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Lecture_10 - Lecture Note 10 Dr Jeff Chak-Fu WONG...

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Lecture Note 10 Dr. Jeff Chak-Fu WONG Department of Mathematics Chinese University of Hong Kong [email protected] MAT 2310 Linear Algebra and Its Applications Fall, 2007 Produced by Jeff Chak-Fu WONG 1
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R EAL V ECTOR S PACES 1. Vector Spaces 2. Subspaces 3. Linear Independence 4. Basis and Dimension 5. Homogeneous Systems 6. The Rank of a Matrix and Applications 7. Coordinates and Change of Basis R EAL V ECTOR S PACES 2
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B ASIS AND D IMENSION B ASIS AND D IMENSION 3
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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 . B ASIS AND D IMENSION 4
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DEFINITION-The 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 9) and distinct and so we write them as a set { v 1 , v 2 , . . . , v k } . B ASIS AND D IMENSION 5
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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. B ASIS AND D IMENSION 6
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Example 2 Show that the set S = { v 1 , v 2 , v 3 , v 4 } , where v 1 = (1 , 0 , 1 , 0) , v 2 = (0 , 1 , - 1 , 2) , v 3 = (0 , 2 , 2 , 1) and v 4 = (1 , 0 , 0 , 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 = 0 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 = 0 c 2 + 2 c 3 = 0 c 1 - c 2 + 2 c 3 = 0 2 c 2 + c 3 + c 4 = 0 , 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. B ASIS AND D IMENSION 7
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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 . B ASIS AND D IMENSION 8
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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 . 1. To show that it spans V , we take any vector in V , that is, a polynomial at 2 + bt + c and must find constants a 1 , a 2 and a 3 such that at 2 + bt + c = a 1 ( t 2 + 1) + a 2 ( t - 1) + a 3 (2 t + 2) = a 1 t 2 + ( a 2 + 2 a 3 ) t + ( a 1 - a 2 + 2 a 3 ) . Since two polynomials agree for all values of t only if the coefficients of respective powers of t agree, we get the linear system a 1 = a a 2 + 2 a 3 = b a 1 - a 2 + 2 a 3 = c.
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