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Unformatted text preview: Lecture Note 42: Feb 5 & 7, 2007 Dr. Jeff ChakFu WONG Department of Mathematics Chinese University of Hong Kong jwong@math.cuhk.edu.hk MAT 2310C Linear Algebra and Its Applications Spring, 2007 Produced by Jeff ChakFu WONG 1 R EAL V ECTOR S PACES 1. Vector Spaces 2. Subspaces 3. Linear Independence 4. Basis and Dimension 5. Coordinates and Change of Basis 6. Homogeneous Systems 7. The Rank of a Matrix and Applications REAL VECTOR SPACES 2 B ASIS AND D IMENSION BASIS AND DIMENSION 3 Our aim is to study the structure of a vector space V by determining a smallest set of vectors in V that completely describes V . BASIS AND DIMENSION 4 DEFINITIONThe vectors v 1 , v 2 ,..., v k in a vector space V are said to form a basis for V if 1. (a) v 1 , v 2 ,..., v k span V , and 2. (b) v 1 , v 2 ,..., v k are linearly independent. Remark If v 1 , v 2 ,..., v k form a basis for a vector space V , then they must be nonzero (see Example 9 in Lecture note 41) and distinct and so we write them as a set { v 1 , v 2 ,..., v k } . BASIS AND DIMENSION 5 Example 1 • The vectors e 1 = (1 , 0) and e 2 = (0 , 1) form a basis for R 2 . • The vectors e 1 , e 2 and e 3 form a basis for R 3 . • In general, the vectors e 1 , e 2 ,..., e n form a basis for R n . Each of these sets of vectors is called the natural basis or standard basis for R 2 , R 3 and R n , respectively. BASIS AND DIMENSION 6 Example 2 Show that the set S = { v 1 , v 2 , v 3 , v 4 } , where v 1 = (1 , , 1 , 0) , v 2 = (0 , 1 , 1 , 2) , v 3 = (0 , 2 , 2 , 1) and v 4 = (1 , , , 1) is a basis for R 4 . Solution 1. To show that S is linearly independent , we form the equation c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = and solve for c 1 ,c 2 ,c 3 and c 4 . Substituting for v 1 , v 2 , v 3 and v 4 , we obtain the linear system (verify) c 1 + c 4 = c 2 + 2 c 3 = c 1 c 2 + 2 c 3 = 2 c 2 + c 3 + c 4 = , which has as its only solution c 1 = c 2 = c 3 = c 4 = 0 (verify), showing that S is linearly independent. Observe that the coefficient matrix of the preceding linear system consists of the vectors v 1 , v 2 , v 3 and v 4 written in column form. BASIS AND DIMENSION 7 2. To show that S spans R 4 , we let v = ( a,b,c,d ) be any vector in R 4 . We now seek constants k 1 ,k 2 ,k 3 and k 4 such that k 1 v 1 + k 2 v 2 + k 3 v 3 + k 4 v 4 = v . Substituting for v 1 , v 2 , v 3 , v 4 and v , we find a solution (verify) for k 1 ,k 2 ,k 3 and k 4 to the resulting linear system for any a,b,c and d . Hence S spans R 4 and is a basis for R 4 . BASIS AND DIMENSION 8 Example 3 Show that the set S = { t 2 + 1 ,t 1 , 2 t + 2 } is a basis for the vector space P 2 . Solution We must show that 1. S spans V , and 2. S is linearly independent ....
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 Spring '11
 JeffWong
 Linear Algebra, Algebra

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