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1. Linear dependence A Denition of linear independence. Pn n n • v depends on S if there are {si }i=1 ⊂ S and {ai }i=1 ⊂ R such that v = i=1 ai si . P n n n • v depends on S if there are {si }i=1 ⊂ S and {ai }i=1 ⊂ R not all zero such that Pnv = i=1 ai si . n n • If v depends on S then there are {si }i=1 ⊂ S and {ai }i=1 ⊂ R such that v = i=1 ai si . 0 1 2 B Linear dependence of the zero vector. The problem is to decide if depends on , 0 0 0 in R2 . 1 2 0 a +b = 0 0 0 a + 2b = 0 a = −2b b = 0 a = 0 dependent Suppose thatthere were a, bsuch that 1 2 0 a +b = 0 0 0 Dependencewould of a, b follow from the existence 1 2 0 such that a +b = 0 0 0 a + 2b 0 This is equivalent to = , 0 0 hence to a + 2b = 0 And hence equivalent to the existence of a, b such that a = −2b. Choosing b = 0, a = 0 this equality holds so such a, b do exist. Then a + 2b = 0 Thus a = −2b. Thus if there were a, b and b = 0 then a = 0 also. 2. Linear maps on Rn A Linearity in a concrete example. Let T : R2 → R3 be a linear map, and suppose that T 3 1 = u and T 5 4 = v . Find an explicit vector x ∈ R2 such that T x = 2u − 3v . 3 5 3 5 −9 • T x = 2u − 3v = 2T − 3T =T 2 −3 =T . Therefore T x = 1 1 4 10 4 −9 −9 T . Therefore x = . 10 10 3 5 3 5 −9 −9 −3 =T so T = 2u−3v . • 2u−3v = 2T −3T =T 2 1 4 1 4 −10 −10 B Linearity of a linear functional. OK C Denition of Kernel. OK D Basis. 5 3 5a + 3b . Then ϕv = 2(5a + 2a • Suppse v ∈ Ker ϕ and that v = a 2 + b 0 = 0 −2 −2b 3b) − 5(2a) + 3(−2b) = 0 . Thus any vector v ∈ Ker ϕ can be written as a linear combination of 5 3 2 , 0 so they span Ker ϕ. 0 −2 1 3. Linear maps on RR For (Ta f )(x) = f (x + a), W = Span {erx | r ∈ R}. B Image of a linear map. Show that Ta W = W . • Ta (er ) = er (a)er ∈ W so Ta W = W . 2