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☰ Search Explore Log in Create new account Upload × 1.7 Linear Independence A homogeneous system represented by the a matrix equation such as 1 2 3 x1 0 3 5 9 x 0 2 5 9 3 x3 0 Can be viewed as a vector equation 1 2 3 0 x1 3 x2 5 x3 9 0 9 5 3 0 This version emphasizes the vectors instead of the solution. Clearly, this has the trivial solution. Does it have any other solutions? Definition: A set of vectors v1 , v 2 ,, v p in Rn is said to be linearly independent if the vector equation x1v1 x2 v 2 x p v p 0 has only the trivial solution. 1 The set v1 , v 2 ,, v p is said to be linearly dependent if there exist weights c1 , c2 ,, c p not all zero, such that c1v1 c2 v 2 c p v p 0 . Example: Let 1 2 3 v1 3 v 2 5 v 3 9 3 5 9 Are they linearly independent? To answer this, we ask the following question: What are the solutions to the equation 2 1 3 0 x1 3 x2 5 x3 9 0 9 5 3 0 ? The corresponding augmented matrix is: 1 2 3 0 1 0 33 0 3 5 9 0 0 1 18 0 5 9 3 0 ~ 0 0 0 0 2 Note: x3 is free, so an infinite number of solutions. The trivial solution is not the only solution, the vectors are not linearly indeptndent. Can we find a linear dependence relation among the vectors? Note: the solution to the homogeneous equation is x1 = -33x3 , x2 = 18x3, and x3 is free. We can choose any non-zero value for x3, say x3 = 1. Then x1 = –33, x2 = 18, x3 = 1 is a non-trivial solution and 2 3 0 1 333 185 1 9 0 9 3 0 5 Write the matrix equation with this non-trivial solution: 3 1 2 3 33 0 3 5 9 18 0 5 9 3 1 0 A x = 0 In this linear dependence relation, each value of x3 gives a non-trivial solution. Fact: The columns of the matrix A are linearly independent iff Ax = 0 has only the trivial solution. Special Cases: 1. A set with one vector: Consider the set containing only one vector {v}. The only solution to x1v = 0 is x1 = 0. Thus, any set w/ only one vector is linearly independent. So, {v} is linearly independent when v ≠ 0. When v = 0, x1v = 0 has infinitely many solutions. 4 2. Set of two vectors Example: 4 2 2 2 u1 u2 v1 v 2 2 , 1 , 3 1 , “Is {u1, u2} linearly independent or linearly dependent?” Notice that u2 = 2u1, so 2u1 + –u2 = 0. This is not the trivial solution, so {u1, u2} is linearly dependent. What about {v1, v2}? Is this set linearly independent or linearly dependent? We need to look at the possible solutions of cv1 + dv2 = 0. Suppose cv1 + dv2 = 0, and c, d ≠ 0. v1 d v2 , c but this is impossible since v1 is not a multiple of v2. 5 c v 2 v1 v Also, we could solve for 2 : but d this is also impossible since v2 is not a multiple of v1. So, the assumption c, d ≠ 0 must be false and c = d = 0. So, {v1, v2} linearly independent. Fact: A set of two vectors is linearly dependent if at least one vector is a multiple of the other. And a set of two vectors is linearly independent iff neither of the vectors is a multiple of the other. Theorem 1.7: An indexed set S v1 , v 2 ,..., v p of two or more vectors is linearly dependent iff at least one of the vectors in S is a linear combination of the others. In fact, if S is linearly dependent and v1 ≠ 0, then some vector vj (j ≥ 2) is a linear combination of the preceding vectors. 6 3. A Set Containing the zero vector 0 is always linearly dependent. This is Theorem 1.9. Theorem 1.9: A set of vectors S v1 , v 2 ,, v p containing the zero vector is linearly dependent. Proof: Let v1 , v 2 ,, v p in Rn be a set containing the zero vector. We can arrange the set so that the zero vector is the first one in the list: 0, v 2 ,, v p Now the equation x1v1 x2 v 2 x p v p 0 becomes x1 0 x2 v 2 x p v p 0 . Thus, x1 can be any real number, so there are infinitely many solutions to the equation. 4. A Set Containing Too Many Vectors Theorem 1.8: If a set contains more vectors than there are entries in each vector, then the set is linearly dependent (i.e. any set v1, v 2 ,, v p is linearly dependent if p > n). 7 Proof: v1 v 2 v p is an n x p matrix. If p > n, then Ax = 0 has more variables than equations, so it has non-trivial solutions. Thus, the columns of A are linearly dependent. Example: With the least amount of work possible, decide which of the following sets is linearly independent, and give a reason. 3 9 a. 2, 6 1 4 This is linearly independent since it is a set of two vectors, and neither is a multiple of the other. 1 6 b. 9 4 2 3 4 5 7 8 9 0 8 7 6 5 3 2 1 8 8 This is linearly dependent by Theorem 1.8, since more column vectors than entries per vector. 3 9 0 c. 2, 6, 0 1 3 0 This is linearly dependent since it contains the zero vector (Theorem 1.9) and v2 = 3v1. 8 2 d. 1 4 This set is linearly independent since any set containing one non-zero vector is linearly independent. 9 Download 1. Math Sec. 1.7 Linear Independence.doc ' POIN TER . CSCI 467 – Assignment Two Jason Detchevery Due: February 11 biopsychology as a neuroscience Document Slides - Sigmobile $doc.title Differentiation and continuity problems–221 fall 2015 72 Day 2 Dividing Radical Expressions Objective: Divide radical expressions. 72 Day 2 Dividing Radical Expressions 2010 February 09, 2010 72 Day 2 Dividing Radical Expressions Objective: Divide radical expressions. 72 Day 2 Dividing Radical Expressions January 30, 2009 72 Day 2 Dividing Radical Expressions Objective: Divide radical expressions. 72 Day 2 Dividing Radical Expressions 2011 March 09, 2011 means Y 45 Expenditure C+I+G+net X studylib © 2017 DMCA Report