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MATH 136
Robert Andre

Wednesday, January 8 − Lecture 2 : Linear independence of vectors in ℝ . n Concepts: 1. Linearlyindependent subset in ℝ . n 2. Characterization of a linearly independent set as one being a set where no vector is a linear combination of the others. 3. A plane in ℝ . A hyperplane in ℝ . n 2.1 Definitions. - Let v 1, v2,.., v ke k vectors in ℝ and suppose the vector 0 represents the zero vector 0 = (0, 0, …, 0). The vectors v , v 1...2,v areksaid to be linearly independentif the only way that can hold true is if α ,1α ,2..., α ake all zeroes. The solution where all the α‘s are i zeros is called the trivial solution of this vector equation. - If v , v ..., v are not linearly independent then they are said to be linearly 1 2, k dependent. Note that a linearly independent set cannot contain the zero-vector, 0. This fact follows from the definition. In class, we will often abbreviate the words “linearly independent” with the letters “L.I.”. 3 2.2 Example − Verify whether the set {v 1, v2, v3} in ℝ where v = (1, 1, 1), v = (02 1, 7) and v 3 (0, 0, 3) is linearly independent. Solution: The given set is linearly independent. m 2.3 Theorem − A subset of a finite linearly independent subset of ℝ is linearly independent. Proof: m - Suppose M = {v , v , 1..,2v } is r linearly independentsubset of ℝ . - Let S = {v , k1, .k2., v } km a non-empty subset of M. We claim that S is linearly independent. - Suppose not. Then there exists α , α , k1 k2 …, α km ,not all equal to zero, such that α k1+k1 v + …k2 k2v = 0. km km - Suppose, without loss of generality, α ≠ 0. Then k1 0v 1 … + (α v + αk1 k1… + αk2 k2+ ... + 0v r where α ≠ k1 - This contradicts the fact that M = {v , v , ..1., 2 } is a rinearly independent subset of m ℝ . - The source of the contradiction is our supposition that S is not linearly independent. - So S = {v k1, vk2 ...., vkm is linearly independent, as required. 2.4 Theorem. An important characterization of linear independence − The vectors M = {v , v , ...., v } of ℝ n with k ≥ 2, are linearly independent ifand only if no vector 1 2 k , in M is a linear combination of the others. Proof of (⇐): Given: None of these vectors in M is a linear combination of the others. n Required to show: That M is linearly independent subset of ℝ . Let’s suppose that M is not linearly independent. - Then there exists α , α 1 ...2 α not akl zeroes such that - Suppose one of the coefficients, sayα is not zerop,Rearrange the order of {v , v , ...., 1 2 vk} so that p = 1, .i.e., α is 1ot zero. - Then - Hence v is1a linear combination of the others. Contradiction. - So M is linearly independent. Proof of (⇒) : Given: M = {v 1, v2, ...., k } is linearly independent. Required to show: That no vector in M is a linear combination of the others. - Suppose one vector of {v , v , 1...2 v } is k linear combination of the others. Say it is v1. We claim that this will lead to a contradiction of our hypothesis. - Then there exists α ...2., α sukh that such that - Then - This contradicts the fact that M is linearly independent. - Then no vector of {v , v ,1....2 v } iska linear combination of the others. 2.5 Remark − The above theorem shows that we may have defined “linearly independent” as follows: n "The set U = {v , v1...2,v } iska linearly independent of ℝ if and only if no vector in U is a linear combination of the other vectors in U." If a vector v jin U is a linear combination of the others we refer to v as
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