MATH 311-504Topics in Applied MathematicsLecture 2-7:Basis and coordinates.IsomorphismDefinition. A linear mapping f : V1→ V2is calledan isomorphism of vector spaces if it is bo thone-to-one and o nto.Two vector spaces V1and V2are called isomorphicif there exists an isomorphism f : V1→ V2.The word “isomorphism” applies when two complexstructures can be mapped o nto each other, in sucha way that to each part of one s tructure there is acorresponding part in the other structure, where“corresponding” means that the two parts playsimilar roles in their respective structures.Examples of i somorphisms• M2,2(R) i s isomorphic to R4.Isomorphism:a bc d7→ (a, b, c, d).• M2,3(R) i s isomorphic to M3,2(R).Isomorphism:a1a2a3b1b2b37→a1b1a2b2a3b3.• The plane z = 0 in R3is i s omorphic to R2.Isomorphism: (x, y, 0) 7→ (x, y).• Pnis i s omorphic to Rn+1.Isomorphism: a0+a1x+ · · · +anxn7→ (a0, a1, . . . , an).Classification problems of linear alg ebraProblem 1 Given vector spaces V1and V2,determine whether they are isomorphic or not.Problem 2 Given a vector space V , determi newhether V is iso morphic to Rnfor some n ≥ 1 .Problem 3 Show that vector spaces Rnand Rmare not isomorphic if m 6= n.Linear independenceDefinition. Let V be a vector space. Vectorsv1, v2, . . . , vk∈ V are called linearly dependent ifthey satisfy a r elationr1v1+ r2v2+ · · · + rkvk= 0,where the coefficients r1, . . . , rk∈ R are not allequal to zero. Otherwise the vectors v1, v2, . . . , vkare call ed linearly independent. T hat is , ifr1v1+r2v2+ · · · +rkvk= 0 =⇒ r1= · · · = rk= 0.An infinite set S ⊂ V is linearly dependent i fthere are som e linearly dependent vectorsv1, . . . , vk∈ S. Otherwis e S is linearlyindependent.Theorem Vectors v1, . . . , vk∈ V are linearlydependent if and onl y if one of them is a linearcombination of the other k − 1 v ectors.Examples of linear independence:• Vectors e1= (1, 0, 0, . . . , 0),e2= (0, 1, 0, . . . , 0),. . . , en= (0, 0, . . . , 0, 1) in Rn.• Matrices1 00 0,0 10 0,0 01 0,0 00 1.• Polynomials 1, x, x2, . . . , xn, . . . .Problem. Show that functions x, ex, and e−xarelinearly independent in C (R).Suppose that ax + bex+ ce−x= 0 for all x ∈ R, wherea, b, c are constants. We have to show that a = b = c = 0.Divide both sides of the identity by ex:axe−x+ b + ce−2x= 0.The left-hand side approaches b as x → +∞ . =⇒ b = 0Now ax + ce−x= 0 for all x ∈ R. For any x 6= 0 divideboth sides of the identity by x:a + cx−1e−x= 0.The left-hand side approaches a as x → +∞. =⇒ a = 0Now ce−x= 0 =⇒ c = 0.Theorem 1 Let λ1, λ2, . . . , λkbe distinct realnumbers. Then the functions eλ1x, eλ2x, . . . , eλkxare linearly independent.Theorem 2 The set of functions{xmeλx| λ ∈ R, m = 0, 1, 2, . . . }is linearly independent.Spanning sets a nd linear dependenceLet v0, v1, . . . , vkbe vectors from a vector space V .Proposition If v0is a linear com bination of vectorsv1, . . . , vkthenSpan(v0, v1, . . . , vk) = Span( v1, . . . , vk).Indeed, if v0= r1v1+ · · · + rkvk, thent0v0+ t1v1+ · · · + tkvk== (t0r1+ t1)v1+ · · · + (t0rk+ tk)vk.Corollary Any s panning set for a vector space isminimal if and only i f it is l inearly independent.BasisDefinition. Let V be a vector space. A l inearlyindependent 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 can 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= . . . = rn− r′n= 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.• Matrices1 00 0,0 10 0,0 01 0,0 00 1form a basis for M2,2(R).a bc d= a1 00 0+ b0 10 0+ c0 01 0+ d0 00 1.• Polynomials 1, x, x2, . . . , xnform a basis forPn= {a0+ a1x + · · · + anxn: ai∈ R}.• The infinite s et {1, x, x2, . . . , xn, . . . } is a basisfor P, the space of all polynomials.Problem Let v1= (2, 5) and v2= (1, 3). Showthat {v1, v2} is a basis f or R2.Linear independence is obvious: v1and v2are not parallel.To show spanning, it is enough to represent vectors e1= (1, 0)and e2= (0, 1) as linear combinations of v1and v2.e1= r1v1+r2v2⇐⇒2r1+ r2= 15r1+ 3r2= 0⇐⇒r1= 3r2= −5e2= r1v1+r2v2⇐⇒2r1+ r2= 05r1+ 3r2= 1⇐⇒r1= −1r2= 2Thus e1= 3v1− 5v2and e2= −v1+ 2v2.Then (x, y ) = xe1+ y e2= x( 3v1− 5v2) + y (−v1+ 2v2)= (3x − y )v1+ (−5x + 2y)v2.Let W be the set of all sol utions of the ODEy′′(x) − y (x) = 0. W is a subspace of the v ectorspace C∞(R) since it is the null-space of the linearoperator L : C∞(R) → C∞(R), L(f ) = f′′− f .W contains functions ex, e−x,hyper bolic sine s inh x =12(ex− e−x), andhyper bolic cosine cosh x =12(ex+ e−x).We have that (sinh x)′= cosh x,(cosh x)′= sinh x, cosh2x − sinh2x = 1.Proposition {ex, e−x} and {cosh x, sinh x} aretwo bases for W .Proposition {ex, e−x} and {cosh x, sinh x} aretwo bases for W .Proof: “Linear independence”: exand e−xare linearlyindependent as shown earlier.Further, cosh 0 = 1 , sinh 0 = 0 , cosh′0 = 0, sinh′0 = 1.It follows that cosh x and sinh x are not scalar multiples ofeach other.“Spanning”: Take any function f ∈ W . Consider a functiong(x) = a cosh x + b sinh x, where a = f (0) , b = f′(0).We have g (0) = a, g′(0) = b.The initial value problem y′′− y = 0, y(0) = a, y′(0) = bhas a unique solution. Therefore f = g .Thus f (x) = a cosh x + b sinh x=a2(ex+ e−x) +b2(ex− e−x) =12(a + b)ex+12(a − b)e−x.Basis and coordinatesIf {v1, v2, . . . , vn} is a basi s for a vector space V ,then any vector v ∈ V has a unique representationv = x1v1+ x2v2+ · · · + xnvn,where xi∈ R. T he coefficients x1, x2, . . . , xnarecalled …