math125-lecture 4.5 - Solve c 1 v 1 + c 2 v 2 = 0 for c 1...

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Unformatted text preview: Solve c 1 v 1 + c 2 v 2 = 0 for c 1 and c 2 → → → Sec 4.5 Saturday, February 06, 2010 11:23 PM Section4dot5 Page 1 Solve c 1 v 1 + c 2 v 2 = 0 for c 1 and c 2 1 1 0 1 0 0 c 1 = 0 and [ v 1 v 2 | 0 ] = rref 3 1 0 0 1 0 c 2 = 0 Therefore, by definition, v 1 and v 2 are linearly independent → → → → → → → → 4.5.1 Saturday, February 06, 2010 11:23 PM Section4dot5 Page 2 Solve c 1 v 1 + c 2 v 2 = 0 for c 1 and c 2 1 1 0 1 0 0 c 1 = 0 and [ v 1 v 2 | 0 ] = rref 3 1 0 0 1 0 c 2 = 0 Therefore, by definition, v 1 and v 2 are linearly independent → → → → → → → → span{ v 1 , v 2 } = { s v 1 + t v 2 | s and t are in R } Let P 1 = - v 1 + 2 v 2 P 2 = v 1- 4 v 2 → → → → → → → → → → 4.5.2 Saturday, February 06, 2010 11:23 PM Section4dot5 Page 3 Solve c 1 v 1 + c 2 v 2 = 0 for c 1 and c 2 1 1 0 1 0 0 c 1 = 0 and [ v 1 v 2 | 0 ] = rref 3 1 0 0 1 0 c 2 = 0 Therefore, by definition, v 1 and v 2 are linearly independent → → → → → → → → span{ v 1 , v 2 } = { s v 1 + t v 2 | s and t are in R } Let P 1 = - v 1 + 2 v 2 P 2 = v 1- 4 v 2 → → → → → → → → → → x y v 1 v 2 → → P 2 P 1 → → 4.5.3 Saturday, February 06, 2010 11:23 PM Section4dot5 Page 4 Solve c 1 v 1 + c 2 v 2 = 0 for c 1 and c 2 1 1 0 1 0 0 c 1 = 0 and [ v 1 v 2 | 0 ] = rref 3 1 0 0 1 0 c 2 = 0 Therefore, by definition, v 1 and v 2 are linearly independent → → → → → → → → span{ v 1 , v 2 } = { s v 1 + t v 2 | s and t are in R } Let P 1 = - v 1 + 2 v 2 P 2 = v 1- 4 v 2 Choosing different values of s and t , we can reach any point in R 2 → → → → → → → → → → x y v 1 v 2 → → P 2 P 1 → → Therefore , span{ v 1 , v 2 } = R 2 We say that v 1 and v 2 span R 2 → → → → 4.5.4 (*) Saturday, February 06, 2010 11:23 PM Section4dot5 Page 5 x y v 3 v 1 v 2 → → → 4.5.5 Monday, February 15, 2010 10:52 PM Section4dot5 Page 6 x y v 3 v 1 v 2 → → → By one of our theorems, three vectors in R 2 must be linearly dependent Indeed we have v 3 = v 1 + v 2 and this gives the dependency equation v 1 + v 2- v 3 = 0 → → → → → → → 4.5.6 Monday, February 15, 2010 10:52 PM Section4dot5 Page 7 x y v 3 v 1 v 2 → → → By one of our theorems, three vectors in R 2 must be linearly dependent Indeed we have v 3 = v 1 + v 2 and this gives the dependency equation v 1 + v 2- v 3 = 0 Any two of these three vectors span R 2 R 2 = span{ v 1 , v 2 , v 3 } = span{ v 1 , v 2 } = span{ v 1 , v 3 } = span{ v 2 , v 3 } → → → → → → → → → → → → → → → → 4.5.7 Monday, February 15, 2010 10:52 PM Section4dot5 Page 8 x y v 3 v 1 v 2 → → → By one of our theorems, three vectors in R 2 must be linearly dependent Indeed we have v 3 = v 1 + v 2 and this gives the dependency equation v 1 + v 2- v 3 = 0 Any two of these three vectors span R 2 R 2 = span{ v 1 , v 2 , v 3 } = span{ v 1 , v 2 } = span{ v 1 , v 3 } = span{ v 2 , v 3 } Observe that v 1 = - 1/3 v 2 These two vectors are linearly dependent...
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This note was uploaded on 06/07/2011 for the course MATH 125 taught by Professor Tom during the Fall '07 term at University of Illinois at Urbana–Champaign.

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math125-lecture 4.5 - Solve c 1 v 1 + c 2 v 2 = 0 for c 1...

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