MATH 304 Linear Algebra Lecture 10 Linear independence Wronskian Spanning set Let S be a subset of a vector space V Definition The span of the set S is the smallest subspace W V that contains S If S is not empty then W Span S consists of all linear combinations r1 v1 r2 v2 rk vk such that v1 vk S and r1 rk R We say that the set S spans the subspace W or that S is a spanning set for W Remark If S1 is a spanning set for a vector space V and S1 S2 V then S2 is also a spanning set for V Linear independence Definition Let V be a vector space Vectors v1 v2 vk V are called linearly dependent if they satisfy a relation r1 v1 r2 v2 rk vk 0 where the coefficients r1 rk R are not all equal to zero Otherwise vectors v1 v2 vk are called linearly independent That is if r1 v1 r2 v2 rk vk 0 r1 rk 0 An infinite set S V is linearly dependent if there are some linearly dependent vectors v1 vk S Otherwise S is linearly independent Examples of linear independence Vectors e1 1 0 0 e2 0 1 0 and e3 0 0 1 in R3 xe1 y e2 ze3 0 x y z 0 x y z 0 1 0 0 1 Matrices E11 E12 0 0 0 0 0 0 0 0 E21 and E22 1 0 0 1 aE11 bE12 cE21 dE22 O a b c d 0 a b c d O Examples of linear independence Polynomials 1 x x 2 x n a0 a1 x a2 x 2 an x n 0 identically ai 0 for 0 i n The infinite set 1 x x 2 x n Polynomials p1 x 1 p2 x x 1 and p3 x x 1 2 a1 p1 x a2 p2 x a3 p3 x a1 a2 x 1 a3 x 1 2 a1 a2 a3 a2 2a3 x a3 x 2 Hence a1 p1 x a2 p2 x a3 p3 x 0 identically a1 a2 a3 a2 2a3 a3 0 a1 a2 a3 0 Problem Let v1 1 2 0 v2 3 1 1 and v3 4 7 3 Determine whether vectors v1 v2 v3 are linearly independent We have to check if there exist r1 r2 r3 R not all zero such that r1 v1 r2 v2 r3 v3 0 This vector equation is equivalent to a system 1 3 4 0 r1 3r2 4r3 0 2 1 7 0 2r r2 7r3 0 1 0r1 r2 3r3 0 0 1 3 0 The vectors v1 v2 v3 are linearly dependent if and only if the matrix A v1 v2 v3 is singular We obtain that det A 0 Theorem The following conditions are equivalent i vectors v1 vk are linearly dependent ii one of vectors v1 vk is a linear combination of the other k 1 vectors Proof i ii Suppose that r1 v1 r2 v2 rk vk 0 where ri 6 0 for some 1 i k Then ri 1 rk vi rr1i v1 ri 1 ri vi 1 ri vi 1 ri vk ii i Suppose that vi s1 v1 si 1 vi 1 si 1 vi 1 sk vk for some scalars sj Then s1 v1 si 1 vi 1 vi si 1 vi 1 sk vk 0 Theorem Vectors v1 v2 vm Rn are linearly dependent whenever m n i e the number of coordinates is less than the number of vectors Proof Let vj a1j a2j anj for j 1 2 m Then the vector equality t1 v1 t2 v2 tm vm 0 is equivalent to the system a t a12 t2 a1m tm 0 11 1 a21 t1 a22 t2 a2m tm 0 a t a t a t 0 n1 1 n2 2 nm m Note that vectors v1 v2 vm are columns of the matrix aij The number of leading entries in the row echelon form is at most n If m n then there are free variables therefore the zero solution is not unique Example Consider vectors v1 1 1 1 v2 1 0 0 v3 1 1 1 and v4 1 2 4 in R3 Two vectors are linearly dependent if and only if they are parallel Hence v1 and v2 are linearly independent Vectors v1 v2 v3 are linearly independent if and only if the matrix A v1 v2 v3 is invertible 1 1 1 1 1 2 6 0 det A 1 0 1 1 1 1 0 1 Therefore v1 v2 v3 are linearly independent Four vectors in R3 are always linearly dependent Thus v1 v2 v3 v4 are linearly dependent Problem Let A 1 1 1 0 Determine whether matrices A A2 and A3 are linearly independent 1 0 0 1 1 1 3 2 A A We have A 0 1 1 1 1 0 The task is to check if there exist r1 r2 r3 R not all zero such that r1 A r2 A2 r3 A3 O This matrix equation is equivalent to a system 1 1 0 0 0 1 0 1 r 0r r 0 1 2 3 1 1 0 0 0 1 1 0 r1 r2 0r3 0 0 1 0 0 1 0 0 0 r r 0r 0 1 2 3 0r r r 0 0 1 1 0 0 0 0 0 1 2 3 The row echelon form of the augmented matrix shows there is a free variable Hence the system has a nonzero solution so that the matrices are linearly dependent one relation is A A2 A3 O Problem Show that functions e x e 2x and e 3x are linearly independent in C R Suppose that ae x be 2x ce 3x 0 for all x R where a b c are constants We have to show that a b c 0 Differentiate this identity twice ae x be 2x ce 3x 0 ae x 2be 2x 3ce 3x 0 ae x 4be 2x 9ce 3x 0 It follows that A x v 0 where x e e 2x e 3x A x e x 2e 2x 3e 3x e x 4e 2x 9e 3x a v b c e x e 2x e 3x A x e x 2e 2x 3e 3x e x 4e 2x 9e 3x det A x e x x 2x 3x e e e 1 e 2x e 3x 1 2e 2x 3e 3x 1 4e 2x 9e 3x 1 1 1 1 2 3 1 4 9 1 1 1 e 6x 0 1 2 0 3 8 e e 6x 6x 1 2 3 8 a v b c e x e 2x 1 1 1 1 2 3 1 4 9 1 1 e 3x 1 2 3e 3x 1 4 9e 3x e 6x 2e 6x 6 0 Since the matrix A x is invertible we obtain A x v 0 v 0 a b c 0 1 1 1 0 1 2 1 4 9 Wronskian Let …
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