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R3 span Hint: We know that R = span(e1, e2, e3).] Use the method of Example 2.23 and Theorem 2.6 to deter- mine if the sets of vectors in Exercises 22-31 are dependent. If, for any of these, the answer can be determined by inspection (i.e., without calculation), state why. For any sets that are linearly in- linearly dependent, find a dependence relation- ship among the vectors. 2 -1 22. 2 - 1 23. 3 3 2 2 3 2 2 24. 2 25. 1 -5 2 "ectors 2 4 3 5 3 26. 27. 4 -1 5 3 21 1 2 3 28. 2 2 ns the of the FENELETA 1913 1 29 ь и + 10 30. 10 3 3 -S = 2 1. hrase this -J span(T) 3 -1 3 31. 1 ation linear 3 that w is erefore In Exercises 32-41, determine if the sets of vectors in the given exercise are linearly independent by converting the LO 2 1 94 Chapter 2 Systems of Linear Equations 95 Section 2.3 Spanning Sets and Linear Independence cheating a bit in this proof. After all, we cannot be sure that v, is a linear combination of the other vectors, nor that ci is nonzero. However, the argument is analogous for some other vector v or for a different scalar , we can just relabel things so that they work out as in the above proof. In a situation like this, a mathematician might begin by saying, "without loss of gen- we may assume that v, is a linear combination of the other vectors and then Note It may appear as if we are The reduced row echelon form is 1 0 010 Cf. Alternatively, 0 1 0 0 0 0 erality. proceed as above. linearly independent. (check this), soC 0, c2= 0 , c3= 0. Thus, the given vectors are (c) A little reflection reveals that 0 Example 2.22 Any set of vectors containing the zero vector is linearly dependent. For if 0, V2. are in R", then we can find a nontrivial combination of the form c 0 + Cv2+ 1 + 0 so the three vectors are linearly dependent. [Set up a linear system as in part (b) to check this algebraically.] Cm Vm 0 by setting c= 1 and c2 c3 =. =Cm= 0. we observe no obvious dependence so we proceed directly to reduce (d) Once again, homogeneous linear system whose augmented matrix has as its columns the given Example 2.23 a Determine whether the following sets of vectors are linearly independent: vectors: Ry+ R2 0 3|0 1 R3- R 0 1 1 | 0 1 R-2R 0 1 1 1 |0 (a) and 2 2 0 1 -2 0 (b) and 0 2 1 4 0 -1 -1 2|0 0 0 0 0 -1 2 0 0 If we let the scalars be ci, c2, and c3, we have (c) 1,and 0 (d) 2 and 4 3c3= 0 -1J 1J 2 C22c3= 0 Solution In answering any question of this type, it is a good idea to see if you can determine by inspection whether one vector is a linear combination of the others. A little thought may save a lot of computation! from which we see that the system has infinitely many solutions. In particular, there must be a nonzero solution, so the given vectors are If we continue, we can describe these solutions exactly: c Thus, for any nonzero value of c3, we have the linear dependence relation linearly dependent. -3c3 and c2 2c3. (a) The only way two vectors can be linearly dependent is if one is a multiple of the other. (Why?) These two vectors are clearly not multiples, so they independent. (b) There is no obvious dependence relation here, so we try to find scalars c, C such that 0 are linearly -3c3 2 +203 +c3 4 0 2 Y (Once again, check that this is correct.) 0 0 C1 + c1+ c3 We summiarize this procedure for testing for linear independence as a theorem. The corresponding linear system is Theorem 2.6 Let vi, V2..., vm be (column) vectors in R" and let A be the n X m matrix [v1 V2 dependent if and only if the homogeneous linear system with augmented matrix [A 0] has a nontrivial solution. with these vectors as its columns. Then v, V C3=0 are linearly = 0 C2 + C3= 0 and the augmented matrix is 1 0 1|0 1 1 0 0 Proof Vi, V2.,Vm are linearly dependent if and only if there are scalars c, C, not all zero, such that cv + c2v2 + + CmVm= 0. By Theorem 2.4, this is equivalent C) LO 1 1|0. to saying that the nonzero vector matrix is [v1 v2 . . Vm 0. Cz is a solution of the system whose augmented , we make the fundamental observation that the columns of the coefficient just the vectors in question! Once again, matrix are m

