Vector Spaces 4.4 Spanning and Independence

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1 Vector Spaces 4.4 and Independence October 18

2 Goals Discuss two important basic concepts: Define linear combination of vectors. Define Span(S) of a set S of vectors. Define linear Independence of a set of vectors.

3 Set theory and set theoretic Notations Borrow (re-introduce) some Set theoretic lingo and notations. A collection S of objects is called a set. Objects in S are called elements of S. We write x S to mean x is in S or x is an element of S. Given two sets, T,S we say T is a subset of S, if each element of T is in S. We write T S to mean T is a subset of S. Read the notation = as implies.

4 Linear Combination Preview Definition. Let V be a vector space and v be a vector in V. Then, v is said to be a linear combination of vectors u 1,u 2,...,u k in V, if v = c 1 u 1 +c 2 u 2 + +c k u k for some scalars c 1,c 2,...,c k R. Reading Assignment: 4.4 Example 1, 2, 3

5 Finding Linear Combination Exercise 2a. Let S = {(1,2, 2),(2, 1,1)} be a set of two vectors in R 3. (a) Write u = (1, 5,5) as a linear combination of the vectors in S, if possible. Solution: Write (1, 5,5) = a(1,2, 2)+b(2, 1,1). So, (1, 5,5) = (a+2b,2a b, 2a+b). So, a +2b = 1 2a b = 5 2a +b = 5

6 The augmented matrix Its reduced row echelon form (use TI-84 rref ) so a = 9 5,b = 7 5. So, (1, 5,5) = 9 5 (1,2, 2)+ 7 5 (2, 1,1).

7 Exercise 2b Exercise2b. Write v = ( 2, 6,6) as a linear combination of the vectors in S, if possible. Solution: Write ( 2, 6,6) = a(1,2, 2)+b(2, 1,1). So, ( 2, 6,6) = (a+2b,2a b, 2a+b). So, a +2b = 2 2a b = 6 2a +b = 6

8 The augmented matrix Its reduced row echelon form (use TI-84 rref ) so a = ,b = So, v = ( 2, 6,6) = 14 5 (1,2, 2)+ 2 5 (2, 1,1).

9 Exercise 2c Exercise2c. Write w = ( 1, 22,22) as a linear combination of the vectors in S, if possible. Solution: Write ( 1, 22,22) = a(1,2, 2)+b(2, 1,1). So, ( 1, 22,22) = (a+2b,2a b, 2a+b). So, a +2b = 1 2a b = 22 2a +b = 22

10 The augmented matrix Its reduced row echelon form (use TI-84 rref ) so a = 9, b = 4. So, w = ( 1, 22,22) = 9(1,2, 2)+4(2, 1,1).

11 Exercise 2d Exercise 2d. Write z = ( 4, 3,3) as a linear combination of the v ectors in S, if possible. Solution: Write ( 4, 3,3) = a(1,2, 2)+b(2, 1,1). So, ( 4, 3,3) = (a+2b,2a b, 2a+b). So, a +2b = 4 2a b = 3 2a +b = 3

12 The augmented matrix Its reduced row echelon form (use TI-84 rref ) so a = 2b = So, z = ( 4, 3,3) = 2(1,2, 2) (2, 1,1).

13 Span of a Sets Definition. Let S = {v 1,v 2,...,v k } be a subset of a vector space V. The span of S is the set of all linear combinations of vectors in S, span(s) = {c 1 v 1 +c 2 v 2 +c k v k : c 1,c 2,,c k are scalars} The span(s) is also denoted by span(v 1,v 2,...,v k ). If V = span(s), we say V is spanned by S. We also say that S is a spanning set of V.

14 Span(S) is a subspace ofv Theorem 4.7 Let S = {v 1,v 2,...,v k } be a subset of a vector space V. Then, span(s) is a subspace of V. In fact, Span(S) is the smallest subspace of V that contains S. That means, if W is a subspace of V then, S W = span(s) W.

