis for R 3. 11

Page 103 1 Find the coordinates of x = (1, 1, 2) with respect to each of the following bases: a. {(1, 0, 0), (0, 1, 0), (0, 0, 1)} [(1, 1, 2)] B = [1, 1, 2] b. {(1, 1, 1), (2, 1, 4), ( 1, 1, 1)} We re looking for constants c i such that (1, 1, 2) = c 1 (1, 1, 1)+c 2 (2, 1, 4)+c 3 ( 1, 1, 1). Solving for the constants we have [(1, 1, 2)] B = [ 3/2, 1, 1/2] c. {(2, 0, 6), (4, 2, 0), (0, 3, 2)} [(1, 1, 2)] B = [45/100, 25/1000, 35/100] 4. Let F = {(1, 2), (8, 3)} a. If [x] F = [1, 0], then x =? b. If [x] F = [1, 1], then x =? c. If [x] F = [0, 0], then x =? d. If x = (1, 1), then [x] F =? x = (1, 2) [x] F = x = (9, 1) x = 0 [ 5 19, 3 ] 19 12

5. Let V = M 2,2. Let F be the set: {[ 0 ] [ 1 1 ] [ 0 1 ] [ 1 1 ]} 1 1 1 1 1 0 1 1 0 a. Show that F is a basis of V. Show that the set F is linearly independent, then note that F contains 4 vectors and dim[m 2,2 ] = 4. Thus, F is a basis of M 22. [ ] 6 4 b. Let x = ; [x] 3 2 F =? We need constants c i such that [ ] [ ] 6 4 0 1 = c 3 2 1 1 1 [ ] [ ] [ ] 1 0 1 1 1 1 + c 2 + c 1 1 3 + c 0 1 4 1 0 Solving this system we get [x] F = [ 11 3, 19 3, 2 3, 1 ] 3 c. If [x] F = [0, 1, 0, 0], then x equals? x = [ ] 1 0 1 1 6. Let V = P 2. Let F = {1, 1 + t, 1 + t + t 2 }. a. Show that F is a basis of V. dim [P 2 ] = 3, there are 3 vectors in F, and they are linearly independent. Thus, F must be a basis for V. b. If [x] F = [ 2, 3, 7], then x equals? x = ( 2)1 + (3)(1 + t) + (7)(1 + t + t 2 ) = 8 + 10t + 7t 2. c. [6 t 2 ] F = [6 t 2 ] F = [6, 1, 1] 13

7. Let S be the standard basis of R 2, and let G = {(1, 6), (2, 3)}. a. Show that G is a basis. There are 2 vectors in G, G is linearly independent, and the dimension of R 2 is two. Hence, G is a basis of R 2. b. Find the change of basis matrix P such that [x] T S = P [x]t G. [ ] 1 2 P = 6 3 c. Find the change of basis matrix Q such that [x] T G = Q[x]T S. Q = P 1 = 1 [ ] 3 2. 9 6 1 d. Compute P Q and QP. Since Q = P 1, we have P Q = QP = I 2. 9. Let F = {( 1, 7), (2, 3)} and G = {(1, 2), (1, 3)}. a. Show that both F and G are bases of R 2. Both sets have 2 vectors, both sets are linearly independent and dim[r 2 ] = 2. b. Find the change of basis matrix P such that [x] T G = P [x]t F. Let P 1 and P 2 be the change of basis matrices defined by [x] T S = P 1 [x] T F, [x] T S = P 2 [x] T G, where S represents the standard basis of R 2. Then [ ] [ ] 1 2 1 1 P 1 = P 7 3 2 = 2 3 Using these two matrices we construct the matrix P. [x] T G = P 1 2 [x] T S = P 1 2 P 1 [x] T F.. Thus, P = P 1 2 P 1 = = [ ] 10 9 9 7 [ 1 1 2 3 ] 1 [ ] 1 2 = 7 3 [ 3 1 2 1 ] [ ] 1 2 7 3 14