R3 span Hint: We know that R = span(e1, e2, e3).] Use the method of Example 2.23 and Theorem 2.6 to deter- mine if the sets of vectors in Exercises 22-31 are dependent. If, for any of these, the answer can be determined by inspection (i.e., without calculation), state why. For any sets that are linearly in- linearly dependent, find a dependence relation- ship among the vectors. 2 -1 22. 2 - 1 23. 3 3 2 2 3 2 2 24. 2 25. 1 -5 2 "ectors 2 4 3 5 3 26. 27. 4 -1 5 3 21 1 2 3 28. 2 2 ns the of the FENELETA 1913 1 29 ь и + 10 30. 10 3 3 -S = 2 1. hrase this -J span(T) 3 -1 3 31. 1 ation linear 3 that w is erefore In Exercises 32-41, determine if the sets of vectors in the given exercise are linearly independent by converting the LO 2 1 94 Chapter 2 Systems of Linear Equations 95 Section 2.3 Spanning Sets and Linear Independence cheating a bit in this proof. After all, we cannot be sure that v, is a linear combination of the other vectors, nor that ci is nonzero. However, the argument is analogous for some other vector v or for a different scalar , we can just relabel things so that they work out as in the above proof. In a situation like this, a mathematician might begin by saying, "without loss of gen- we may assume that v, is a linear combination of the other vectors and then Note It may appear as if we are The reduced row echelon form is 1 0 010 Cf. Alternatively, 0 1 0 0 0 0 erality. proceed as above. linearly independent. (check this), soC 0, c2= 0 , c3= 0. Thus, the given vectors are (c) A little reflection reveals that 0 Example 2.22 Any set of vectors containing the zero vector is linearly dependent. For if 0, V2. are in R", then we can find a nontrivial combination of the form c 0 + Cv2+ 1 + 0 so the three vectors are linearly dependent. [Set up a linear system as in part (b) to check this algebraically.] Cm Vm 0 by setting c= 1 and c2 c3 =. =Cm= 0. we observe no obvious dependence so we proceed directly to reduce (d) Once again, homogeneous linear system whose augmented matrix has as its columns the given Example 2.23 a Determine whether the following sets of vectors are linearly independent: vectors: Ry+ R2 0 3|0 1 R3- R 0 1 1 | 0 1 R-2R 0 1 1 1 |0 (a) and 2 2 0 1 -2 0 (b) and 0 2 1 4 0 -1 -1 2|0 0 0 0 0 -1 2 0 0 If we let the scalars be ci, c2, and c3, we have (c) 1,and 0 (d) 2 and 4 3c3= 0 -1J 1J 2 C22c3= 0 Solution In answering any question of this type, it is a good idea to see if you can determine by inspection whether one vector is a linear combination of the others. A little thought may save a lot of computation! from which we see that the system has infinitely many solutions. In particular, there must be a nonzero solution, so the given vectors are If we continue, we can describe these solutions exactly: c Thus, for any nonzero value of c3, we have the linear dependence relation linearly dependent. -3c3 and c2 2c3. (a) The only way two vectors can be linearly dependent is if one is a multiple of the other. (Why?) These two vectors are clearly not multiples, so they independent. (b) There is no obvious dependence relation here, so we try to find scalars c, C such that 0 are linearly -3c3 2 +203 +c3 4 0 2 Y (Once again, check that this is correct.) 0 0 C1 + c1+ c3 We summiarize this procedure for testing for linear independence as a theorem. The corresponding linear system is Theorem 2.6 Let vi, V2..., vm be (column) vectors in R" and let A be the n X m matrix [v1 V2 dependent if and only if the homogeneous linear system with augmented matrix [A 0] has a nontrivial solution. with these vectors as its columns. Then v, V C3=0 are linearly = 0 C2 + C3= 0 and the augmented matrix is 1 0 1|0 1 1 0 0 Proof Vi, V2.,Vm are linearly dependent if and only if there are scalars c, C, not all zero, such that cv + c2v2 + + CmVm= 0. By Theorem 2.4, this is equivalent C) LO 1 1|0. to saying that the nonzero vector matrix is [v1 v2 . . Vm 0. Cz is a solution of the system whose augmented , we make the fundamental observation that the columns of the coefficient just the vectors in question! Once again, matrix are m

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.3: Spanning Sets And Linear Independence
Problem 30EQ
