6.4 Basis and Dimension

Transcription

1 6.4 Basis and Dimension DEF ( p. 263) AsetS ={v 1, v 2, v k } of vectors in a vector space V is a basis for V if (1) S spans V and (2) S is linearly independent. MATH 316U (003) (Basis and Dimension) / 1

2 EXAMPLE 1 ( EXAMPLE 1 from the previous lecture) Let S ={i, j, k }={(1,0,0),(0,1,0),(0,0,1)} (1) we ve shown that S spans R 3 (2) c 1 i + c 2 j + c 3 k = 0 corresponds to the homogeneous system with the augmented matrix The solution is unique: c 1 = c 2 = c 3 = 0(the trivial solution). Answer: S is a basis for R 3. Example 1 p. 263 ( n = 2, general n) MATH 316U (003) (Basis and Dimension) / 2

3 EXAMPLE 2 Is the set S ={(1,1),(1, 1)} abasisforr 2? (1) Does S span R 2? 1 Solve c c = x y 1c 1 + 1c 2 = x 1c 1 1c 2 = y 1 1 x 1 1 y r 2 r 1 r x 0 2 y x This system is consistent for every x and y, therefore S spans R 2. MATH 316U (003) (Basis and Dimension) / 3

4 (2) Is S linearly independent? 1 1 Solve c 1 + c = 0 0 1c 1 + 1c 2 = 0 1c 1 1c 2 = r 2 r 1 r The system has a unique solution c 1 = c 2 = 0 (trivial solution). Therefore S is linearly independent. Consequently, S is a basis for R 2. MATH 316U (003) (Basis and Dimension) / 4

5 EXAMPLE 3 Is S ={(1,2,3),(0,1,2),( 1,0,1)} abasisforr 3? It was already shown ( EXAMPLE 3 from the previous lecture) that S does not span R 3. Therefore S is not a basis for R 3. EXAMPLE 4 Is S ={(1,0),(0,1),( 2,5)} abasis for R 2? It was already shown ( EXAMPLE 4 from the previous lecture) that S is linearly dependent. Therefore S is not a basis for R 2. Example 2 p.263. MATH 316U (003) (Basis and Dimension) / 5

6 EXAMPLE 5 S ={ , , , } v 1 is a basis for the vector space M 22. v 2 v 3 v 4 (1) c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = v = is equivalent to: c 1 c 2 c 3 c 4 = which is consistent for every a,b, c, andd. Therefore S spans M 22. a c b d a c b d MATH 316U (003) (Basis and Dimension) / 6

7 (2) c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = 0 is equivalent to: c 1 c 2 c 3 c 4 = The system has only the trivial solution S is linearly independent. Consequently, S is a basis for M 22. MATH 316U (003) (Basis and Dimension) / 7

8 EXAMPLE 6 Is S ={1, t,t 2,t 3 } abasisforp 3? (1) c 1 (1)+c 2 (t)+c 3 (t 2 )+c 4 (t 3 ) = a + bt + ct 2 + dt 3 has a solution for every a,b,c, andd : c 1 = a, c 2 = b, c 3 = c,c 4 = d. Therefore S spans P 3. (2) c 1 (1)+c 2 (t)+c 3 (t 2 )+c 4 (t 3 )=0 can only be solved by c 1 = c 2 = c 3 = c 4 = 0. Therefore S is linearly independent. Consequently, S is a basis for P 3. Example 3 p.264. MATH 316U (003) (Basis and Dimension) / 8

9 THEOREM ( Th. 6.5 p. 265) Let S ={v 1, v 2, v k } be a set of nonzero vectors in a vector space V. The following statements are equivalent: (A) S is a basis for V, (B) every vector in V can be expressed as a linear combination of the vectors in S in a unique way. MATH 316U (003) (Basis and Dimension) / 9

10 Proof (A) (B) Every vector in V can be expressed as a linear combination of vectors in S because S spans V. Suppose v can be represented as a linear combination of vectors in S in two ways: v = c 1 v c k v k Subtract: v = d 1 v d k v k 0 =(c 1 d 1 )v 1 + +(c k d k )v k Since S is linearly independent, then c 1 d 1 = = c k d k = 0 so that c 1 = d 1 c k = d k The representation is unique. MATH 316U (003) (Basis and Dimension) / 10

