Exercise Sheet 8 Linear Algebra I

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1 Fachbereich Mathematik Martin Otto Achim Blumensath Nicole Nowak Pavol Safarik Winter Term 2008/2009 (E8.1) [Morphisms] Exercise Sheet 8 Linear Algebra I Let V be a finite dimensional F-vector space and ϕ : V V a linear map. Show the equivalence of the following: a) ϕ is injective. b) ϕ is surjective. c) ϕ is bijective. (*) Can you give examples of linear mappings from an infinite dimensional vector space to itself that are injective and not surjective, or surjective and not injective? Let B = {b 1,b 2,...,b n } be a basis of V. Notice that it is sufficient to show the equivalence of a) and b), since c) is simply their conjunction. a) b) If ϕ is injective, the set ϕ(b) = {ϕ(b 1 ), ϕ(b 2 ),...,ϕ(b n )} consists of n distinct vectors. It spans image(ϕ) (Lemma 3.1.9) and is linearly independent: 0 = i λ iϕ(b i ) implies that ϕ( i λ ib i ) = 0, and as ker(ϕ) = {0} by injectivity (Lemma 3.1.7), i λ ib i = 0, which implies that the linear combination is trivial. So image(ϕ) has dimension n, hence is equal to V. So ϕ is surjective. b) a) Assume that ϕ is surjective. The set ϕ(b) contains at most n vectors, but at the same time it spans the entire space V, which has dimension n. Therefore, the set ϕ(b) must consist of n linearly independent vectors. To show that ϕ is injective (or, equivalently, that ker(ϕ) = {0}) consider v = n i=1 λ ib i such that ϕ(v) = 0. From ϕ(v) = ϕ( n i=1 λ ib i ) = n i=1 λ iϕ(b i ) = 0 and from the linear independence of ϕ(b) it follows that all λ i are 0 and v = 0. So ϕ is injective. Remark: Using the dimension formula for linear maps (to be covered later) one directly gets this equivalence. (*) For example: see (E8.3). The function that increases any exponent in a polynimial by 1 is injctive but not surjective. The constant functions are not reached. (E8.2) [Images and preimages of affine subspaces] Let F be a field. (i) Consider a linear map ϕ : F n F n. Show that the image under ϕ of a 1-dimensional affine subspace (i.e., a line) is either a point or again a 1-dimensional affine subspace. Discuss the same problem for the image of a 2-dimensional affine subspace (i.e., a plane). Does it make a difference if ϕ is assumed to have an inverse?

2 (ii) Let V,W F-vector spaces and ϕ : V W a linear map. Prove that if B is an affine subspace of W, its preimage ϕ 1 (B)(= {x V : ϕ(x) B}) is an affine subspace of V or empty. (i) The image under ϕ of an affine subspace S = v + U is ϕ(s) = ϕ(v) + ϕ(u). If B is a basis for U, then ϕ(b) spans ϕ(u), so dim(ϕ(u)) dim(u). Therefore dim(ϕ(s)) dim(s). If the dimension of S is 1, the dimension of ϕ(s) is either 0 or 1 (a point or a line, respectively). If the dimension of S is 2, the dimension of ϕ(s) can be any one of 0, 1 or 2. Of course, when ϕ is invertible it is an isomorphisms. Then U and ϕ(u) are also isomorphic and have the same dimension. (ii) Suppose that B = b + U, for a linear subspace U of W, is an affine subspace of W and that ϕ 1 (B) is non-empty. That means there are a V and u U such that ϕ(a) = b + u. We prove that ϕ 1 (B) = a + ϕ 1 (U). First we prove ϕ 1 (U) is a linear subspace: Since ϕ is a linear map and U ist a linear subspace, we have 0 ϕ 1 (U). Further, by linearity of ϕ we obtain closure of ϕ 1 (U) under vector addition and scalar multiplication. : Let x ϕ 1 (B). Then there is a u U with ϕ(x) = b + u. For v := x a we get ϕ(v) = ϕ(x) ϕ(a) = b + u (b + u) = u u U. This shows that v ϕ 1 (U) and therefore x = a + v a + ϕ 1 (U). : If x a + ϕ 1 (U) then there is a v ϕ 1 (U) with x = a + v. Now ϕ(x) = ϕ(a) + ϕ(v) b + U follows. Therefore x ϕ 1 (B). We may now conclude: Since ϕ 1 (U) is a linear subspace the claim then follows. (E8.3) [Derivative of Polynomials] Consider the map ϕ : Pol(R) Pol(R), p p, where p is the derivative of the polynomial function p. [For p = n i=0 a ix i we have p = n i=1 ia ix i 1.] From the tutorial (T8.3) we know that this is a linear map. (i) What is the kernel of ϕ? (ii) What is the image of ϕ? (iii) Is ϕ injective? Is it surjective? (i) The kernel of ϕ is the subset in Pol(R) formed by the functions p with derivative p = 0 (the constant function with value 0). These are exactly the constant (polynomial) functions.