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R3 span Hint: We know that R = span(e1, e2, e3).] Use the method of Example 2.23 and Theorem 2.6 to deter- mine if the sets of vectors in Exercises 22-31 are dependent. If, for any of these, the answer can be determined by inspection (i.e., without calculation), state why. For any sets that are linearly in- linearly dependent, find a dependence relation- ship among the vectors. 2 -1 22. 2 - 1 23. 3 3 2 2 3 2 2 24. 2 25. 1 -5 2 "ectors 2 4 3 5 3 26. 27. 4 -1 5 3 21 1 2 3 28. 2 2 ns the of the FENELETA 1913 1 29 ь и + 10 30. 10 3 3 -S = 2 1. hrase this -J span(T) 3 -1 3 31. 1 ation linear 3 that w is erefore In Exercises 32-41, determine if the sets of vectors in the given exercise are linearly independent by converting the LO 2 1
Transcribed Image Text:R3 span Hint: We know that R = span(e1, e2, e3).] Use the method of Example 2.23 and Theorem 2.6 to deter- mine if the sets of vectors in Exercises 22-31 are dependent. If, for any of these, the answer can be determined by inspection (i.e., without calculation), state why. For any sets that are linearly in- linearly dependent, find a dependence relation- ship among the vectors. 2 -1 22. 2 - 1 23. 3 3 2 2 3 2 2 24. 2 25. 1 -5 2 "ectors 2 4 3 5 3 26. 27. 4 -1 5 3 21 1 2 3 28. 2 2 ns the of the FENELETA 1913 1 29 ь и + 10 30. 10 3 3 -S = 2 1. hrase this -J span(T) 3 -1 3 31. 1 ation linear 3 that w is erefore In Exercises 32-41, determine if the sets of vectors in the given exercise are linearly independent by converting the LO 2 1
94 Chapter 2 Systems of Linear Equations 95 Section 2.3 Spanning Sets and Linear Independence cheating a bit in this proof. After all, we cannot be sure that v, is a linear combination of the other vectors, nor that ci is nonzero. However, the argument is analogous for some other vector v or for a different scalar , we can just relabel things so that they work out as in the above proof. In a situation like this, a mathematician might begin by saying, "without loss of gen- we may assume that v, is a linear combination of the other vectors and then Note It may appear as if we are The reduced row echelon form is 1 0 010 Cf. Alternatively, 0 1 0 0 0 0 erality. proceed as above. linearly independent. (check this), soC 0, c2= 0 , c3= 0. Thus, the given vectors are (c) A little reflection reveals that 0 Example 2.22 Any set of vectors containing the zero vector is linearly dependent. For if 0, V2. are in R", then we can find a nontrivial combination of the form c 0 + Cv2+ 1 + 0 so the three vectors are linearly dependent. [Set up a linear system as in part (b) to check this algebraically.] Cm Vm 0 by setting c= 1 and c2 c3 =. =Cm= 0. we observe no obvious dependence so we proceed directly to reduce (d) Once again, homogeneous linear system whose augmented matrix has as its columns the given Example 2.23 a Determine whether the following sets of vectors are linearly independent: vectors: Ry+ R2 0 3|0 1 R3- R 0 1 1 | 0 1 R-2R 0 1 1 1 |0 (a) and 2 2 0 1 -2 0 (b) and 0 2 1 4 0 -1 -1 2|0 0 0 0 0 -1 2 0 0 If we let the scalars be ci, c2, and c3, we have (c) 1,and 0 (d) 2 and 4 3c3= 0 -1J 1J 2 C22c3= 0 Solution In answering any question of this type, it is a good idea to see if you can determine by inspection whether one vector is a linear combination of the others. A little thought may save a lot of computation! from which we see that the system has infinitely many solutions. In particular, there must be a nonzero solution, so the given vectors are If we continue, we can describe these solutions exactly: c Thus, for any nonzero value of c3, we have the linear dependence relation linearly dependent. -3c3 and c2 2c3. (a) The only way two vectors can be linearly dependent is if one is a multiple of the other. (Why?) These two vectors are clearly not multiples, so they independent. (b) There is no obvious dependence relation here, so we try to find scalars c, C such that 0 are linearly -3c3 2 +203 +c3 4 0 2 Y (Once again, check that this is correct.) 