15 Proof. First we show that span(s) is a subspace of V. By theorem 4.5 we need to show that span(s) is closed under addition and scalar multiplication. Let u,v Span(S) and c be a scalar. Then u = c 1 v 1 +c 2 v 2 + +c k v k, v = d 1 v 1 +d 2 v 2 + +d k v k where c 1,c 2,...,c k,d 1,d 2,...,d k are scalars. Then u+v = (c 1 +d 1 )v 1 +(c 2 +d 2 )v 2 + +(c k +d k )v k cu = (cc 1 )v 1 +(cc 2 )v 2 + +(cc k )v k So, both u+v,cu are linear combination of elments in S. So u+v,cu span(s). So span(s) is a subspace of V.

16 Now, we prove that span(s) is the smallest subspace W, of V, that contains S. Suppose W is a subspace of V and S W. Let u span(s). we will have to show u W. Then, u = c 1 v 1 +c 2 v 2 + +c k v k, where c 1,c 2,...,c k are scalars. Now, v 1,v 2,...,v k W. Since W is closed under scalar multiplication c i v i W. Since W is closed under addition u W. The proof is complete.

17 More Problems on spanning sets of Sets Reading Assignment: 4.4 Example 4-6. Example (4a): Most obvious and natural spanning set of the 3 space R 3 is S = {(1,0,0),(0,1,0),(0,0,1)} Because for any vector u = (a,b,c) R 3 w e have u = (a,b,c) = a(1,0,0)+b(0,1,0)+c(0,0,1) Similarly, Most obvious and natural spanning set of the real plane R 2 is S = {(1,0),(0,1)}. Remark. If S is spanning set of V and T is a bigger set (i.e. S T) than T is also a spanning set of V.

18 4.4 Exercise 6. Preview More Problems on spanning sets Let S = {(1, 1),(2,1)}. Is S a spanning set of R 2? Solution. Yes, it is a spanning set of R 2. We need to show that any vector (x,y) R 2 is a linear combination of elements is S. { a +2b = x So, a(1, 1)+b(2,1) = (x,y) OR a +b = y must have at least one solution, for any (x,y). In the matrix form [ ][ ] [ ] 1 2 a x = 1 1 b y

19 More Problems on spanning sets Use TI-84 [ ] a = b [ ] 1 [ x y ] [ 1 2 = ][ x y ] Since the systems have solution for all (x,y) S is a spanning set R 2.

20 4.4 Exercise 10. Preview More Problems on spanning sets Let S = {(1,1)}. Is S a spanning set of R 2? Solution. No. Because span(s) = {c(1,1) : c R} = {(c,c) : c R}. is only the line through the origin and (1,1). It is strictly smaller that R 2. For example, (1,2) / span(s).

21 4.4 Exercise 20. Preview More Problems on spanning sets Let S = {(1,0,1),(1,1,0),(0,1,1)}. Is S a spanning set of R 3? Solution. Yes, it is a spanning set of R 3. We need to show that any vector (x,y,z) R 3 is a linear combination of elements is S. So, a(1,0,1)+b(1,1,0)+c(0,1,1) = (x,y,z) a +b = x OR b +c = y a +c = z must have at least one solution, for any (x,y,z).

22 More Problems on spanning sets In the matrix form a Use TI-84 b = c x y = z a b c = x y z x y z Showing that these systems have solutions of all (x,y,z).

23 on Linearly Independent sets Liniear Independence Definition. Let S = {v 1,v 2,...,v k } be a subset of a vector space V. The set S is said to be linearly independent, if c 1 v 1 +c 2 v 2, +c k v k = 0 = c 1 = c 2 = = c k = 0. That means, the equation on the left has only the trivial solution. If S is not linearly independent, we say that S is linearly dependent.

24 on Linearly Independent sets Comments (1) Let S = {v 1,v 2,...,v k } be a subset of a vector space V. If 0 S then, S is linearly dependent. Proof.For simplicity, assume v 1 = 0. Then, 1v 1 +0v v k = 0 So, S is not linearly indpendent. (2)Methods to test Independence: We will mostly be working with vector in R 2,R 3, or n spaces R n. Gauss-Jordan elimination (with TI-84) will be used to check if a set is independent.

