MATH 304 Linear Algebra Lecture 10: Linear independence. Wronskian.

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 r 1 v 1 + r 2 v 2 + + r k v k such that v 1,...,v k S and r 1,...,r k R. We say that the set S spans the subspace W or that S is a spanning set for W. Remark. If S 1 is a spanning set for a vector space V and S 1 S 2 V, then S 2 is also a spanning set for V.

Linear independence Definition. Let V be a vector space. Vectors v 1,v 2,...,v k V are called linearly dependent if they satisfy a relation r 1 v 1 + r 2 v 2 + + r k v k = 0, where the coefficients r 1,...,r k R are not all equal to zero. Otherwise vectors v 1,v 2,...,v k are called linearly independent. That is, if r 1 v 1 +r 2 v 2 + +r k v k = 0 = r 1 = = r k = 0. An infinite set S V is linearly dependent if there are some linearly dependent vectors v 1,...,v k S. Otherwise S is linearly independent.

Examples of linear independence Vectors e 1 = (1, 0, 0), e 2 = (0, 1, 0), and e 3 = (0, 0, 1) in R 3. xe 1 + ye 2 + ze 3 = 0 = (x, y, z) = 0 = x = y = z = 0 ( ) 1 0 Matrices E 11 =, E 0 0 12 = ( ) ( ) 0 0 0 0 E 21 =, and E 1 0 22 =. 0 1 ae 11 + be 12 + ce 21 + de 22 = O = = a = b = c = d = 0 ( a b c d ( ) 0 1, 0 0 ) = O

Examples of linear independence Polynomials 1, x, x 2,...,x n. a 0 + a 1 x + a 2 x 2 + + a n x n = 0 identically = a i = 0 for 0 i n The infinite set {1, x, x 2,...,x n,... }. Polynomials p 1 (x) = 1, p 2 (x) = x 1, and p 3 (x) = (x 1) 2. a 1 p 1 (x) + a 2 p 2 (x) + a 3 p 3 (x) = a 1 + a 2 (x 1) + a 3 (x 1) 2 = = (a 1 a 2 + a 3 ) + (a 2 2a 3 )x + a 3 x 2. Hence a 1 p 1 (x) + a 2 p 2 (x) + a 3 p 3 (x) = 0 identically = a 1 a 2 + a 3 = a 2 2a 3 = a 3 = 0 = a 1 = a 2 = a 3 = 0

Problem Let v 1 = (1, 2, 0), v 2 = (3, 1, 1), and v 3 = (4, 7, 3). Determine whether vectors v 1,v 2,v 3 are linearly independent. We have to check if there exist r 1, r 2, r 3 R not all zero such that r 1 v 1 + r 2 v 2 + r 3 v 3 = 0. This vector equation is equivalent to a system r 1 + 3r 2 + 4r 3 = 0 2r 1 + r 2 7r 3 = 0 0r 1 + r 2 + 3r 3 = 0 1 3 4 0 2 1 7 0 0 1 3 0 The vectors v 1,v 2,v 3 are linearly dependent if and only if the matrix A = (v 1,v 2,v 3 ) is singular. We obtain that det A = 0.

Theorem The following conditions are equivalent: (i) vectors v 1,...,v k are linearly dependent; (ii) one of vectors v 1,...,v k is a linear combination of the other k 1 vectors. Proof: (i) = (ii) Suppose that r 1 v 1 + r 2 v 2 + + r k v k = 0, where r i 0 for some 1 i k. Then v i = r 1 ri v 1 r i 1 r i v i 1 r i+1 r i v i+1 r k ri v k. (ii) = (i) Suppose that v i = s 1 v 1 + + s i 1 v i 1 + s i+1 v i+1 + + s k v k for some scalars s j. Then s 1 v 1 + + s i 1 v i 1 v i + s i+1 v i+1 + + s k v k = 0.