0 0 C1 + c1+ c3 We summiarize this procedure for testing for linear independence as a theorem. The corresponding linear system is Theorem 2.6 Let vi, V2..., vm be (column) vectors in R" and let A be the n X m matrix [v1 V2 dependent if and only if the homogeneous linear system with augmented matrix [A 0] has a nontrivial solution. with these vectors as its columns. Then v, V C3=0 are linearly = 0 C2 + C3= 0 and the augmented matrix is 1 0 1|0 1 1 0 0 Proof Vi, V2.,Vm are linearly dependent if and only if there are scalars c, C, not all zero, such that cv + c2v2 + + CmVm= 0. By Theorem 2.4, this is equivalent C) LO 1 1|0. to saying that the nonzero vector matrix is [v1 v2 . . Vm 0. Cz is a solution of the system whose augmented , we make the fundamental observation that the columns of the coefficient just the vectors in question! Once again, matrix are m
Transcribed Image Text:94 Chapter 2 Systems of Linear Equations 95 Section 2.3 Spanning Sets and Linear Independence cheating a bit in this proof. After all, we cannot be sure that v, is a linear combination of the other vectors, nor that ci is nonzero. However, the argument is analogous for some other vector v or for a different scalar , we can just relabel things so that they work out as in the above proof. In a situation like this, a mathematician might begin by saying, "without loss of gen- we may assume that v, is a linear combination of the other vectors and then Note It may appear as if we are The reduced row echelon form is 1 0 010 Cf. Alternatively, 0 1 0 0 0 0 erality. proceed as above. linearly independent. (check this), soC 0, c2= 0 , c3= 0. Thus, the given vectors are (c) A little reflection reveals that 0 Example 2.22 Any set of vectors containing the zero vector is linearly dependent. For if 0, V2. are in R", then we can find a nontrivial combination of the form c 0 + Cv2+ 1 + 0 so the three vectors are linearly dependent. [Set up a linear system as in part (b) to check this algebraically.] Cm Vm 0 by setting c= 1 and c2 c3 =. =Cm= 0. we observe no obvious dependence so we proceed directly to reduce (d) Once again, homogeneous linear system whose augmented matrix has as its columns the given Example 2.23 a Determine whether the following sets of vectors are linearly independent: vectors: Ry+ R2 0 3|0 1 R3- R 0 1 1 | 0 1 R-2R 0 1 1 1 |0 (a) and 2 2 0 1 -2 0 (b) and 0 2 1 4 0 -1 -1 2|0 0 0 0 0 -1 2 0 0 If we let the scalars be ci, c2, and c3, we have (c) 1,and 0 (d) 2 and 4 3c3= 0 -1J 1J 2 C22c3= 0 Solution In answering any question of this type, it is a good idea to see if you can determine by inspection whether one vector is a linear combination of the others. A little thought may save a lot of computation! from which we see that the system has infinitely many solutions. In particular, there must be a nonzero solution, so the given vectors are If we continue, we can describe these solutions exactly: c Thus, for any nonzero value of c3, we have the linear dependence relation linearly dependent. -3c3 and c2 2c3. (a) The only way two vectors can be linearly dependent is if one is a multiple of the other. (Why?) These two vectors are clearly not multiples, so they independent. (b) There is no obvious dependence relation here, so we try to find scalars c, C such that 0 are linearly -3c3 2 +203 +c3 4 0 2 Y (Once again, check that this is correct.) 0 0 C1 + c1+ c3 We summiarize this procedure for testing for linear independence as a theorem. The corresponding linear system is Theorem 2.6 Let vi, V2..., vm be (column) vectors in R" and let A be the n X m matrix [v1 V2 dependent if and only if the homogeneous linear system with augmented matrix [A 0] has a nontrivial solution. with these vectors as its columns. Then v, V C3=0 are linearly = 0 C2 + C3= 0 and the augmented matrix is 1 0 1|0 1 1 0 0 Proof Vi, V2.,Vm are linearly dependent if and only if there are scalars c, C, not all zero, such that cv + c2v2 + + CmVm= 0. By Theorem 2.4, this is equivalent C) LO 1 1|0. to saying that the nonzero vector matrix is [v1 v2 . . Vm 0. Cz is a solution of the system whose augmented , we make the fundamental observation that the columns of the coefficient just the vectors in question! Once again, matrix are m
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