Math 115AH HW 4 Selected Solutions

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1 Math 115AH HW 4 Selected Solutions Yehonatan Sella Which of the following functions T from R 2 to R 2 are linear transformations? a) T (x 1, x 2 ) = (1 + x 1, x 2 ). This is not a linear transformation. For example, T (, ) = (1, ). But a linear transformation must always map the vector to the zero vector. b) T (x 1, x 2 ) = (x 2, x 1 ). This is a linear transformation. Indeed, suppose we are given (x 1, x 2 ) R 2 and (x 1, x 2) R 2. Then T ((x 1, x 2 ) + (x 1, x 2)) = T (x 1 + x 1, x 2 + x 2) = (x 2 + x 2, x 1 + x 1) = (x 2, x 1 ) + (x 2, x 1) = T (x 1, x 2 ) + T (x 1, x 2) Also, for any scalar c R and vector (x 1, x 2 ) R 2, we have T (c(x 1, x 2 )) = T (cx 1, cx 2 ) = (cx 2, cx 1 ) = c(x 2, x 1 ) = ct (x 1, x 2 ) Let V be the space of all polynomial functions from F to F, and let D : V V be the formal differentiation operator, as in Example 2. Then the null space of D consists exactly of the constant functions, since the derivative of a polynomial is if and only if it is the constant polynomial. On the other hand, the range of D is all of V, since every polynomial has an antiderivative. Now let V be the space of all continuous functions from R to R and let T : V V be integration, given by (T (f))(x) = x f(x)dx, as in Example 5. We know by the fundamental theorem of calculus that d T (f) = f. So whenever T (f) = it follows that f = d T (f) equals as well. Thus ker(t ) = {}. dx dx Consider the subspace W V consisting of all functions g : R R which are differentiable with continuous derivative, and such that g() =. We claim that the range of T is W. 1

2 Indeed, we know by the fundamental theorem of calculus that, for any f V, T (f) is differentiable with derivative f, which is continuous since it belongs to V. Moreover, (T (f))() = f(x) =. Thus the range of T is contained in W. On the other hand, let g W. Then by definition of W, g is differentiable and f := g (x) is continuous. So f V. We now note that, again by the fundamental theorem of calculus, (T (f))(x) = x f(x) = g(x) g() = g(x), where we use the fact that g() = since g W. This shows that W is contained in the range of T. So W equals the range of T Is there a linear transformation T : R 3 R 2 such that T (1, 1, 1) = (1, ) and T (1, 1, 1) = (, 1)? Yes. Let v 1 = (1, 1, 1) and let v 2 = (1, 1, 1). Then v 1 and v 2 are linearly independent since they are two vectors neither of which is a multiple of the other. So v 1, v 2 can be extended to a basis of R 3, which we denote v 1, v 2, v 3. Let w 1 = (1, ), w 2 = (, 1) and let w 3 be any third fixed vector in R 2. By theorem 3.1.1, there is a linear map T : R 3 R 2 such that T (v i ) = w i for all i = 1, 2, 3. In particular, T (v 1 ) = w 1 and T (v 2 ) = w Let V be a vector space and T a linear transformation from V to V. Prove that the following two statements about T are equivalent: a) The intersection of the null space of T (a.k.a. ker(t )) and the range of T (a.k.a. im(t )) is the subspace. b) If T (T (α)) =, then T (α) =. Proof: We must show that a) implies b) and b) implies a). a) = b): Suppose ker(t ) im(t ) = {}. We want to show that if T (T (α)) =, then T (α) =. So let α V and suppose T (T (α)) =. We want to show that T (α) =. Indeed, since T (T (α)) =, it follows that T (α) ker(t ). But also, by definition of image, T (α) im(t ). Thus T (α) ker(t ) im(t ), which by our assumption is {}. Thus T (α) =, as desired. b) = a): Suppose T (T (α)) = implies T (α) =. We want to show ker(t ) im(t ) = {}. Of course, {} ker(t ) im(t ), since kernel and image contain. So it suffices to show ker(t ) im(t ) {}. So let v ker(t ) im(t ). We want to show v =. Indeed, since v im(t ), we can write v = T (w) for some w V. On the other hand, since v ker(t ), we see that T (T (w)) = T (v) =. But by our assumption, 2

3 this implies T (w) =. So v = T (w) =, as desired Let T be the linear operator on R 3 defined by x 1 3x 1 T x 2 x 3 = x 1 x 2 2x 1 + x 2 + x 3 Is T invertible? If so, find a rule which defines T 1 like the one which defines T. We do this with the aid of matrices. The linear operator T can be rephrased x 1 3 x 1 using matrices as follows: T x 2 = 1 1 x 2 x x 3 3 So, letting A = 1 1, we see that T (v) = Av for all vectors v R To show that T is invertible, it suffices to show that the matrix A is invertible. Indeed, in that case the linear transformation given by S(v) = A 1 v will be inverse to T, since T (S(v)) = AA 1 v = Iv = v, and S(T (v)) = A 1 Av = Iv = v. So we just need to find A 1, which we do using row-reduction. I will skip the steps We generalize the claim. Let T : V W be a linear transformation between vector spaces such that dim(v ) > dim(w ), and let U : W V be another linear transformation. Then the linear transformation U T cannot possibly be invertible. Proof: Note that im(ut ) im(u). Indeed, any vector in im(ut ) is of the form UT (v) = U(T (v)) for some v V, so it is in the image of U. Thus On the other hand, by rank-nullity, rank(ut ) rank(u) rank(u) + nullity(u) = dim(w ) So in particular rank(u) dim(w ). Combining this with the above inequality and our asumption that dim(w ) < dim(v ), we see that rank(ut ) < dim(v ) But UT is a map from V to V. Since its rank is strictly smaller than the dimension of V, it cannot possibly be surjective. But invertible maps are always surjective, 3

