Unit20

# Unit20 - Unit 20 Linear Dependence and Independence The...

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Unformatted text preview: Unit 20 Linear Dependence and Independence The idea of dimension is fairly intuitive. Consider any vector in Rfractur m , ( a 1 , a 2 , a 3 , ..., a m ). Each of the m components is independent of the others. That is, choosing a value for a single component in no way affects the avail- able choices of values for the other components, in order to have the choice end up being a vector in Rfractur m . For this reason we say that Rfractur m has m degrees of freedom or equivalently has dimension m . Thus, Rfractur 2 has dimension 2, Rfractur 3 has dimension 3, and so forth. Consider now the “subspace” (soon to be defined) of Rfractur 3 consisting of all vectors ( x, y, z ) which have x + y + z = 0. We may now only choose values for 2 of the components freely and these will determine the value of the third component. For instance, if we choose x = 1 and y = 2, then we must have z =- 3, in order to satisfy x + y + z = 0. Therefore this subspace is said to have dimension 2. The rest of this course will be concerned with this idea of dimension and related concepts. In this unit, we lay the foundation with the concepts of linear dependence and independence. Definition 20.1. A linear combination of a set of vectors, vectorv 1 ,vectorv 2 , ..,vectorv k in Rfractur m is an expression of the form c 1 vectorv 1 + c 2 vectorv 2 + ... + c k vectorv k , where each c i is a scalar. Example 1 . Let vectorv 1 = (1 , , 1), vectorv 2 = (- 1 , 1 , 0) and vectorv 3 = (1 , 2 , 3). Express vectorv 3 as a linear combination of vectorv 1 and vectorv 2 . Solution: We must find scalars c 1 and c 2 so that vectorv 3 = c 1 vectorv 1 + c 2 vectorv 2 . 1 Using our knowledge of scalar multiplication and addition of vectors, from Unit 13, we set (1 , 2 , 3) = c 1 (1 , , 1) + c 2 (- 1 , 1 , 0) ⇒ (1 , 2 , 3) = ( c 1 , , c 1 ) + (- c 2 , c 2 , 0) ⇒ (1 , 2 , 3) = ( c 1- c 2 , c 2 , c 1 ) Equating corresponding components we see that what we have is a SLE: 1 = c 1- c 2 2 = c 2 3 = c 1 Since c 1 = 3 and c 2 = 2 does satisfy c 1- c 2 = 1, we see that this is the solution to the SLE. Thus, we see that (1 , 2 , 3) = 3(1 , , 1) + 2(- 1 , 1 , 0), so we have found the required linear combination to be vectorv 3 = 3 vectorv 1 + 2 vectorv 2 . We now give two definitions for the same thing. Each definition is (of course) equivalent to the other, but sometimes the use of one definition is more suited to a given task than the other. Definition 20.2. A set of vectors in Rfractur m is said to be linearly dependent if there is at least one vector in the set which can be expressed as a linear combination of the others. If a set is not linearly dependent then it is said to be linearly independent . Definition 20.3. A set of vectors { vectorv 1 ,vectorv 2 , ...,vectorv n } in Rfractur m is linearly dependent if the vector equation c 1 vectorv 1 + c 2 vectorv 2 + ... + c n vectorv n = vector 0 has a solution in which at least one c i negationslash = 0....
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## This note was uploaded on 11/03/2009 for the course MATH 0110A taught by Professor Olds,vicky during the Fall '09 term at UWO.

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Unit20 - Unit 20 Linear Dependence and Independence The...

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