If t4 0 wed be able to solve for h showing it is in

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Unformatted text preview: the vector space is not important, this result is true in any vector space. Indeed, assume t1 1 + t2(1 + f ) + t3( f + g) + t4( g + h) = 0. Then we have t4 h = − (t1 + t2) 1 − (t2 + t3) f − (t3 + t4) g . If t4 0 we’d be able to solve for h showing it is in span(S ), therefore we must have t4 = 0. Then, by independence of {1, f , g }, we get t1 + t2 = t2 + t3 = t3 = 0, from which it easily follows that t1 = t2 = t3 = 0 . b) It can’t be onto since rank A < m = dimension of the target. It also can’t be oneto-one since rank A = m − 1 < n = dimension of the domain. c) The zero matrix of size 3 × 1 work...
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This note was uploaded on 01/19/2010 for the course MAT MAT223 taught by Professor Uppal during the Spring '09 term at University of Toronto.

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