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**Unformatted text preview: **Math 104 Fall 2009 Homework 1: solution set We use the notations below to ease readability. Matrices are bold capital, vectors are bold lowercase and scalars or entries are not bold. For instance, A is a matrix and a ij its ( i,j )th entry. Likewise x is a vector and x j its j th component. The linear span generated by a group of vecotors v 1 , v 2 ,..., v n is denoted by span( v 1 , v 2 ,..., v n ). Problem 1 (a) Since v 3 = v 1 + v 2 , we have v 3 span( v 1 , v 2 ). Thus span( v 1 , v 2 , v 3 ) = span( v 1 , v 2 ). Suppose a,b R , such that a v 1 + b v 2 = 0. Then we have a- b = 0, b = 0 and- a = 0, which implies a = b = 0. Hence, v 1 and v 2 are linearly independent and, therefore, v 1 , v 2 is a basis of span( v 1 , v 2 ) = span( v 1 , v 2 , v 3 ). In conclusion dim(span( v 1 , v 2 , v 3 )) = 2. (b) Since v 1 + v 2- v 3 = 0, v 1 , v 2 , v 3 are linearly dependent, they cannot form a basis....

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