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

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Unformatted text preview: 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 , ,- 1) , u 3 = (3 , 4 , 1 , 0) , and u 4 = (1 , 2 , 1 , 3) . Lets find the basis and a dimension of their span. To do 1 this well 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 1 2 1 3 interchange 1st and 2nd 2 1 3 4 1 1 0- 1 2 3 1 1 3 4 1 1 2 1 3 subtract 1st from others 2 1 3 4 1 1 0- 1 0 1 1 3 0 1 1 3 0 1 1 4 subtract 2nd from others...
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This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.

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

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