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8. Linear Maps

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8. Linear Maps 176 8. Linear Maps In this chapter, we study the notion of a linear map of abstract vector spaces. This is the abstraction of the notion of a linear transformation on Rn. 8.1. Basic definitions Definition 8.1. Let U, V be vector spaces. A linear map L : U V (reads L from U → to V ) is a rule which assigns to each element U an element in V , such that L preserves addition and scaling. That is, L(u1 + u2)=L(u1)+L(u2),L(cu)=cL(u). Notation: we sometimes write L : u v,ifL(u)=v,andwecallv the image of u under → L. The two defining properties of a linear map are called linearity.Wheredoesa linear map L : U V send the zero element O in U? By linearity, we have → L(O)=L(0O)=0L(O). But 0L(O) is the zero element, which we also denote by O,inV .Thus L(O)=O. Example. For any vector space U, we have a linear map Id : U U given by U → Id : u u. U → 177 This is called the identity map. For any other vector space V , we also have a linear map O : U V given by → O : u O. → This is called the zero map. Example. Let A be an m n matrix, and define × L : Rn Rm,L(X)=AX. A → A We have seen, in chapter 3, that this map is linear. We have also seen that every linear map L : Rn Rm is represented by a matrix A,ie.ifL = L for some m n matrix A. → A × Example. Let A an m n matrix, and define × L : M(n, k) M(m, k),L(B)=AB. A → A This map is linear because A(B1 + B2)=AB1 + AB2,A(cB)=c(AB) for any B ,B ,B M(n, k) and any number c (chapter 3). 1 2 ∈ Exercise. Is L : Rn R defined by L(X)=X X, linear? → · Exercise. Fix a vector A in Rn.IsL : Rn R defined by L(X)=A X, linear? → · Example. (With calculus) Let C∞ be the vector space of smooth functions, and let D(f)=f .ThenD : C∞ C∞ is linear. differentiation → Example. (With calculus) Continuing with the preceding example, let V = Span 1,t,..,tn n { } k k 1 in C∞.SinceD(t )=kt − for k =0, 1, 2,.., it follows that D(f) lands in Vn 1 for any − function f in Vn. Thus we can regard D as a linear map D : Vn Vn 1. → − Theorem 8.2. Let U, V be vector spaces, S be a basis of U. Let f : S V be any map. → Then there is a unique linear map L : U V such that L(u)=f(u) for every element u → 178 in S. Proof: If u1,..,uk are distinct elements in S,thenwedefine k k L( aiui)= aif(ui) i=1 i=1 for any numbers ai.Sinceeveryelementw in U can be expressed uniquely in the form k a u , this defines a map L : U V . Moreover, L(u)=l(u) for every u in S.Soit i=1 i i → remains to verify that L is linear. Let w1,w2,w be elements in U, and c be a scalar. We can write k w1 = ai ui i=1 k w2 = aiui i=1 k w = aiui i=1 where u1,..,uk are distinct elements in S, and the a’s are numbers. Then L(w1 + w2)=L[ (ai + ai)ui] = (ai + ai)f(ui) = ai f(ui)+ aif(ui) = L(w1)+L(w2) L(cw)=L[c aiui] = L[ (cai)ui] = (cai)f(ui) = c aif(ui) = cL(w). Thus L is linear. For the rest of this section, L : U V will be a linear map. → Definition 8.3. The set Ker L = u U L(u)=O { ∈ | } 179 is called the kernel of L. The set Im L = L(u) u U { | ∈ } is called the image of L. More generally, for any subspace W U,wewrite ⊂ L(W )= L(w) w W . { | ∈ } Lemma 8.4. Ker L is linear subspace of U. Im L is a linear subspace of V . Proof: If u ,u Ker L,then 1 2 ∈ L(u1 + u2)=L(u1)+L(u2)=O + O = O. Thus u + u Ker L.Ifc is any number and u Ker L,then 1 2 ∈ ∈ L(cu)=cL(u)=cO = O. Thus cu Ker L.ThusKer L is closed under addition and scaling in U, hence is a linear ∈ subspace of U. The proof for Im L is similar and is left as an exercise. Example. Let A an m n matrix, and L : Rn Rm with L (X)=AX.Then × A → A t Ker LA = Null(A), and Im LA = Row(A ). This shows that the notions of kernel and image are abstractions of the notions of null space and row space. Example. (With calculus) Consider D : C∞ C∞ with D(f)=f . Ker D consists → of just constant functions. Any smooth function is the derivative of a smooth function (fundamental theorem of calculus). Thus Im D = C∞. n Example. (With calculus) As before, let V = Span 1,t,..,t in C∞, and consider n { } k k 1 D : Vn Vn 1. As before, Ker D consists of the constant functions. Since D(t )=kt − → − n 1 for k =0, 1, 2,.., it follows 1,t,..,t − are all in Im D. Thus any of their linear combination is also in Im D.SowehaveIm D = Vn 1. − Example. Fix a nonzero vector A in Rn, and define L : Rn R,L: X A X. → → · 180 Then Ker L = V ⊥ where V is the line spanned by A, and Im L = R. Example. Let L : U V be a linear map, and W be a linear subspace of U.Wedefine → a new map L : W V as follows: |W → L (w)=L(w). |W This map is linear. L is called the restriction of L to W . |W 8.2. A dimension relation Throughout this section, L : U V will be a linear map of finite dimensional → vector spaces. Lemma 8.5. Suppose that Ker L = O and that u ,..,u is a linearly independent { } { 1 k} set in U. Then L(u1),..,L(uk) form a linearly independent set. Proof: Let k xiL(ui)=O. i=1 By linearity of L, this reads k L( xiui)=O. i=1 Since Ker L = O , this implies that { } k xiui = O. i=1 Since u ,..,u is linearly independent, it follows that the x’s are all zero. Thus { 1 k} L(u1),..,L(uk) form a linearly independent set. Theorem 8.6. (Rank-nullity relation) dim U = dim (Ker L)+dim (Im L). Proof: If Im L is the zero space, then Ker L = U, and the theorem holds trivially. Suppose Im L is not the zero space, and let v ,..,v be a basis of Im L. Let L(u )=v . { 1 s} i i 181 Let w ,..,w be a basis of Ker L U. We will show that the elements u ,..,u ,w ,..,w { 1 r} ⊂ 1 s 1 r form a linearly independent set that spans U. Linear independence: Let • x u + + x u + y w + + y w = O. 1 1 ··· s s 1 1 ··· r r Applying L to this, we get x v + + x v = O. 1 1 ··· s s Since v ,..,v is linearly independent, the x’s are all zero, and so { 1 s} y w + + y w = O. 1 1 ··· r r Since w ,..,w is linearly independent, the y’s are all zero. This shows that { 1 r} u1,..,us,w1,..,wr form a linearly independent set. Spanning property: Let u U.Since v ,..,v spans Im L,wehave • ∈ { 1 s} L(u)=a v + + a v 1 1 ··· s s for some a . This gives L(u)=a L(u )+ + a L(u ). By linearity of L,thisyields i 1 1 ··· s s L(u a u a u )=0, − 1 1 −···− s s which means that u a u a u Ker L.So − 1 1 −···− s s ∈ u a u a u = b w + + b w − 1 1 −···− s s 1 1 ··· r r because w ,..,w spans Ker L. This shows that { 1 r} u = a u + + a u + b w + + b w . 1 1 ··· s s 1 1 ··· r r This completes the proof. Example. Let A be an m n matrix, and consider L : Rn Rm, L (X)=AX.Since × A → A t t Ker LA = Null(A),ImLA = Row(A ),rankA= dim Row(A ), The preceding theorem yields n = rank A + dim Null(A). We have proven this in chapter 4 by using an orthogonal basis. Note that the preceding theorem does not involve using any inner product. 182 n Example. (With calculus) As before, let V = Span 1,t,..,t in C∞, and consider n { } D : Vn Vn 1. Note that dim Vn = n + 1. We have seen that Ker D consists of the → − constant functions, and that Im D = Vn 1.Sodim(Ker D) = 1 and dim(Im D)=n. − Thus, in this case, dim Vn = dim(Ker D)+dim(Im D) as expected. Corollary 8.7. Let U, W be finite dimensional linear subspaces of V . Then dim(U)+dim(W )=dim(U + W )+dim(U W ). ∩ Proof: Define a linear map L : U W V ,(u, w) u w.Then ⊕ → → − Ker L = (u, u) u U, u W ,ImL= U + W.
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