Mathematical Economics: Lecture 6
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1 Mathematical Economics: Lecture 6 Yu Ren WISE, Xiamen University October 10, 2012
2 Outline Chapter 11 Linear Independence 1 Chapter 11 Linear Independence
3 New Section Chapter 11: Linear Independence
4 Linear Independence Vector V 1,, V k in R n are linear dependent if and only if c 1,, c k not all zero, s.t. c 1 V c k V k = 0.
5 Linear Independence Vector V 1,, V k in R n are linear independent if and only if c 1 V c k V k = 0 implies c 1 = c 2 = = c k = 0
6 Chapter 11 Linear Independence Linear Independence; Example Example 11.2 The vectors w 1 = 1 2 3, w 2 = 4 5 6, and w 3 = are linearly dependent in R 3, since = as can easily be verified.
7 Linear Independence Theorem 11.1: Vectors v 1,, v k in R n are linearly dependent if and only if the linear system A = (v 1,, v k ), Ac = 0 has nonzero solution (c 1,, c k ).
8 Linear Independence Theorem 11.2: A set of n vectors v 1,, v n in R n is linearly independent if and only if det(v 1,, v n ) 0
9 Linear Independence Theorem 11.3 If k > n, any set of k vectors in R n is linearly dependent.
10 Spanning Sets the line generated or spanned by V : L[V ] {rv : r R 1 }. Figure 11.1 (page 238)
11 Spanning Sets Set generated or spanned by V 1,, V k : L[V 1,, V k ] {c 1 V c k V k : c 1, c k R 1 }.
12 Spanning Sets: Example Example 11.4 The x 1 x 2 -plane in R 3 is the span of the unit vectors e 1 = (1, 0, 0) and e 2 = (0, 1, 0), because any vector (a, b, 0) in this plane can be written as a 1 0 b = a 0 + b
13 Spanning R k any linear independent vector V 1,, V k can span R k. This means any vector U R k, c 1,, c k, s.t. c 1 V c k V k = U. To judge whether B belongs to the space spanned by {V 1,, V k }, check whether Vc = B has a solution c. (Theorem 11.4) A set of vectors that span R n must contain at least n vectors. (Theorem 11.6)
14 Spanning R k any linear independent vector V 1,, V k can span R k. This means any vector U R k, c 1,, c k, s.t. c 1 V c k V k = U. To judge whether B belongs to the space spanned by {V 1,, V k }, check whether Vc = B has a solution c. (Theorem 11.4) A set of vectors that span R n must contain at least n vectors. (Theorem 11.6)
15 Spanning R k any linear independent vector V 1,, V k can span R k. This means any vector U R k, c 1,, c k, s.t. c 1 V c k V k = U. To judge whether B belongs to the space spanned by {V 1,, V k }, check whether Vc = B has a solution c. (Theorem 11.4) A set of vectors that span R n must contain at least n vectors. (Theorem 11.6)
16 Spanning R k : Example Example 11.1 The vectors e 1 =.,..., e 0 n =. 0 1 in R n are linearly independent
17 Spanning R k : Example because if c 1,..., c n are scalars such that c 1 e 1 + c 2 e c n e n = 0, c 1 0 c 1. +c c 0 n. = 0. = c c n The last vector equation implies that c 1 = c 2 = c n = 0.
18 Basis and dimension in R n Definition: Let V 1,, V k be a fixed set of k vectors in R n. let V be the set L[V 1,, V k ] spanned by V 1,, V k. Then if V 1,, V k are linearly independent, V 1,, V k is called a basis of V. More generally, let W 1,, W k be a collection of vectors in V. Then W 1,, W k forms a basis if (a) W 1,, W k span V and (b) W 1,, W k are linearly independent.
19 Basis and dimension in R n : Example Example 11.7 v 1 = (1, 1, 1), v 2 = (1, 1, 1), and v 3 = (2, 0, 0) v 3 = v 1 + v 2 linear combination of v 1, v 2, and v 3 = linear combination of just v 1, and v 2
20 Basis and dimension in R n : Example w = av 1 + bv 2 + cv 3 = av 1 + bv 2 + c(v 1 + v 2 ) = (a + c)v 1 + (b + c)v 2. The set {v 1, v 2 } is a more efficient spanning set than is the set {v 1, v 2, v 3 }.
21 Basis and dimension in R n : Example Example 11.8 We conclude from Examples 11.1 and 11.5 that the unit vectors e 1 =.,..., e 0 n =. 0 1 form a basis of R n. Since this is such a natural basis, it is called the canonical basis of R n.
22 Basis and dimension in R n V 1,, V k are linearly independent V 1,, V k span R k. V 1,, V k form a basis of R k the determinant of [V 1,, V k ] is nonzero. (Theorem 11.8) we call the number of basis of a space as the dimension.
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