DEF 1 Let V be a vector space and W be a nonempty subset of V. If W is a vector space w.r.t. the operations, in V, then W is called a subspace of V.

Transcription

1 6.2 SUBSPACES DEF 1 Let V be a vector space and W be a nonempty subset of V. If W is a vector space w.r.t. the operations, in V, then W is called a subspace of V. HMHsueh 1

2 EX 1 (Ex. 1) Every vector space has at least two subspaces, itself and {0}, the zero subspace. Proof. The zero subspace is a subspace of any vector space (α) If u, v {0}, then u v = 0 0 = 0 {0}. (β) If u {0} and c R, then c u = c 0 = 0 {0}. Moreover, the properties (a)-(h) can be verified easily. HMHsueh 2

3 EX 2 (Ex. 2) Check if the following W is a subspace of R 3, with the operations of the usual vector addition and scalar multiplication, Proof. W = {(a, b, 0) : a, b R}. (α) If u = (a 1, b 1, 0), v = (a 2, b 2, 0) W, then u v = (a 1 + a 2, b 1 + b 2, 0) = (a, b, 0) W, since a = a 1 + a 2, b = b 1 + b 2 R. HMHsueh 3

4 (β) If u = (a 1, b 1, 0) W and c R, then c u = (ca 1, cb 1, c0) = (a, b, 0) W since a = ca 1, b = cb 1 R. (a) For u = (a 1, b 1, 0), v = (a 2, b 2, 0) W, since u v = (a 1 + a 2, b 1 + b 2, 0) and v u = (a 2 + a 1, b 2 + b 1, 0), Thus u v = v u. Properties (b)-(h) can be verified easily. Thus, W is a subspace of R 3. HMHsueh 4

5 THM 1 (Thm 6.2) Let V be a vector space with operations,. Let W be a nonempty subset of V. Then W is a subspace of V if and only if (α) If u, v W, then u v W. (β) For any real c and u W, c u W. Note: According to this theorem, one only needs to check the conditions (α), (β) for W to be a subspace of V. HMHsueh 5

6 Proof.(T.1) : Straightforward. : If (α) and (β) are true, then for scalars c, d and u, v, w W, properties (a) u v = v u, (b) u (v w) = (u v) w, (e) (f) c (u v) = (c u) (c v), (c + d) u = (c u) (d u), (g) c (d u) = (cd) u, (h) 1 u = u, are true, since W V, thus u, v, w W implies u, v, w V. HMHsueh 6

7 Further, by Theorem 6.1(a), (d), since V is a vector space, for any u, 0 = 0 u and u = ( 1) u. By taking c = 0 and c = 1 in the property (β), one can find the zero vector and the negative of any vector in W, i.e. 0 = 0 u W, u = ( 1) u W. HMHsueh 7

8 Remarks: The subspace {0} is a nonempty subspace. If a subset W of V does not contain the zero vector, it is not a subspace of V. A nonempty subset W of a vector space V is a subspace of V if and only if (a u) (b v) W for any u, v W and a, b R. HMHsueh 8

9 Proof.(T.2) (α), (β) W, a, b R. (a u) (b v) W, for u, v For any u, v W, a, b R, by (β), a u W, b v W. By (α), (a u) (b v) W. HMHsueh 9

10 For any u, v W and a, b R, Then taking b = 0, and for any a R, (a u) (b v) W. b v = 0 (a u) (b v) = (a u) 0 = a u W, thus (β) holds. Further, taking a = 1, b = 1, since a u = 1 u = u, b v = 1 v = v, thus (a u) (b v) = u v W, property (α) holds. HMHsueh 10

11 EX 3 (Ex. 4) Which of the following subsets of R 2 with the operations of the usual vector addition and scalar multiplication are subspaces? (a) W 1 = {(x, y) : x 0, y R} (b) W 2 = {(x, y) : x 0, y 0} (c) W 3 = {(x, y) : x = 0, y R}. HMHsueh 11

12 (a) Consider a vector u = (x, y) W 1 and x > 0, taking c < 0, c u = (cx, cy) not in W, since cx < 0. Thus, (β) fails and W 1 is not a subspace of R 2. (b) Same as (a). W 2 is not a subspace. (c) For any u = (0, b 1 ), v = (0, b 2 ) W 3 and a, c R, (a u) (c v) = (0, ab 1 ) (0, cb 2 ) = (0, ab 1 + cb 2 ) = (0, b ) W 3 since b = ab 1 + cb 2 R. W 3 is thus a subspace of R 2. HMHsueh 12