Theorem Vectors v 1,v 2,...,v m R n are linearly dependent whenever m > n (i.e., the number of coordinates is less than the number of vectors). Proof: Let v j = (a 1j, a 2j,...,a nj ) for j = 1, 2,...,m. Then the vector equality t 1 v 1 + t 2 v 2 + + t m v m = 0 is equivalent to the system a 11 t 1 + a 12 t 2 + + a 1m t m = 0, a 21 t 1 + a 22 t 2 + + a 2m t m = 0, a n1 t 1 + a n2 t 2 + + a nm t m = 0. Note that vectors v 1,v 2,...,v m are columns of the matrix (a ij ). 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 v 1 = (1, 1, 1), v 2 = (1, 0, 0), v 3 = (1, 1, 1), and v 4 = (1, 2, 4) in R 3. Two vectors are linearly dependent if and only if they are parallel. Hence v 1 and v 2 are linearly independent. Vectors v 1,v 2,v 3 are linearly independent if and only if the matrix A = (v 1,v 2,v 3 ) is invertible. 1 1 1 det A = 1 0 1 1 0 1 = 1 1 1 1 = 2 0. Therefore v 1,v 2,v 3 are linearly independent. Four vectors in R 3 are always linearly dependent. Thus v 1,v 2,v 3,v 4 are linearly dependent.

Problem. Let A = ( 1 1 1 0 ). Determine whether matrices A, A 2, and A 3 are linearly independent. We have A = ( ) 1 1, A 1 0 2 = ( 0 1 1 1 ), A 3 = ( ) 1 0. 0 1 The task is to check if there exist r 1, r 2, r 3 R not all zero such that r 1 A + r 2 A 2 + r 3 A 3 = O. This matrix equation is equivalent to a system r 1 + 0r 2 + r 3 = 0 r 1 r 2 + 0r 3 = 0 r 1 + r 2 + 0r 3 = 0 0r 1 r 2 + r 3 = 0 1 0 1 0 1 1 0 0 1 1 0 0 1 1 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 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 + A 2 + A 3 = 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 A(x) = ex e 2x e 3x e x 2e 2x 3e 3x, v = a b. e x 4e 2x 9e 3x c

e x e 2x e 3x a A(x) = e 2e 2x 3e 3x, v = b. e x 4e 2x 9e 3x c 1 e 2x e 3x det A(x) = e x 1 2e 2x 3e 3x 1 4e 2x 9e 3x = 1 1 e 3x ex e 2x 1 2 3e 3x 1 4 9e 3x 1 1 1 = e x e 2x e 3x 1 2 3 1 4 9 = 1 1 1 e6x 1 2 3 1 4 9 = 1 1 1 e6x 0 1 2 1 4 9 1 1 1 = e 6x 0 1 2 0 3 8 = e6x 1 2 3 8 = 2e6x 0. Since the matrix A(x) is invertible, we obtain A(x)v = 0 = v = 0 = a = b = c = 0

Wronskian Let f 1, f 2,...,f n be smooth functions on an interval [a, b]. The Wronskian W[f 1, f 2,...,f n ] is a function on [a, b] defined by W[f 1, f 2,...,f n ](x) = f 1 (x) f 2 (x) f n (x) f 1(x) f 2(x) f n(x)....... f (n 1) 1 (x) f (n 1) 2 (x) f n (n 1) (x) Theorem If W[f 1, f 2,...,f n ](x 0 ) 0 for some x 0 [a, b] then the functions f 1, f 2,...,f n are linearly independent in C[a, b].

Theorem Let λ 1, λ 2,...,λ k be distinct real numbers. Then the functions e λ 1x, e λ 2x,...,e λ kx are linearly independent. W[e λ1x, e λ2x,...,e λkx ](x) = = e (λ 1+λ 2 + +λ k )x e λ 1x e λ 2x e λ kx λ 1 e λ 1x λ 2 e λ 2x λ k e λ kx...... λ k 1 1 e λ 1x λ k 1 2 e λ 2x λ k 1 1 1 1 λ 1 λ 2 λ k...... λ k 1 1 λ k 1 2 λ k 1 k 0 k e λ kx since the latter determinant is the transpose of the Vandermonde determinant.