HW1_Sol - Math 104 Fall 2009 Homework 1 solution set We use...

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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. (c) There are many possible answers here; one of them is the 1 × 3 matrix A = 111 . Of course, using this example we could construct many other examples, for example for any integer m we can consider an m by 3 matrix all of whose rows are scalar multiples of (111). Such a matrix also has the
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