A. MillerM340May 97Linear TransformationsDefinition 1 A linear transformation is a function L : V → W from onevector space V to another W which satisfies:1. L(u + v) = L(u) + L(v)2. L(au) = aL(u)for every u, v ∈ V and scalar a.Exercise 2 Let L : V → W be a function from a vector space V to a vectorspace W . Prove that L is a linear transformation iffL(au + bv) = aL(u) + bL(v)for any u, v ∈ V and scalars a and b.Theorem 3 Every linear transformation L : Rn→ Rmis determined by amatrix A ∈ MATm,n:L(X) = AXfor every X ∈ Rnproof:Given A sinceA(X + Y ) = AX + AY and A(aX) = a(AX)it is clear that L(X) = AX is a linear transformation.For the converse, assume L : Rn→ Rmis a linear transformation. Thecanonical basis for Rnis the sequence of columns of the n×n identity matrix,In×n. So let ei= coli(In×n). Note for any vectorx1x2...xn= x1e1+ x2e2+ · · · + xnen.1Let A be the matrix such thatL(ei) = coli(A)for every i = 1, . . . , n. ThenLx1x2...xn= L(x1e1+ x2e2+ · · · + xnen)= x1L(e1) + x2L(e2) + · · · + xnL(en)= x1col1(A) + x2col2(A) + · · · + xncoln(A)= Ax1x2...xnQEDDefinition 4 For L : V → W a linear transformation, define the null spaceof L, null(L), and the range space of L, range(L) as follows:1. null(L) = {v ∈ V : L(v) = z}2. range(L) = {w ∈ W : there exists v ∈ V such that L(v) = w}Theorem 5 null(L) and range(L) are subspaces of V and W , respectively.proof:null(L): Suppose u, v ∈ null(L) and a, b scalars. Then L(u) = z andL(v) = z andL(au + bv) = aL(u) + bL(v) = az + bz = z + z = z.Hence au + bv ∈ null(L)range(L): Suppose w1, w2∈ range(L) and a1, a2are scalars. Then thereexists v1, v2∈ V such that L(v1) = w1and L(v2) = w2. ThenL(a1v1+ a2v2) = a1L(v1) + a2L(v2) = a1w1+ a2w2and so a1w1+ a2w2∈ range(L).QED2Definition 6 A function L : V → W is one-to-one iff L(u) = L(v) impliesu = v for every u, v ∈ V .Definition 7 A function L : V → W is onto iff for every w ∈ W thereexists v ∈ V such that L(v) = w.Definition 8 A function L : V → W is an isomorphism iff L is a one-to-one, onto, linear transformation.Theorem 9 A linear transformation L : V → W is an isomorphism iffnull(L) = {z} and range(L) = W .proof:It is clear from the definitions that L is onto iff range(L) = W .Claim L is one-to-one iff null(L) = {z}.proof:Assume L is one-to-one. Then L(u) = z and L(z) = z implies u = z andso null(L) = {z}.Conversely, assume null(L) = {z}. Then if L(u) = L(v), then L(u − v) =L(u) − L(v) = z and so u − v ∈ null(L). Hence u − v ∈ {z} and so u − v = zand therefore u = v. So L(u) = L(v) implies u = v and therefore L isone-to-one. This proves the Claim.QEDTheorem 10 Suppose A is an invertible n × n matrix. If L : Rn→ Rnisdefined by L(X) = AX, then L is an isomorphism.proof:We already know from Theorem 3 that L is a linear transformation. Tosee that it is one-to-one, assume L(X) = L(Y ). Then AX = AY impliesA−1AX = A−1AY implies X = Y . To see that it is onto, suppose Y ∈ Rnand let X = A−1Y , then L (X) = AX = AA−1Y = Y .QEDTheorem 11 A linear transformation L : V → W is an isomorphism iffthere exists a linear transformation L−1: W → V such that L ◦ L−1= IdWand L−1◦ L = IdV. (IdVis the identity map on V , i.e., IdV(u) = u allu ∈ V .)3proof:First suppose that such an L−1exists. To see that L is one-to-one, supposeL(u) = L(v). Then L−1(L(u)) = L−1(L(v)), butu = L−1(L(u)) = L−1(L(v)) = vand so u = v. To see that L is onto, suppose w ∈ W . Let v = L−1(w), thenL(v) = L(L−1(w)) = w. Hence for every w ∈ W there exists v ∈ V suchthat L(v) = w and so L is onto.For the converse, suppose that L is an isomorphism. This means that Lis both one-to-one and onto. It f ollows that for every w ∈ W there existsa unique v ∈ V such that L(v) = w. Define L−1: W → V by the ruleL−1(w) = v iff L(v) = w. It follows that v = L−1(w) = v = L−1(L(v))for every v ∈ V and so L−1◦ L = IdV; and also w = L(v) = L(L−1(w)for every w ∈ W and so L ◦ L−1= IdW. Finally we prove the L−1isa linear transformation. Suppose w1, w2∈ W and a1, a2are scalars. Letu = L−1(a1w1+ a2w2) and let v = a1L−1w1+ a2L−1(w2), thenL(u) = L(L−1(a1w1+ a2w2)) = a1w1+ a2w2andL(v) = L(a1L−1w1+a2L−1(w2)) = a1L(L−1(w1)+a2L(L−1(w2) = a1w1+a2w2hence L(u) = L(v) and since L is one-to-one we know that u = v and soL−1(a1w1+ a2w2) = u = v = a1L−1w1+ a2L−1(w2)QEDExercise 12 Suppose L : V → W is a is one-to-one linear transformationand v1, . . . , vnare linearly independent vectors in V .Prove that L(v1), . . . , L(vn) are linearly independent vectors in W .Exercise 13 Suppose L : V → W is a linear transformation and v1, . . . , vnare vectors in V . Prove thatL(span({v1, . . . , vn})) = span({L(v1), . . . , L(vn)})4Definition 14 V and W are isomorphic iff there exists an isomorphismL : V → W .Theorem 15 V is isomorphic to W iff dim(V ) = dim(W ).proof:This is true in general, but we only proof it in case the spaces have finitedimension.Suppose that V is isomorphic to W . Let v1, . . . , vnbe a basis for V anddefine w1, . . . , wnby L(vi) = wifor i = 1, . . . , n. By exercise 12 w1, . . . , wnare linearly independent (since L is one-to-one) and by exercise 13 they spanW since L is onto.For the converse suppose that dim(V ) = dim(W ) and let v1, . . . , vnbe abasis for V and w1, . . . , wnbe a basis for W . Given any v ∈ V there exists aunique sequence of scalars a1, . . . , ansuch thatv = a1v1+ · · · + anvnThey exists because v1, . . . , vnspan V and they are unique becausev = a1v1+ · · · + anvnand v = b1v1+ · · · + bnvnimpliesz = (a1− b1)v1+ · · · + (an− bn)vnand so by linear independence a1= b1, . . . , an= bn. Now define L : V → WbyL(a1v1+ · · · + anvn) = a1w1+ · · · + anwnNow we check that L is linear. Suppose v = a1v1+ · · · + anvnandw = b1v1+ …