MATH 304Linear AlgebraLecture 15:Wronskian.The Vandermonde determinant.Basis of a vector space.Linear independenceDefinition. Let V be a vector space. Vectorsv1, v2, . . . , vk∈ V are called linearly dependent if theysatisfy a relationr1v1+ r2v2+ · · · + rkvk= 0,where the coefficients r1, . . . , rk∈ R are not all equal to zero.Otherwise the vectors v1, v2, . . . , vkare called linearlyindependent. That is, ifr1v1+r2v2+ · · · +rkvk= 0 =⇒ r1= · · · = rk= 0.An infinite set S ⊂ V is linearly dependent if there aresome linearly dependent vectors v1, . . . , vk∈ S. Otherwise Sis linearly independent.Remark. If a set S (finite or infinite) is linearly independentthen any subset of S is also linearly independent.Theorem Vectors v1, . . . , vk∈ V are linearlydependent if and only if one of them is a linearcombination of the other k − 1 vectors.Examples of linear independence.• Vectors e1= (1, 0, 0), e2= (0, 1, 0), ande3= (0, 0, 1) in R3.• Matrices E11=1 00 0, E12=0 10 0,E21=0 01 0, and E22=0 00 1.• Polynomials 1, x, x2, . . . , xn, . . .Problem. Show that functions ex, e2x, and e3xare linearly independent in C∞(R).Suppose that aex+ be2x+ ce3x= 0 for all x ∈ R, wherea, b, c are constants. We have to show that a = b = c = 0.Differentiate this identity twice:aex+ be2x+ ce3x= 0,aex+ 2be2x+ 3ce3x= 0,aex+ 4be2x+ 9ce3x= 0.It follows that A(x)v = 0, whereA(x) =exe2xe3xex2e2x3e3xex4e2x9e3x, v =abc.A(x) =exe2xe3xex2e2x3e3xex4e2x9e3x, v =abc.det A(x) = ex1 e2xe3x1 2e2x3e3x1 4e2x9e3x= exe2x1 1 e3x1 2 3e3x1 4 9e3x= exe2xe3x1 1 11 2 31 4 9= e6x1 1 11 2 31 4 9= e6x1 1 10 1 21 4 9= e6x1 1 10 1 20 3 8= e6x1 23 8= 2e6x6= 0.Since the matrix A(x) is invertible, we obtainA(x)v = 0 =⇒ v = 0 =⇒ a = b = c = 0WronskianLet f1, f2, . . . , fnbe smooth functions on an interval[a, b]. The Wronskian W [f1, f2, . . . , fn] is afunction on [a, b] defined byW [f1, f2, . . . , fn](x) =f1(x) f2(x) · · · fn(x)f′1(x) f′2(x) · · · f′n(x)............f(n−1)1(x) f(n−1)2(x) · · · f(n−1)n(x).Theorem If W [f1, f2, . . . , fn](x0) 6= 0 for somex0∈ [a, b] then the fu nctions f1, f2, . . . , fnarelinearly independent in C [a, b].Theorem Let λ1, λ2, . . . , λkbe distinct realnumbers. Then the functions eλ1x, eλ2x, . . . , eλkxare linearly independent.W [eλ1x, eλ2x, . . . , eλkx](x) =eλ1xeλ2x· · · eλkxλ1eλ1xλ2eλ2x· · · λkeλkx............λk−11eλ1xλk−12eλ2x· · · λk−1keλkx= e(λ1+λ2+···+λk)x1 1 · · · 1λ1λ2· · · λk............λk−11λk−12· · · λk−1k.The Vandermonde determinantDefinition. The Vandermonde determinant isthe determinant of the following matrixV =1 x1x21· · · xn−111 x2x22· · · xn−121 x3x23· · · xn−13...............1 xnx2n· · · xn−1n,where x1, x2, . . . , xn∈ R. Equivalently,V = (aij)1≤i,j≤n, where aij= xj−1i.Examples.•1 x11 x2= x2− x1.•1 x1x211 x2x221 x3x23=1 x101 x2x22− x1x21 x3x23− x1x3=1 0 01 x2− x1x22− x1x21 x3− x1x23− x1x3=x2− x1x22− x1x2x3− x1x23− x1x3= (x2− x1)1 x2x3− x1x23− x1x3= (x2− x1)(x3− x1)1 x21 x3= (x2− x1)(x3− x1)(x3− x2).Theorem1 x1x21· · · xn−111 x2x22· · · xn−121 x3x23· · · xn−13...............1 xnx2n· · · xn−1n=Y1≤i<j≤n(xj− xi).Corollary The Vandermonde determinant is notequal to 0 if and only if the numbers x1, x2, . . . , xnare distinct.Let x1, x2, . . . , xnbe distinct real numbers.Theorem For any b1, b2, . . . , bn∈ R there exists aunique polynomial p(x) = a0+a1x+ · · · +an−1xn−1of degree less than n such that p(xi) = bi,1 ≤ i ≤ n.a0+ a1x1+ a2x21+ · · · + an−1xn−11= b1a0+ a1x2+ a2x22+ · · · + an−1xn−12= b2· · · · · · · · · · · ·a0+ a1xn+ a2x2n+ · · · + an−1xn−1n= bna0, a1, . . . , an−1are unknowns. The coefficientmatrix is the Vandermonde matrix.BasisDefinition. Let V be a vector space. A linearlyindependent spanning set for V is called a basis.Suppose that a set S ⊂ V is a basis for V .“Spanning set” means that any vector v ∈ V ca n berepresented as a linear combinationv = r1v1+ r2v2+ · · · + rkvk,where v1, . . . , vkare distinct vectors from S andr1, . . . , rk∈ R. “Linearly independent” implies that the aboverepresentation is unique:v = r1v1+ r2v2+ · · · + rkvk= r′1v1+ r′2v2+ · · · + r′kvk=⇒ (r1− r′1)v1+ (r2− r′2)v2+ · · · + (rk− r′k)vk= 0=⇒ r1− r′1= r2− r′2= . . . = rk− r′k= 0Examples. • Standard basis for Rn:e1= (1, 0, 0, . . . , 0, 0), e2= (0, 1, 0, . . . , 0, 0),. . . ,en= (0, 0, 0, . . . , 0, 1).Indeed, (x1, x2, . . . , xn) = x1e1+ x2e2+ · · · + xnen.• Matrices1 00 0,0 10 0,0 01 0,0 00 1form a basis for M2,2(R).a bc d= a1 00 0+ b0 10 0+ c0 01 0+ d0 00 1.• Polynomials 1, x, x2, . . . , xn−1form a basis forPn= {a0+ a1x + · · · + an−1xn−1: ai∈ R}.• The infinite set {1, x, x2, . . . , xn, . . . } is a basisfor P, the space of all
View Full Document
Unlocking...