11 Proof (B) (A) (B) Every vector in V is in span S. Zero vector in V can be represented in a unique way as a linear combination of vectors in S: 0 = c 1 v c k v k This unique way must be: c 1 = = c k = 0. Therefore S is linearly independent. Consequently, S is a basis for V. MATH 316U (003) (Basis and Dimension) / 11

12 Back to EXAMPLE 2: S ={(1,1), (1, 1)} Instead of showing that c 1 v 1 + c 2 v 2 = v has a solution, and c 1 v 1 + c 2 v 2 = 0 has a unique solution, we can show c 1 v 1 + c 2 v 2 = v has a unique solution. 1 1 x 1 1 y r 2 r 1 r x 0 2 y x Unique solution for every x and y S is a basis for R 2. MATH 316U (003) (Basis and Dimension) / 12

13 TH 6.6 ( p. 266) Let S ={v 1, v 2, v k } be a set of nonzero vectors in a vector space V. Some subset of S is a basis for W = span S. Procedure p. 268 MATH 316U (003) (Basis and Dimension) / 13

14 EXAMPLE 7 Find a basis for span{ (1,2,3), ( 1, 2, 3), (0,1,1), (1,1,2)}. v 1 v 2 v 3 v 4 Set c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = 0. The corresponding system has augmented matrix: which is equivalent (r 2 2r 1 r 2 ; r 3 3r 1 r 3 ; r 3 r 2 r 3 ) to MATH 316U (003) (Basis and Dimension) / 14

15 Can set c 2 and c 4 arbitrary. For example If c 2 = 1,c 4 = 0 then v 2 can be expressed as a linear combination of v 1 and v 3. If c 2 = 0,c 4 = 1 then v 4 can be expressed as a linear combination of v 1 and v 3. Therefore, every vector in span S can be expressed as a linear combination of v 1 and v 3. Also note that v 1 and v 3 are linearly independent. Consequently, they form a basis for span S. Summarizing: The vectors corresponding to the columns with leading entries form a basis for W. Different initial ordering of vectors, e.g., {v 2, v 1, v 3, v 4 } may change the basis obtained by the procedure above (in this case: v 2, v 3 ). MATH 316U (003) (Basis and Dimension) / 15

16 TH 6.7 ( p. 269) Let S ={v 1, v 2, v k } span V and let T ={w 1,w 2, w n } be a linearly independent set of vectors in V. Then n k. COROLLARY 6.1 ( p. 270) Let S ={v 1, v 2, v k } and T ={w 1,w 2, w n } both be bases for V. Then n = k. DEF ( p. 270) The dimension of a vector space V, denoted dim V, is the number of vectors in a basis for V. dim({ 0 }) = 0. dim(r n )=n ( Example 6 p. 270) dim(p n )=n + 1( Example 7 p. 270) dim(m mn )=mn MATH 316U (003) (Basis and Dimension) / 16

17 TH 6.8 ( p. 271) If S is a linearly independent set of vectors in a finite-dimensional vector space V, then there exists a basis T for V, which contains S. EXAMPLE 8 ( Example 9 p. 271) Find a basis for R 4 that contains the vectors v 1 =(1,0,1,0) and v 2 =( 1,1, 1,0). Solution: The natural basis for R 4 : {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} e 1 e 2 Follow the procedure of EXAMPLE 7 to determine a basis of span{v 1, v 2, e 1, e 2, e 3, e 4 }. e e 4 MATH 316U (003) (Basis and Dimension) / 17

18 has the reduced row echelon form: Answer: {v 1, v 2, e 1, e 4 }. TH 6.9 ( p. 272) Let V be an n-dimensional vector space, and let S ={v 1, v 2, v n } beasetofn vectors in V. (a) If S is linearly independent then it is a basis for V. (b) If S spans V then it is a basis for V. MATH 316U (003) (Basis and Dimension) / 18