3 (ii) For p = n i=0 a ix i Pol(R) we have that ϕ ( n a i i=0 xi+1) = p (of course one may i+1 add a constant term). So Im(ϕ) = Pol(R). (iii) The function ϕ is not injective, since its kernel is not {0}, by (i). However ϕ is surjective due to (ii). (E8.4) [Direct Products and Sums] Let F be a field, V an F-vector space and U, W V subspaces of V, with V = U W. (i) Show that ϕ : V V/U v v + U is a linear map. (ii) Determine ker(ϕ). (iii) Show that ϕ : V V/U V/W v (v + U,v + W) is an isomorphism. (i) Since ϕ(v 1 + v 2 ) = (v 1 + v 2 ) + U = (v 1 + U) + (v 2 + U) = ϕ(v 1 ) + ϕ(v 2 ) and ϕ(λv 1 ) = (λv 1 ) + U = λ(v 1 + U) = λϕ(v 1 ) for all v 1,v 2 V, λ F holds, we have that ϕ is linear. (ii) Since 0 + U = U the equivalence class [0] = U in the quotient space. By definition ker(ϕ) = {v V : ϕ(v) = 0 = U}. Thus ker(ϕ) = U, since v +U = U is equivalent to v U. (iii) In order to show that ϕ : V V/U V/W, v (v + U,v + W) is an isomorphism, we still have to show that ϕ is injective and surjective. That ϕ is linear follows since ϕ(v) = (ϕ 1 (v), ϕ 2 (v)) for linear maps ϕ 1 : V V/U and ϕ 2 : V V/W according to (i). Since ker(ϕ) = {v V : (v + U,v + W) = (U, W)} we get ker(ϕ) U W = {0}. Let now (v 1 +U,v 2 +W) (V/U, V/W). Since V = U W we find u i U,w i W such that v i = u i +w i for i = 1, 2. Thus (v 1 + U,v 2 + W) = (w 1 + U,u 2 + W) and ϕ(w 1 + u 2 ) = (v 1 + U,v 2 + W). This gives us surjectivity of ϕ. Note that, in the finite-dimensional case, we could also reason with injectivity and equality of dimensions: dim(v/u) = dim(v ) dim(u), dim(v/w) = dim(v ) dim(w) and dim(u W) = dim(u) + dim(w) together imply that dim(v ) = dim(v/u V/W). (E8.5) [Bonus Problem: Direct Products] Consider a (possibly infinite) family (V i ) i I of F-vector spaces. The direct product of this family is the F-vector space V whose vectors are all functions (families)

4 f : I i i I V i f(i) with f(i) V i for all i I, equipped with point-wise vector addition and scalar multiplication in the natural manner. (i) Verify that V as defined above is an F-vector space, and that in the special case that I = {1, 2} it is isomorphic with the direct product V 1 V 2 as defined in the lecture. (ii) Show that for each i I, we have V i ˆV i, where ˆV i is the subspace of V consisting of those f V for which f(j) = 0 V j for all j i. (iii) How/when does the direct product V differ from the direct sum of all its subspaces ˆV i, which is naturally defined as the span of the union of these ˆV i, span( i I ˆV i )? (i) First one has to show that the given operations on V are well defined. Let f, g V. It holds that (f + g)(i) = f(i) + g(i) V i, as V i is a vector space. Take f V and λ K. Since V i is a vector space (λf)(i) = λ f(i) V i. The vector space inherits commutativity and the distiributivity laws from the field F. With these facts one can prove all vector space axioms. Take I = {1, 2}. We want to show V V 1 V 2. Consider the following function ϕ: ϕ : V V 1 V 2 f (f(1), f(2)) The definition of pointwise addition and scalar multiplication yields linearity of ϕ. Surjectivity: For every element (v 1, v 2 ) V 1 V 2 we can define the function f with f(1) = v 1 and f(2) = v 2. This function we have ϕ(f) = (v 1, v 2 ). Injectivity: ϕ(f) = ϕ(g) (f(1), f(2)) = (g(1), g(2)) (f(1) = g(1)) and (f(2) = g(2)) f = g. Hence, ϕ is an isomorphism. (ii) For a fixed i I we consider the function ψ: ψ : V i ˆV i v f v where f v is zero everywhere except at the point i where we define f v (i) := v. Linearity follows from the fact, that the function f v+w can be written as f v + f w. Surjectivity: Since for every function f in ˆV i it holds that f(i) V i, any function is reached by ψ. Injectivity: ψ(v) = ψ(w) f v = f w f v (i) = f w (i) v = w. Hence, ψ is an isomorphism. (iii) A difference only occurs in the infinite dimensional case. Consider the vector space of real sequences F(N, R). Take B = {u i : i N}, where u i is the sequence, that is zero everywhere except at the i-th entry where it has value 1. In this case the direct product V is isomorphic to F(N, R) and the subspaces ˆV i = span(u i ). Considering

5 the direct sum of the ˆV i we obtain U := span(b) F(N, R). Hence, as discussed in example 2.5.3, one has to keep in mind that the span only contains finite linear combinations of vectors in B. Therefore, U contains only sequences that are nonzero at finitely many points. This yields U V. In this example the direct sum is a much smaller set than the direct product. Every Christmas comes with an advent calender Even in the math building you can find many doors that you can open on the right day. Behind them cookies can be found. (Attention: The physicists do not yet occupy the whole first floor.)