13 EX 4 (Ex. 5) Check whether the following W is a subspace of R 3 w.r.t. the operations of the usual vector addition and scalar multiplication, W = {(a, b, 1) : a, b R}. For any u = (a 1, b 1, 1), v = (a 2, b 2, 1) W, c 1, c 2 R, (c 1 u) (c 2 v) = (c 1 a 1, c 1 b 1, c 1 ) (c 2 a 2, c 2 b 2, c 2 ) = (c 1 a 1 + c 2 a 2, c 1 b 1 + c 2 b 2, c 1 + c 2 ) = (a, b, c ) is not in W if c = c 1 + c 2 1. Thus W is not a subspace of R 3. HMHsueh 13

14 EX 5 (Ex. 6) P n = {all polynomials of degree n and the zero polynomial}, P = {all polynomials}. Recall that for p(t) = a n t n + a n 1 t n a 1 t + a 0 q(t) = b n t n + b n 1 t n b 1 t + b 0, p(t), q(t) P n, consider the operations p(t) q(t) = (a n + b n )t n + (a n 1 + b n 1 )t n 1 + +(a 1 + b 1 )t + (a 0 + b 0 ) c p(t) = (ca n )t n + (ca n 1 )t n 1 + +(ca 1 )t + (ca 0 ). Show that P n is a subspace of P n+1 and that P n is a subspace of P w.r.t. the two operations. HMHsueh 14

15 Proof. P n is a subspace of P n+1 For any p(t) P n, p(t) P n+1. P n is a subset of P n+1. For p(t), q(t) P n, c 1, c 2 R, (c 1 p(t)) (c 2 q(t)) = {(c 1 a n )t n + (c 1 a n 1 )t n (c 1 a 1 )t + (c 1 a 0 )} {(c 2 b n )t n + (c 2 b n 1 )t n (c 2 b 1 )t + (c 2 b 0 )} = (c 1 a n + c 2 b n )t n + (c 1 a n 1 + c 2 b n 1 )t n 1 + +(c 1 a 1 + c 2 b 1 )t + (c 1 a 0 + c 2 b 0 ) = a nt n + a n 1 tn a 1 t + a 0 W, since a i = c 1a i + c 2 b i, i = 1,, n R. Hence P n is a subspace of P n+1. HMHsueh 15

16 EX 6 (Ex. 7) Let V be the set of all polynomials of degree 2, i.e., V = {p(t) = a 2 t 2 + a 1 t + a 0, a 2 0, a 1, a 0 R}. Show that V is a subset of P 2 but not a subspace of P 2. Proof, Since for p(t) V, p(t) P 2, V is a subset of P 2. However, consider c 1, c 2 R, p(t), q(t) V, where p(t) = a 2 t 2 + a 1 t + a 0, q(t) = b 2 t 2 + b 1 t + b 0, a 2, b 2 0. HMHsueh 16

17 Then (c 1 p(t)) (c 2 q(t)) = {(c 1 a 2 )t 2 + (c 1 a 1 )t + (c 1 a 0 )} {(c 2 b 2 )t 2 + (c 2 b 1 )t + (c 2 b 0 )} = (c 1 a 2 + c 2 b 2 )t 2 + (c 1 a 1 + c 2 b 1 )t + (c 1 a 0 + c 2 b 0 ) = a 2 t2 + a 1 t + a 0, will not be in V if a 2 = (c 1a 2 + c 2 b 2 ) = 0. For example, if c 1 = 1, c 2 = (a 2 /b 2 ), (c 1 p(t)) (c 2 q(t)) = a 1 t + a 0 is reduced to be a polynomial of degree one and thus not in V. HMHsueh 17

18 EX 7 (Ex. 9) Consider the homogeneous system Ax = 0, where A is an m n matrix. A solution consists of a vector x R n. Let W = {x : Ax = 0} = the set of all solutions to the system. Show that W is a subspace of R n w.r.t. the usual vector addition and scalar multiplication. HMHsueh 18

19 Since A0 = 0, 0 W and thus W is nonempty. Further, for x, y W and a, b R, since by the properties of Thm. 1.2 and 1.3 in Section 1.4, A{(a x) (b y)} = {A(a x)} {A(b y)} = {a (Ax)} {b (Ay)} = (a 0) (b 0) = 0 0 = 0, thus, (a x 1 ) (b x 2 ) is also a solution of the homogenous system and thus it is in W. W is a subspace of R n. HMHsueh 19

20 Remarks: W is called the solution space of the homogenous system, or the null space of A. The set of all solutions to the linear system Ax = b, where A is m n, is not a subspace of R n if b 0. Why? T.3. HMHsueh 20

21 EX 8 (Ex. 10) A simple way to construct a subspaces in a vector space V : 1. Start with two fixed vectors v 1, v 2 in V. 2. Let W = {w : w = (a 1 v 1 ) (a 2 v 2 ), for all a 1, a 2 R}. Then W is a subspace of V. HMHsueh 21

