6.4 BASIS AND DIMENSION (Review) DEF 1 Vectors v 1, v 2,, v k in a vector space V are said to form a basis for V if. (a) v 1,, v k span V and

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1 6.4 BASIS AND DIMENSION (Review) DEF 1 Vectors v 1, v 2,, v k in a vector space V are said to form a basis for V if (a) v 1,, v k span V and (b) v 1,, v k are linearly independent. HMHsueh 1

2 Natural Basis or Standard Basis of R n, P n : {e 1, e 2,, e n } is the natural basis for R n. {t n, t n 1,, t, 1} is the natural basis for P n. HMHsueh 2

3 EX 1 (Ex. 2) Show that S = {v 1, v 2, v 3, v 4 } is a basis for R 4, where v 1 = (1, 0, 1, 0), v 2 = (0, 1, 1, 2), v 3 = (0, 2, 2, 1), v 4 = (1, 0, 0, 1). Proof. (a) Show that S spans R 4. For any vector v = (a, b, c, d) R 4, find the scalars k 1, k 2, k 3, k 4 such that k 1 v 1 + k 2 v 2 + k 3 v 3 + k 4 v 4 = v. For simplicity, consider to express the vectors v 1,, v 4 and v in column vectors. HMHsueh 3

4 Define the matrix A and the column vector k, Since A = ( v 1 v 2 v 3 v 4 ), k = Ak = ( v 1 v 2 v 3 v 4 ) The equation is equivalent to k 1 k 2 k 3 k 4 k 1 k 2 k 3 k 4. = v 1 k 1 + v 2 k 2 + v 3 k 3 + v 4 k 4. Ak = v, where v is the vector v expressed in column. HMHsueh 4

5 Since the matrix A A = = ( v 1 v 2 v 3 v 4 ), is row equivalent to an identity matrix, I 4, the solution exists and is equal to k 1 = 4a + 5b 3c 4d, k 2 = 2a + 3b 2c 2d, k 3 = a b + c + d, k 4 = 3a 5b + 3c + 4d. That is, any vector v in R 4 is a linear combination of v 1, v 2, v 3, v 4, and thus span S = V. HMHsueh 5

6 (2)Show that S is linearly independent. Again, if the zero vector is expressed by c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = 0, then according to previous results, it is equivalent to solve Ac = 0, where c = Since A is row equivalent to I 4, only the trivial solution is found, Thus, S is linearly independent. S is a basis for R 4. c 1 c 2 c 3 c 4. c 1 = c 2 = c 3 = c 4 = 0. HMHsueh 6

7 REMARKS: Determine whether S = {v 1,, v k } is a basis of V = R n. Step 0. Written the vectors in S in column vectors and form the coefficient matrix A = (v 1,, v k ). Step 1. S spans V. For any v V, check the existence of a solution k for Ak = v. Step 2. S is linearly independent. Check if the homogenous linear system Ak = 0 has only the trivial solution. HMHsueh 7

8 A vector space V is called finite-dimensional if a basis of V which is a finite subset of V exists. If there is no such finite subset of V, then V is called infinitedimensional. For example,r 2, R 4, P 2, P 4 are finite-dimensional. But P = {all polynomials}, C(, ) = {all continuous functions, f : R R } are infinite-dimensional. HMHsueh 8

9 V : A vector space and S = {v 1, v 2,, v n } V. THM 1 (Thm. 6.5) S is a basis of V if and only if every vector in V can be written in unique way as a linear combination of the vectors in S. THM 2 (Thm. 6.6) Let W = span S. Then some subset T of S is a basis for W. T=? HMHsueh 9

10 Find the subset T when V = R m (P306) If V = R m and S = {v 1,, v n }, each v i R m, n m. Let W = span S. Find a basis T of W which is a subset of S. SOL. Step 1: If S is linearly independent, S is a basis of W. Step 2: If S is not linearly independent, find and remove the set of vectors which are linear combinations of other vectors. The remaining set T is shown to span W. Step 3: Show that T is linearly independent. T is thus a basis of W. HMHsueh 10

11 S is linearly independent? Solve the homogeneous linear system of m equations in n unknowns c i s, c 1 v 1 + c 2 v c n v n = 0. Expressing each v i as a column vector, form the coefficient matrix A m n = (v 1 v 2 v n ), and the linear system can be written by Ac = 0, where c = If the trivial solution is the only solution, then S is linearly independent and S itself is a basis of W. c 1. c n. HMHsueh 11

12 Now since n m, the number of unknowns is no less than the number of equations, nontrivial solutions exist. (Step 1 fails and Step 2 continues). HMHsueh 12

13 Assume the reduced row echelon form of A has r(r m) nonzero rows. Without loss of generality, assume that the r leading 1s in the nonzero rows occur in the first r columns. Let the reduced row echelon form of A be B = b 1 r+1 b 1 n b 2 r+1 b 2 n b r r+1 b r n then c 1,, c r can be solved in terms of c r+1,, c n., HMHsueh 13