12. Let F = {(1, 1), ( 1, 2)}, G = {( 1, 2), (1, 1)}. a. Find the change of basis matrix P such that [x] T G = P [x]t F. Notice that the bases F and G contain the same vectors, but in a different order. What this means as far as coordinates are concerned is that if [x] T G = [c 1, c 2 ], then [x] T F = [c 2, c 1 ]. So the change of basis matrix P has to switch components of a vector. Thus, [ ] 0 1 P =. 1 0 b. If x = (1, 1), find [x] F and [x] G. The vector (1, 1) is the first basis vector in F and the second basis vector in G. Thus, [x] F = [1, 0], [x] G = [0, 1]. c. Show [x] T F = P 1 [x] T G. P 1 [x] T G = [ ] [ ] 0 1 0 1 0 1 = [ ] 1 = [x] 0 F. Note that P = P 1. 18. Let V = P 3. Let F = { 2, t + t 2, t 2 1, t 3 + 1}. Let S denote the standard basis of V. a. Verify that F is a basis of V. Let f i, for 1 i 4 denote the basis vectors of F. Then f 1 generates all of the constant polynomials, a linear combination of f 1 and f 3 equals t 2, a linear combination of f 1 and f 4 equals t 3, and f 2 t 2 = t. Thus, the span of F equals P 3. Since there are four vectors in F and the dimension of P 3 is 4, F is a basis. b. Find the change of basis matrix P such that [x] T S = P [x]t F. 2 0 1 1 P = 0 1 0 0 0 1 1 0. 0 0 0 1 15

d. Compute the coordinates of t 3, t 2, t, and 1 with respect to the basis F. The matrix P 1 is the change of basis matrix, which converts coordinates with respect to the standard basis to coordinates with respect to the basis F. 1 1 1 1 P 1 = 1 0 2 0 0 2 0 2 2 0 0 0 0 2 1 1 1 1 1 1/2 1/2 [1] F = 1 0 2 0 0 0 2 0 2 2 0 0 = 0 0, [t] F = 1 1 0 0 0 2 0 0 0 [t 2 ] F = 1/2 0 1 0, [t3 ] F = 1/2 0 0 1 22. Find the coordinates of (2,6) with respect to each of the following bases of R 2 : a. {(1, 0), (0, 1)} b. {(0, 1), (1, 0)} c. {(2, 6), (3, 0)} d. {(3, 0), (2, 6)} a. [(2, 6)] = [2, 6] b. [(2, 6)] = [6, 2] c. [(2, 6)] = [1, 0] d. [(2, 6)] = [0, 1]. 16

Page 106 5. Let V be the vector space consisting of all real-valued functions defined on [0,1]. Which of the following subsets are subspaces, and which of these are finite-dimensional? a. All polynomials of degree 4 This set is a subspace. In fact, it s P 4, so it s finite dimensional. b. All f(t) for which f( 1 2 ) = 0 This is an infinite dimensional subspace. c. All f(t) such that f(0) = f(1) This is an infinite dimensional subspace. d. All f(t) such that f(0) + f(1) = 1 Not a subspace. The set is not closed under vector addition or scalar multiplication. e. All functions of the form ce rt for some constants c and r Not a subspace. The set is not closed under vector addition. e t and e 2t are in the set, but there are no constants c and r such that e t + e 2t = ce rt. To see that this is the case. Suppose such constants exist. Then c = 2, and r = 3/2 in order that both sides of the equality have the same value at t = 0 and their derivatives have the same value at t = 0, but then the second derivatives will not be equal at t = 0. 14. For which numbers x are the vectors (2, 3) and (1, x) linearly independent? We need a condition on x which will guaranty that if a linear combination of these two equals the zero vector, then it must be the trivial linear combination. So suppose c 1 and c 2 are such that c 1 (2, 3) + c 2 (1, x) = (0, 0) = 2c 1 + c 2 = 0 6c 1 + 3c 2 = 0 3c 1 + xc 2 = 0 6c 1 + 2xc 2 = 0 This tells us that (2x 3)c 2 = 0. To force c 2 = 0 we must have x 3/2. 17