22 DEF 2 Let v 1,, v k V. A vector v V is called a linear combination of v 1,, v k if there exist c 1,, c k R, such that v = (c 1 v 1 ) (c 2 v 2 ) (c k v k ) = c 1 v 1 + c 2 v c k v k. See Figure 6.4 for a linear combination of v 1, v 2 R 2. HMHsueh 22

23 EX 9 (Ex. 11) In R 3, check if v = (2, 1, 5) is a linear combination of v 1 = (1, 2, 1), v 2 = (1, 0, 2), v 3 = (1, 1, 0). Sol. Check the existence of c 1, c 2, c 3 such that v = c 1 v 1 + c 2 v 2 + c 3 v 3. HMHsueh 23

24 Since if c 1 v 1 + c 2 v 2 + c 3 v 3 = v c 1 (1, 2, 1) + c 2 (1, 0, 2) + c 3 (1, 1, 0) = (2, 1, 5) c 1 +c 2 +c 3 = 2 2c 1 +c 3 = 1 c 1 +2c 2 = 5. Solving the linear system, one obtains c 1 = 1, c 2 = 2, c 3 = 1 and thus v = v 1 + 2v 2 v 3, a linear combination of v 1, v 2, v 3. HMHsueh 24

25 DEF 3 If S = {v 1, v 2,, v k } V, then span S = span {v 1, v 2,, v k } = {v : v = (c 1 v 1 ) (c 2 v 2 ) (c k v k ), c 1,, c k R}. EX 10 Figure 6.6 show that in R 3, if v 1, v 2 are not collinear, i.e. v 1 = kv 2, then span {v 1, v 2 } is a plane that passes through the origin and contains the vectors v 1 and v 2. HMHsueh 25

26 EX 11 Consider the set of 2 3 matrices given by {( ) ( ) ( ) ( S =,,, Then span S is the set in M 23 consisting of all vectors of the form ( ) ( ) ( ) ( a b c d ( ) a b 0 = 0 c d where a, b, c, d R. )}. ) HMHsueh 26

27 THM 2 (Thm. space V. Then 6.3) Let S = {v 1,, v k } be a set in a vector span S is a subspace of V. Proof. If w span S and w = (c 1 v 1 ) (c 2 v 2 ) (c k v k ), since v 1,, v k V and V is a vector space, then w V. That is, span S is a subset of V. HMHsueh 27

28 If w 1, w 2 span S, w 1 = (c 1 v 1 ) (c 2 v 2 ) (c k v k ), w 2 = (d 1 v 1 ) (d 2 v 2 ) (d k v k ), for some c 1,, c k, d 1,, d k R. Then (a w 1 ) (b w 2 ) = a {(c 1 v 1 ) (c 2 v 2 ) (c k v k )} b {(d 1 v 1 ) (d 2 v 2 ) (d k v k )} = {[(ac 1 ) v 1 ] [(ac 2 ) v 2 ] [(ac k ) v k ]} +{[(bd 1 ) v 1 ] [(bd 2 ) v 2 ] [(bd k ) v k ]} = [(ac 1 + bd 1 ) v 1 ] [(ac k + bd k ) v k ] = (c 1 v 1) (c k v k) span S, since c 1 = (ac 1 + bd 1 ),, c k = (ac k + bd k ) R. Hence span S is a subspace of V. HMHsueh 28

29 EX 12 (Ex. 13) In P 2 let v 1 = 2t 2 + t + 2, v 2 = t 2 2t, v 3 = 5t 2 5t + 2, v 4 = t 2 3t 2. Determine if the vector u = t 2 + t + 2 belongs to span {v 1, v 2, v 3, v 4 }. Sol. If one can find c 1, c 2, c 3, c 4 such that u = (c 1 v 1 ) (c 2 v 2 ) (c 3 v 3 ) (c 4 v 4 ) then u span {v 1,, v 4 }. HMHsueh 29

30 Since (c 1 v 1 ) (c 2 v 2 ) (c 3 v 3 ) (c 4 v 4 ) = u c 1 (2t 2 + t + 2) c 2 (t 2 2t) c 3 (5t 2 5t + 2) c 4 ( t 2 3t 2) = t 2 + t c 1 +c 2 +5c 3 c 4 = 1 c 1 2c 2 5c 3 3c 4 = 1 2c 1 +2c 3 2c 4 = 2. Solving the linear equation, one finds the reduced row echelon form of the augmented matrix, which indicates that the system is inconsistent and it has no solution. Hence u does not belong to span {v 1, v 2, v 3, v 4 }. HMHsueh 30

31 EXERCISE. 3, 5, 9, 16, 18, 23, 25(a), 26(a), 27(a), T3, T5, T6, T7, T8(Hint: T13), T9, T10, T11, T13 HMHsueh 31