4 so UT is not invertible We first prove a lemma about functions (not necessarily linear transformations between vector spaces). Lemma: Let X and Y be sets. Let f : X Y be a function and let g : Y X be a function which is a right-inverse of f, that is, f g(y) = y for all y y. Then a) f is a surjection. b) If f is in addition an injection, then g is also a left-inverse of f, that is, g f(x) = x for all x X. Proof: a) Let y Y. Then y = f g(y) = f(g(y)), so indeed y is in the image of f. b) Suppose f is an injection. We want to show that g f(x) = x for all x X. Indeed, since g is a right-inverse of f, we have f(x) = f g(f(x)) = f(g f(x)) Since f is an injection, we conclude from the above that x = g f(x) for all x X, as desired. We now use the lemma in our situation: Let T : V V be a linear operator on the finite-dimensional vector space V. Suppose there is a linear operator U : V V such that T U = I. Prove that T is invertible and U = T 1. Proof: Since U is a right-inverse of T, it follows from part a) of the lemma that T is a surjection. So rank(t ) = dim(v ). But by rank-nullity, we have rank(t ) + nullity(t ) = dim(v ) So it follows nullity(t ) =, so ker(t ) = {}, from which it follows that T is injective. By part b) of the lemma, it follows that U is also a left-inverse of T. Since U is both a right-inverse and left-inverse of T, then by definition T is invertible with inverse U. This property fails for infinite-dimensional spaces. For example, consider the differentiation operator D on the space V of all real polynomials. Let U : V V denote integration (U(p))(x) = x p(t)dt. Then DU is the identity on V. However UD is not the identity on V. For example, for any constant function c, we have UD(c) = U() = c. 4

5 Let V be a finite-dimensional vector space and let T be a linear operator on V. Suppose that rank(t 2 ) = rank(t ). Prove that im(t ) ker(t ) =. Proof: What we want to prove is exactly condition a) in exercise Since we showed conditions a) and b) are equivalent in exercise , it suffices to prove condition b): if T (T (α)) =, then T (α) =. Rephrased, this is saying that ker(t 2 ) ker(t ). Note we always have ker(t ) ker(t 2 ), so what we really want to prove is ker(t 2 ) = ker(t ). By rank-nullity, nullity(t ) = dim(v ) rank(t ) nullity(t 2 ) = dim(v ) rank(t 2 ) But by assumption rank(t ) = rank(t 2 ), so the two right-hand sides above are equal. Therefore the left-hand sides are equal, so nullity(t ) = nullity(t 2 ). In other words, ker(t ) and ker(t 2 ) have the same dimension. But since ker(t ) is a subspace of ker(t 2 ) of the same dimension, it must be equal to it. So indeed ker(t ) = ker(t 2 ), as desired. Extra questions: Let g : X Y and f : Y Z be functions. (1) If f g is surjective, then g is surjective. Not necessarily. Let X = {1}, Y = {1, 2}, Z = {1}. Define g : X Y by g(1) = 1. Define f : Y Z by f(1) = 1, f(2) = 1. Then f g : X Z is surjective since 1, the only element of Z, is in the image of f g. Indeed, f g(1) = f(g(1)) = f(1) = 1. However, g is not surjective since 2 Y is not in the image of g. (4) If f g is injective, then g is injective. This is true. Let x 1, x 2 X and suppose g(x 1 ) = g(x 2 ). Applying f to both sides, we get f g(x 1 ) = f g(x 2 ). But since f g is injective, it follows that x 1 = x 2. Since g(x 1 ) = g(x 2 ) implies x 1 = x 2, it follows that g is injective. (1) Let S = N be the set of natural numbers. Let T S be the set of positive natural numbers. T is a proper subset of S since is in S but not in T. We can define a map f : S T by f(n) = n + 1. This map is surjective: for any number m T, we know m > so m 1 N and m = (m 1) + 1 = f(m 1). Also f is injective, since if f(n) = f(n ) for n, n N, then n + 1 = n + 1, so n = n. Thus f is a bijection. 5

6 Footnote: this example is the setting of the story of Hilbert s Hotel. There is a hotel with infinitely many rooms. The hotel is one long corridor that stretches forever, with room numbers 1, 2, 3, and on and on. The sign outside says there are no vacancies. But a very special guest comes into the hotel and insists on getting a room. The receptionist figures out a solution: he tells every guest in the hotel to leave their room and move to the room a number up from theirs (so the guest in room 1 moves to room 2, the guest in room 2 moves to room 3, and so on). This leaves room 1 vacant, and the satisfied guest moves in there. 6