Math 4153 Exam 1 Review

Transcription

1 The syllabus for Exam 1 is Chapters 1 3 in Axler. 1. You should be sure to know precise definition of the terms we have used, and you should know precise statements (including all relevant hypotheses) for the main theorems proved. Know how to do all of the homework problems. 2. Outline of subjects for Exam 1: Vector spaces Subspaces Sums and direct sums of subspaces Linearly independent and spanning lists Bases and dimension Linear maps; nullspace, range, and invertibility Matrices of linear maps 3. Definitions: The complex numbers C (in terms of real numbers R) A vector space F n, where F = R or F = C. P n (F) and P(F) Mat(m, n, F) A subspace U V A sum of subspaces U + W A direct sum of subspaces V = U W A linearly independent list A spanning list A basis A finite-dimensional vector space V The dimension dim V of a finite-dimensional vector space A linear map T : V W The vector space L(V, W ) of all linear maps from V to W Injectivity, surjectivity, and invertibility of linear maps Nullspace and range of linear maps Isomorphisms and inverses of linear maps 4. Computations involving subspaces and direct sums: 1 1

2 Recognize whether something is a subspace or not. Compute intersections and sums of subspaces. Are unions of subspaces a subspace? Recognize whether a sum of subspaces is direct or not. 5. Computations involving linearly independent and spanning lists: Determine if a list is linearly independent, a spanning list, or a basis. How to extend linearly independent lists to bases and reduce spanning lists to bases Show that if U V and V is finite dimensional, then U = {0} if and only if dim U = 0 and U = V if and only if dim U = dim V. Otherwise 0 < dim U < dim V. 6. Linear maps: What does a linear transformation from F n F m look like? Know how to compute the matrix of a linear map. 7. Major results. Know the statements, but don t memorize them word for word know what they say mathematically. Especially important ones are in bold. How to prove properties of vector spaces: Propositions : don t memorize any proofs or statements! Just know how to do something like this and be able to know if a given statement is true or false. Propositions 1.8 and 1.9. Theorem 2.6, and how it implies Theorem Theorem 2.12 and how it implies Theorem 2.13 Maximal linearly independent lists span Minimal spanning lists are linearly independent Theorem 2,18, and Proposition Theorem 3.4. How Theorem 3.4 implies corollaries 3.5 and 3.6, and results such as Exercises 9 and 10 (Chapter 3). Theorem 3.18 Theorem 3.21 Generalizing Theorem 3.21: Let V and W be finite-dimensional of the same dimension. then T L(V, W ) is invertible if and only if it is injective if and only if it is surjective. 2 2

3 Review Exercises 1. Determine if each of the following statements is true or false. (a) If the list (v 1,..., v m ) contains the vector 0, then the list is linearly dependent. Solution. True. If v i = 0 then a nontrivial linear dependence relation is given by 0 v v i v n = 0. (b) If the list (v 1,..., v 6 ) is linearly independent, then the list (v 1, v 3, v 5 ) is linearly independent. Solution. True Any linear dependence relation on the sublist is a linear dependence relation on the larger list by taking a 2 = a 4 = a 6 = 0. (c) If the list (v 1,..., v 6 ) is linearly dependent, then the list (v 1, v 3, v 5 ) is linearly dependent. Solution. False A counterexample is the list (e 1, 0, e 2, 0, e 3, 0) in F 3 where (e 1, e 2, e 3 ) is the standard basis of F 3. (d) If the list (v 1, v 2, v 3 ) is linearly dependent, then v i is a scalar multiple of v j for some i and j with i = j. Solution. False A counterexample is ((1, 0), (0, 1), (1, 1)). 2. Let U and V be subspaces of R 405 of dimensions 400 and 300 respectively. (a) List all possibilities for dim(u V ). Solution. dim(u V ) = dim U +dim V dim(u +V ) = dim(u +V ). Since U + V contains both U and V and is a subset of R 405, we have that 400 dim(u + V ) 405. Thus, the possible values for dim(u + V ) are 400, 401, 402, 403, 404, and 405, which then give the following values for dim(u V ): 300, 299, 298, 297, 296, and 295. (b) Is it possible to choose U and V so that every element of R 405 can be written uniquely as u + v with u U and v V? (Naturally, reasons are expected.) Solution. This is not possible. If every element of R 405 can be written uniquely as u + v with u U and v V, then this means that R 405 = U V, which is true if and only if U + V = R 405 and U V = {0}. The second condition cannot be true since, from part (a), dim(u V ) > (a) Define a subspace of a vector space. 3 3

