MAT223H1 Chapter 2.3: Chapter 2.3 Linear Independence
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If the only solution to the vector equation x1u1 + x2u2 + . + xmum = 0 (1) is the trivial solution then the set x1 = x2 = . xm = 0, (cid:8)u1, . If there are nontrivial solutions, then the set is linearnly de- pendent. Note that testing linearly independence is equivalent: show that 0 belongs to the span{u1, . , um}: the system u x = 0 has only one solution and is where u =(cid:2) u1. 1. the set is linearly dependent. x = 0, (cid:3). um (cid:9) is a set of vectors in rn. Another form of having a linearly dependent set when they are colinear. Think in r2 and consider the equation x1u1 + x2u2 = 0 (2) If x2 (cid:54)= 0, then the vector u2 = x1 x2 u1. For the case of sets with three or more vectors, nding out if the vectors are linearly independent or not requires solv- ing (1).
