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lecture_21 - MA 265 LECTURE NOTES WEDNESDAY FEBRUARY 27...

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MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 27 Linear Independence (cont’d) Example. Let V = M 22 be the real vector space corresponding to the 2 × 2 matrices. Consider the following three vectors: v 1 = 2 1 0 1 , v 2 = 1 2 1 0 , and v 3 = 0 - 3 - 2 1 . We determine whether these vectors are linearly independent. We consider the equation a 1 v 1 + a 2 v 2 + a 3 v 3 = 0 = 2 a 1 + a 2 a 1 + 2 a 2 - 3 a 3 a 2 - 2 a 3 a 1 + a 3 = 0 0 0 0 . This corresponds to homogeneous linear system 2 a 1 + a 2 = 0 a 1 + 2 a 2 - 3 a 3 = 0 a 2 - 2 a 3 = 0 a 1 + a 3 = 0 The augmented matrix and its reduced row echelon form are 2 1 0 0 1 2 - 3 0 0 1 - 2 0 1 0 1 0 = 1 0 1 0 0 1 - 2 0 0 0 0 0 0 0 0 0 This corresponds to the system a 1 + a 3 = 0 a 2 - 2 a 3 = 0 = a 1 a 2 a 3 = - r 2 r r . Since there are infinitely many solutions, there is at least one nontrivial solution. In particular, when r = 1 we have the linear combination ( - 1) 2 1 0 1 + (2) 1 2 1 0 + (1) 0 - 3 - 2 1 = 0 0 0 0 . This shows that the vectors are not linearly independent. More Criteria For Determining Linear Independence. Let ( V, + , · ) be a real vector space, and consider a set S = { v 1 , v 2 , . . . , v k } of k vectors in V . We focus on the case where V = R n . We can form an n × k matrix where the vectors are its columns: A = a 11 a 12 · · · a 1 k a 21 a 22 · · · a 2 k . . . . . . . . . . . . a n 1 a n 2 · · · a nk where v j = a 1 j a 2 j .
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