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03-03-03 - Lecture 13 Andrei Antonenko March 3 2003 1...

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Lecture 13 Andrei Antonenko March 3, 2003 1 Dimension and basis of the span Last lecture we formulated the problem of finding the basis and the dimension of the span of given vectors. This lecture we will give the algorithm to determine these characteristics of the span. 1.1 Algorithm Step 1. Write the given vectors as a rows of a matrix. Step 2. Reduce this matrix to REF keeping track which row in the matrix corresponds to which vector, i.e. initially, i -th row corresponds to u i – so, label i -th row as i , and after Type 1 elementary operation we interchange row labels. Step 3. The number of nonzero rows is a dimension of a span. The labels of nonzero rows are subscripts of vectors in basis. Moreover, nonzero rows form a basis for a span as well. Example 1.1. Consider the following 4 vectors in R 4 : u 1 = (2 , 3 , 1 , 1) , u 2 = (1 , 1 , 0 , - 1) , u 3 = (3 , 4 , 1 , 0) , and u 4 = (1 , 2 , 1 , 3) . Let’s find the basis and a dimension of their span. To do 1
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this we’ll form a matrix with labels and reduce it to REF. 1 2 3 4 2 3 1 1 1 1 0 - 1 3 4 1 0 1 2 1 3 interchange 1st and 2nd ˆ 2 1 3 4 1 1 0 - 1 2 3 1 1 3 4 1 0 1 2 1 3 subtract 1st from others ˆ 2 1 3 4 1 1 0 - 1 0 1 1 3 0 1 1 3 0 1 1 4 subtract 2nd from others ˆ 2 1 3 4
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