14 Thus c 1 = b 1 r+1 c r+1 b 1 n c n c 2 = b 2 r+1 c r+1 b 2 n c n. c r = b r r+1 c r+1 b r n c n, where c r+1,, c n are arbitrary real values. Q: Which vectors are linear combinations of others? A: The vectors v r+1,, v n corresponding to the non-leading 1 columns. HMHsueh 14

15 Show that v r+1 is a linear combination of v 1,, v r. Consider the solution c r+1 = 1, c r+2 = 0,, c n = 0, c 1 = b 1 r+1, c 2 = b 2 r+1,, c r = b r r+1, for the linear system, we have c 1 v 1 + c 2 v c n v n = 0 b 1 r+1 v 1 b 2 r+1 v 2 + b r r+1 v r + v r+1 = 0, b 1 r+1 v 1 + b 2 r+1 v b r r+1 v r = v r+1. v r+1 is a linear combination of v 1,, v r. Thus span {v 1,, v r, v r+2,, v n } = span S = W, and v r+1 can be deleted from S. HMHsueh 15

16 Show that v r+2 is a linear combination of v 1,, v r. Consider the solution c r+1 = 0, c r+2 = 1,, c n = 0, and c 1 = b 1 r+2, c 2 = b 2 r+2,, c r = b r r+2 for the linear system, we have c 1 v 1 + c 2 v c n v n = 0 b 1 r+2 v 1 b 2 r+2 v 2 + b r r+2 v r + v r+2 = 0, b 1 r+2 v 1 + b 2 r+2 v b r r+2 v r = v r+2, v r+2 is a linear combination of v 1,, v r. Thus span {v 1,, v r, v r+3,, v n } = span S = W, and v r+2 can be deleted from S as well. HMHsueh 16

17 Continuing the process, we found that v r+1,, v n combinations of {v 1,, v r } and are linear span {v 1,, v r } = span S = W, that is, T = {v 1,, v r } is a spanning set of W. (End of Step 2) HMHsueh 17

18 Show that T = {v 1,, v r } is linearly independent. Let A D be the matrix formed by v 1,, v r, A D = (v 1 v 2 v r ). Now consider to solve the following homogenous linear system, c 1 v c r v r = 0 A D c = 0. Then according to previous results, A D is row equivalent to the matrix B D which consists of the first r columns of B, B D = HMHsueh 18

19 Then the homogeneous systems A D c = 0 and B D c = 0 are equivalent and have only the trivial solution, c 1 = = c r = 0. Thus, T = {v 1,, v r } are linearly independent. We have found a subset T of S which is a basis of span S. HMHsueh 19

20 EX 2 (Ex. 5) Let S = {v 1,, v 5 } R 4, where v 1 = (1, 2, 2, 1), v 2 = ( 3, 0, 4, 3), v 3 = (2, 1, 1, 1) v 4 = ( 3, 3, 9, 6), v 5 = (9, 3, 7, 6). Find a subset of S that is a basis for W = span S. SOL. Form the coefficient matrix A, A = (v 1 v 2 v 3 v 4 v 5 ) = HMHsueh 20

21 Since the reduced row echelon form of A, B = has leading 1 in columns 1 and 2., So {v 1, v 2 } is a basis for W = span S. HMHsueh 21

22 REMARK: The basis for W depends on the order of the vectors in the original set S. (Ex. 5 Cont.) Let S = {v 4, v 3, v 2, v 1, v 5 }, then the matrix A and its reduced row echelon form B are A = (v 4 v 3 v 2 v 1 v 5 ), B = Thus {v 4, v 3 } is also a basis for W = span S.. HMHsueh 22

23 REMARK: A nonzero real vector space always has infinitely many bases. Corollary. (T.9) If {v 1, v 2,, v n } is a basis for a vector space V, then {cv 1, v 2,, v n } is also a basis if c 0. HMHsueh 23

24 THM 3 (Thm.6.7) If S = {v 1,, v n } is a basis for a vector space V and T = {w 1,, w r } V is a linearly independent set, then r n. HMHsueh 24

25 PROOF. Let T 1 = {w 1, v 1,, v n }. Then span S = span T 1 = V. Since w 1 is a linear combination of S, T 1 is linearly dependent. By Thm. 6.4, some v j is a linear combination of the preceding vectors in T 1. Deleting the v j in T 1, we have S 1 = {w 1, v 1,, v j 1, v j+1,, v n }. Since any vector in V can be expressed by a linear combination of S 1, S 1 spans V. HMHsueh 25

26 Next, let T 2 = {w 2, w 1, v 1,, v j 1, v j+1,, v n }. Then since w 2 is a linear combination of S 1, T 2 is linearly dependent. Some vector is a linear combination of preceding vectors in T 2. Since w 1, w 2 from T are linearly independent, the vector cannot be w 2 and must be some v i, i j. Deleting the v i from T 2, we have S 2 and S 2 spans V. Repeat this process. HMHsueh 26

27 If n r, one obtains and S n spans V. Then the set S n = {w n, w n 1,, w 1 }, T n+1 = {w n+1, w n, w n 1,, w 1 } should be linearly dependent which contradicts the assumption of independence for T. Thus r n. HMHsueh 27

