Lecture6

# Lecture6 - Lecture 6 Generating Sets and Bases Let V be the...

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Lecture 6 Generating Sets and Bases Let V be the vector space R 2 and consider the vectors (1 , 0) , (0 , 1). Then, every vector ( x,y ) R 2 can be written as a combination of those vectors. That is: ( x,y ) = x (1 , 0) + y (0 , 1) . Similarly, the two vectors (1 , 1) and (1 , 2) do not belong to the same line, and every vector in R 2 can be written as a combination of those two vectors. 6.1 Introduction In particular: ( x,y ) = a (1 , 1) + (1 , 2) gives us two equations a + b = x and a + 2 b = y Thus, by substituting the ﬁrst equation to the second, we get b = - x + y Inserting this into the ﬁrst equation we get a = 2 x - y Take for example the point (4 , 3). Then: (4 , 3) = 5(1 , 1) + ( - 1)(1 , 2) = 5(1 , 1) - (1 , 2) 35

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36 LECTURE 6. GENERATING SETS AND BASES We have similar situation for R 3 and all of the spaces R n . In the case of R 3 , for example, every vector can be written as combinations of (1 , 0 , 0) , (0 , 1 , 0) and (0 , 0 , 1), i.e., ( x,y,z ) = x (1 , 0 , 0) + y (0 , 1 , 0) + z (0 , 0 , 1) . Or, as a combination of (1 , - 1 , 0) , (1 , 1 , 1) and (0 , 1 , - 1), that is: ( x,y,z ) = a (1 , - 1 , 0) + b (1 , 1 , 1) + c (0 , 1 , - 1) . The latter gives three equation: a + b = x (1) - a + b + c = y (2) b - c = z (3) . (2) + (3) gives: - a + 2 b = y + z (4) (4) + (1) gives: 3 b = x + y + z or b = x + y + z 3 . Then (1) gives: a = x - b = x - x + y + z 3 = 2 x - y - z 3 . Finally, (3) gives: c = b - z = x + y - 2 z 3 Hence, we get: ( x,y,z ) = 2 x - y - z 3 (1 , - 1 , 0) + x + y + z 3 (1 , 1 , 1) + x + y - 2 z 3 (0 , 1 , - 1) .
6.2. IN GENERAL 37 6.2 In general Notice that we get only one solution, so there is only one way that we can write a vector in R 3 as a combination of those vectors. In general, if we have k vectors in R 3 , then the equation: x = ( x 1 ,x 2 ,...,x n ) = c 1 v 1 + c 2 v 2 + ... + c k v k ( * ) gives n -equations involving the n -coordinates of

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Lecture6 - Lecture 6 Generating Sets and Bases Let V be the...

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