25 on Linearly Independent sets A Property of Linearly Dependent Sets Theorem 4.8 Let S = {v 1,v 2,...,v k } be a subset of a vector space V. Assume S has at least 2 elements (k 2). Then, S is linearly dependent if and only if one of the vectors v j can be written as a linear combination of rest of the vectors in S. Proof. Again, we have to prove two statments.

26 on Linearly Independent sets First, we prove if part. We assume that one of the vectors v j can be written as a linear combination of rest of the vectors in S. For simplicity, we assume that v 1 is a linear combination of rest of the vectors in S. So, v 1 = c 2 v 2 +c 3 v 3 + +c k v k So, ( 1)v 1 +c 2 v 2 +c 3 v 3 + +c k v k = 0 This has at least one coefficient 1 that is nonzero. This establishes that S is a linearly dependent set.

27 on Linearly Independent sets Now, we prove the only if part. We assume that S is linearly dependent. So, there are scalars c 1,,c k, at least one non-zero, such that c 1 v 1 +c 2 v 2 +c 3 v 3 + +c k v k = 0 Without loss of generality (i.e. for simplicity) assume c 1 0. ( So, v 1 = c ) ( 2 v 2 + c ) ( ) 3 ck v v k c 1 c 1 c 1 Therefore, v 1 is a linear combination of the rest. The proof is complete.

28 on Linearly Independent sets of Linearly Independent sets 1. Reading Assignment: 4.4 Example 7, 8, 9, 10, Again, most obvious and natural example of linearly independent set in 3 space R 3 is S = {(1,0,0),(0,1,0),(0,0,1)} Because a(1,0,0)+b(0,1,0)+c(0,0,1) = (0,0,0) = a = b = c = Similarly, most obvious and natural example of linearly independent set in the real plane R 2 is S = {(1,0),(0,1)}. 4. Remark. If S is a linearly independent set set in V and R is a smaller set (i.e. R S) then R is also a linearly independent set of V.

29 Exercise 24. Preview on Linearly Independent sets Let S = {( 2,4),(1, 2)}. Is it linearly independent or dependent? Solution. We can see a non-trivial linear combination 1 ( 2,4)+2 (1, 2) = (0,0). So, S is not linearly independent. Alternately, if we want to prove it formally, suppose a( 2,4)+b(1, 2) = (0,0). { 2a +b = 0 Then, 4a 2b = 0

30 on Linearly Independent sets Add 2 times the first equation to the second: { 2a +b = 0 0 = 0 { a = t So, solution to this system is b = 2t So, there are lots of non-zero a,b. where t R.

31 on Linearly Independent sets Exercise 31. Let S = {( 4, 3,4),(1, 2,3),(6,0,0)}. Is it linearly independent or dependent? Solution. Let a( 4, 3,4)+b(1, 2,3)+c(6,0,0) = (0,0,0) We will have to see, there are non-zero solutions or not. So, 4a +b +6c = 0 3a 2b = 0 4a +3b = 0

32 on Linearly Independent sets The augmented matrix Its reduced row echelon form (use TI-84 rref ): So, the only solution is the the zero-solution: a = 0,b = 0,c = 0. We conclude S is linearly independent.

33 on Linearly Independent sets Exercise 44. Let S = {x 2 1,2x +5} be a set of polynomials. Is it linearly independent or dependent? Solution. Write a(x 2 1)+b(2x +5) = 0. So, ax 2 +2bx +( a+5b) = 0, Equating coefficients of x 2,x and the constant terms: a = 0,2b = 0, a+5b = 0 or a = b = 0. We conclude, S is linearly independent.

34 Preview : 4.4 Exercise 2a, 2b, 14, 15, 16, 32, 33, 34, 35, 41, 43, 51, 59 Need NOT do geometric interpretation.

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