28 REMARKS: If S = {v 1,, v n } and T = {w 1,, w r } are subsets of a vector space V. (1) If S is a basis and r > n, then T is a linearly dependent set. (2) If T is linear independent and r > n, then S cannot be a basis of V. HMHsueh 28

29 COROLLARY 6.1 If S = {v 1, v 2,, v n } and T = {w 1, w 2,, w m } are bases for a vector space, then n = m. PROOF. Since S is a basis and T is linearly independent, Thm. 6.7 implies that m n. Similarly, since T is a basis and S is linearly independent, thus n m. Thus, m = n. HMHsueh 29

30 DIMENSION DEF 2 The dimension of a nonzero vector space V is the number of vectors in a basis for V. For the zero vector space {0}. The space is spanned by the set {0} which is linearly dependent and is not a basis. The zero vector space is thus set to have dimension 0. HMHsueh 30

31 EX 3 (Ex. 6, 7) The dimension of R 2 is 2; the dimension of R 3 is 3; in general, the dimension of R n is n. (Hint: The standard basis of R n?) The dimension of P 2 is 3; the dimension of P 3 is 4; in general, the dimension of P n is n + 1. (Hint: The standard basis of P n?) HMHsueh 31

32 EX 4 (Ex. 8) In Ex. 5, S = {v 1,, v 5 } R 4, where v 1 = (1, 2, 2, 1), v 2 = ( 3, 0, 4, 3), v 3 = (2, 1, 1, 1) v 4 = ( 3, 3, 9, 6), v 5 = (9, 3, 7, 6), and W = span S. It s found that the basis of the subspace W = span S is {v 1, v 2 }. Thus W has dimension 2. HMHsueh 32

33 REMARKS: (T.2) If V is a finite-dimensional vector space, then every nonzero subspace W of V has a finite basis and dim W dim V. Hint: The basis of W is linearly independent in V. HMHsueh 33

34 (T.3, T.4)If a vector space V has dimension n,then any set of n + 1 vectors in V is linearly dependent. (Hint:Simply by Thm. 6.7) no set of n 1 vectors can span V. (Hint: If there exists such a set, dim V < n. Contradiction.) HMHsueh 34

35 THM 4 (Thm 6.8)(T.5) If S is a linearly independent set in a finite-dimensional vector space V, then there is a basis T for V, which contains S. PROOF. (Using the similar approach of Thm 6.7) Assume that dim V = n and a basis of V is T 0 = {v 1,, v n }. Further S = {w 1,, w r }, then by Thm.6.7, r n. Consider the set T 1 = {w 1, v 1,, v n }. T 1 spans V and is linearly dependent. Then there exists one v i which is a linear combination of the preceding vectors. Deleting v i, the resulting set S 1 is a basis of V. HMHsueh 35

36 Further consider the set T 2 = {w 2, w 1, v 1,, v i 1, v i+1,, v n }, again, T 2 spans V but is linearly dependent. Since w 2, w 1 are linear independent, there exists one v j which is a linear combination of the preceding vectors. Deleting v j, the resulting set S 2 is a basis of V. Repeat the process until S is contained in the basis. Then one has found a basis which contains S. HMHsueh 36

37 EX 5 (Ex. 9) Find a basis for R 4 that contains the vectors v 1 = (1, 0, 1, 0), and v 2 = ( 1, 1, 1, 0). SOL. Consider the natural basis of R 4, e 1 = (1, 0, 0, 0), e 2 = (0, 1, 0, 0) e 2 = (0, 0, 1, 0) e 4 = (0, 0, 0, 1). Let S = {v 1, v 2, e 1, e 2, e 3, e 4 }. Then S spans V but is linearly dependent. Now use the strategy on P306 to find a subset of S which is a basis for R 4 = span S. HMHsueh 37

38 Form the coefficient matrix A = (v 1 v 2 e 1 e 2 e 3 e 4 ) = Then it can be found that the reduced row echelon form of A, has leading 1s in columns 1,2,3,6., Thus, {v 1, v 2, e 1, e 4 } is a basis for R 4 containing v 1 and v 2. HMHsueh 38

39 THM 5 (Thm. 6.9)(T.6) Let V be an n-dimensional vector space, and let S = {v 1, v 2,, v n } be a set of n vectors in V. (a) If S is linearly independent, then it is a basis for V. (Hint: By Thm.6.8) (b) If S spans V, then it is a basis for V. (Hint: If S is linear dependent, dim < n. Contradiction!) HMHsueh 39

40 EX 6 (Ex. 10) In Ex. 5, S R 4 and W = span S, thus dim W 4. Since S contains 5 vectors, S can t be linearly independent and thus is not a basis for W. In Ex. 2, S R 4 and S contains 4 vectors. Thus S is possibly a basis for R 4. By Thm 6.9, only one of the conditions, S is linearly independent or S spans R 4, is necessary to check. HMHsueh 40

41 PART I: EXERCISE 2(a),4(a),6,7,9,14,18,21,30,32,35 T1,T8,T9,T10,T12,T13 PART II: EXERCISE 11, 13,15,28 T1, T2, T3, T4, T6, T7 HMHsueh 41