4 Solution. A subset U of a vector space V is a subspace if U is a vector space using the vector addition and scalar multiplication of V. (b) Suppose that V is a finite dimensional vector space and that U V is a subspace such that dim U = dim V. Prove carefully that U = V. Solution. Let (u 1,..., u m ) be a basis of U. Thus m = dim U = dim V. Since (u 1,..., u m ) is a basis of U, it is, in particular, a linearly independent list of vectors in U V. Since m = dim V, Proposition 2.17 shows that this linearly independent list is in fact a basis of V. In particular, span(u 1,..., u m ) = V. But (u 1,..., u m ) is a basis of U so we also have U = span(u 1,..., u m ). Therefore, U = V. 4. Let T : F 4 F 7 be a linear map. Show that range(t ) cannot be the subspace U = { (x 1,..., x 7 ) F 7 : x x 7 = 0 }. Solution. Note that U = null(s) where S : F 7 F is the linear map S(x 1,..., x 7 ) = x x 7. Since range(s) = F it follows that dim U = dim null(s) = dim F 7 dim F = 7 1 = 6. However, 4 = dim F 4 = dim null(t ) + dim range(t ) so that dim range(t ) 4 < 6 = dim U. Therefore, it is not possible for range(t ) = U. 5. If (u, v, w) is a linearly independent list, show that (u v, v w, w + v) is also a linearly independent list. Solution. Suppose there is a linear dependence relation a 1 (u v) + a 2 (v w) + a 3 (w + v) = 0. Rewrite this as a linear dependence relation on the list (u, v, w): a 1 u + (a 2 a 1 + a 3 )v + (a 3 a 2 )w = 0. Since the list (u, v, w) is linearly independent, this means that the coefficients of u, v, and w must all be 0. That is, a 1 = 0 a 2 a 1 + a 3 = 0 a 3 a 2 = 0. Solving this system of equations gives a 1 = a 2 = a 3 = 0, which implies that the list (u v, v w, w + v) is linearly independent. 6. (a) Find a basis for the subspace U = { (y 1, y 2, y 3, y 4 ) F 4 : y 1 = y 2 and y 3 = y 4 }, and prove that the list you give is in fact a basis of U. 4 4

5 Solution. If u U, then u = (y 1, y 2, y 3, y 4 ) = (y 1, y 1, y 4, y 4 ) = y 1 (1, 1, 0, 0) + y 4 (0, 0, 1, 1). This shows that U = span((1, 1, 0, 0), (0, 0, 1, 1)) and since neither of two vectors in this list is a scalar multiple of the other, it follows that the list is linearly independent and hence a basis. (b) Suppose that W is another subspace of F 4 such that U + W = F 4. What can you say about dim W? Solution. Since 4 = dim F 4 = dim(u +W ) = dim U +dim W dim(u W ) = 2 + dim W dim(u W ), it follows that dim W = 2 + dim(u W ) 2. (c) Suppose instead that U W = F 4. Now what can you say about dim W? Solution. In this case U W = {0} so we must have dim W = 2 + dim(u W ) = 2 7. Let T : V W be a linear map. Suppose that (v 1,..., v n ) is a linearly independent list in V and T is injective. Show that the list (T v 1,..., T v n ) is linearly independent. Solution. Suppose that there is a linear dependence relation a 1 T v a n T v n = 0. Since T is a linear map, this implies that we have T (a 1 v a n v n ) = 0. But T is assumed to be injective, so this means that a 1 v a n v n = 0, and since (v 1,..., v n ) is a linearly independent list in V, this implies that a 1 = = a n = 0 and we conclude that the list (T v 1,..., T v n ) is linearly independent. 8. (a) Show that a linear map T : V W is injective if and only if null(t ) = {0}. Solution. This is Proposition 3.2, page 43. (b) Define T : P 3 (F) P 3 (F) by T (p) = (z 2 + z)p where p denotes the second derivative of p. Describe null(t ) and range(t ). What are their dimensions. 5 5

6 Solution. Since T (p) = (z 2 + z)p and z 2 + z = 0 it follows that p null(t ) if and only if p = 0. But the second derivative of a polynomial is 0 if and only if the polynomial has the form az + b. That is, it is a polynomial of degree at most 1. Thus null(t ) = P 1 (F) P 3 (F). Hence, dim null(t ) = dim P 1 (F) = 2 and thus, dim range(t ) = 4 2 = 2. To describe the range of T, note that the range of the second derivative map is just P 1 (F), that is, the polynomials of degree at most 1. Thus, the range of T consists of all polynomials of degree at most 3 of the form (z 2 + z)(az + b) = az 3 + (a + b)z 2 + bz. That is, a polynomial of degree at most 3 is in the range of T provided that there is no constant term and the coefficient of z 2 is the sum of the coefficients of z and z 3. (c) Find the matrix M(T ) that represents T with respect to the standard basis (1, z, z 2, z 3 ) of P 3 (F). Solution. T (1) = 0, T (z) = 0, T (z 2 ) = 2z 2 + 2z, and T (z 3 ) = 6z 3 + 6z 2. Therefore the matrix of T with respect to the standard basis is M(T ) =