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Lecture 20

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Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter

Description
Wednesday, February 26 − Lecture 20: Linear independence Concepts: 1. Linearly independent set. 2. Recognize that subsets of linearly independent sets are linearly independent. 3. Characterize a linearly independent set as one being a set where no vector isa linear combination of the others. 20.1 Definition – Generalization of the definition of linear independence. Let v , v ...., v 1 2, k be k vectors in a vector space V. The vectors v , v ..., v are said to be linearly 1 2, k independent if and only if the only way that α 1 1 α v 2 2.. + α v = 0k k can hold true is if α ,1α ,2....., α kre all zeroes. If v1, v2,.., vkare not linearly independent (i.e. there existsα , α , .1...2 α not alk zero such that α v1 1α v +2 2... + α v = k k then they are said to be linearly dependent. x 20.1.1 Example – We know that the two functions e and sinx are vectors in the set F described above. Then S = Span{e , sinx} = { αe + βsinx : α, β belong to ℝ } is a x subspace of F. Show that the set {e , sinx} is linearly independent. Solution: Let us first recall that two functions f and g are equal on their domain Dif and only if f(x) = g(x) for all x in D. Suppxse αe + βsinx = 0(x) for all x (where 0(x) denotes the zero function; it maps every x in the domain to 0.). x - Suppose there exists α ≠ 0 such that αe + βsinx = 0(x). x - Then e = (–β/α)sinx for all x. This includes the value x = π. - But 0 ≠ e = (–β/α)sinπ = 0, a contradiction. - Then α must be 0. x x - Suppose β ≠ 0 is such that, 0e + βsinx = 0(x). Then (–0/β)e = sinx for all x. - Since sin(π/2) =1 ≠ 0, we have a contradiction. - So β must also be 0. - So {e , sinx} is linearly independent. 20.1.2 Proposition − The vectors M = {v , v , ..1, v2} with k , 2, are linearly independent if and only if no vector in M is a linear combination of the other vectors in M. Proof : This is similar to the previously given proof for the case of V = ℝ . n (⇐) - Suppose none of these vectors is a linear combination of the others. - Suppose M is not linearly independent. - Then there exists α , α 1...2., α not kll zeroes such that α v + α v +1 1..+ α2 2= k k 0. - Suppose α is not zero. Rearrange the order of {v , v , ...., v } so that p = 1, .i.e., p 1 2 k α 1s not zero. - Then v = 1 (α /α )2v +1− (2 /α ) v +3, 1.., 3 − (α /α )v . k 1 k - Hence v is a linear combination of the others. Contradiction. 1 (⇒) - Suppose {v , v1, .2.., v } ks linearly independent. - Suppose one vector of {v , v , .1..,2v } is k linear combination of the others. Say it is v . 1 - Then there exists α ....2, α suck that such that v = α v + 1.... 2 2 v . k k - Then v − 1 v − 2 v2+ , ..3, 3 α v . = 0. k k - Since the coefficient of v is n1t zero this contradicts our hypothesis. - Then no vector of {v , v ,1...2, v } iska linear combination of the others. If a vector in U is a linear combination of the others we will refer to it as being “redundant”. It doesn't contribute anything to its span. 20.2 Definition – Let {v 1, v2,..., vk} be a spanning family for a vector space V. That is, Span{v , v ..., v } = V. Then for every vector v in V there exists scalars α , α , ....., α 1 2, k 1 2 k such that v = α v1 1α v +2 2... + α v . k k If for each vectorv this set of scalars α , α , ..., α satisfying v = α v + α v + ..... + α v 1 2 k 1 1 2 2 k k is unique then we say that Span{v 1, v2,.., v k satisfies the unique representation property with respect to its spanning family. 20.2.1 Example – The spanning family Span{(1, 0), (0,1)} = Span{ e , e } = ℝ 1 2 2 satisfies the unique representation property since for an arbitrary vector v= (a, b) in ℝ , a and b are the unique scalars associated to v so that v = ae + be . 1 2 20.2.2 Theorem − Let S = {v , v , .1..,2v } be a k
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