2.2. Show that U 0 is a vector space. For each α 0 in F, show by example that U α does not satisfy closure.

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1 Hints for Exercises 1.3. This diagram says that f α = β g. I will prove f injective g injective. You should show g injective f injective. Assume f is injective. Now suppose g(x) = g(y) for some x, y A. Then β(g(x)) = β(g(y)). Therefore f(α(x)) = f(α(y). Since f is injective, this implies α(x) = α(y). Since α is injective, this implies x = y. Therefore g is injective Show that U 0 is a vector space. For each α 0 in F, show by example that U α does not satisfy closure Use proof by contradiction: Let U 1 and U 2 be subspaces of V such that U 1 U 2 = V, but suppose U 1 V and U 2 V. First show that U 1 = U 2 is impossible. Second, argue that one can find x U 1 \U 2 and one can find y U 2 \U 1. Third, argue that x + y must belong to U 1 or U 2. Fourth, argue that either outcome yields a contradiction using the closure properties of U 1 and U 2. Give brief but accurate reasons for each of the four steps outlined above. 2. Let the elements of the finite field be α 1, α 2, α 3,..., α n, where α 1 = 0 and α 2 = 1. Let i 3. You must show that α i has a multiplicative inverse, i.e. that α i α j = 1 for some j. Complete these steps: (a) Prove that the elements α i α 1, α i α 2,..., α i α n are distinct (no two on this list are the same). (b) Given (a), prove that every element in the finite field is contained in the list α i α 1, α i α 2,..., α i α n. (c) Given (b), prove that α i α j = 1 for some j Let {u 1, u 2,..., u p } be a basis for U 1. Expand this to a basis {u 1, u 2,..., u p, u p+1,..., y n } for V. Let U 2 = L F (u p+1,..., u n ). Prove that U 2 is complementary to U 1. 1

2 3. To avoid dealing with trivial cases, assume n 1. Let V be a vector space over C and let v 1, v 2..., v n V. The book describes two vector spaces over R, W r = L R (v 1, v 2,..., v n ) and W c = L C (v 1, v 2,..., v n ), and we wish to compare dim R (W r ) to dim R (W c ). In order to do this we must compare the two sets W r and W c. Complete the steps below. (a) Prove that W c = L R (v 1, iv 1, v 2, iv 2,..., v n, iv n ). Be sure to prove both subset inclusions. (b) Prove that if the vectors v 1, v 2,..., v n are linearly independent over C, then the vectors v 1, iv 1, v 2, iv 2,..., v n, iv n are linearly independent over R. (c) Given (a) and (b), prove that 0 dim R (W r ) n and 0 dim R (W c ) 2n. (d) Let n = 3 and V = C 2. Find v 1, v 2, v 3 V such that W r has dimension 3 over R and W c has dimension 4 over R. Make sure to specify the coordinates of v 1, v 2, and v 3. Hint: let v 1 and v 2 be two vectors which are linearly independent over C, and let v 3 L C (v 1, v 2 ). Make the simplest possible choices. 3.3P. (a) Explain geometrically why the shortest distance between a line g and a point p is along a path from p to g which is perpendicular to g. A diagram may help. (b) Given (a), explain why the shortest distance between two non-intersecting lines g 1 and g 2 in R 3 is along a path which is perpendicular to both. (c) Given the previous two statements, explain how solving the system of three equations in the three unknowns t 1, t 2, t 3, (p 1 + t 1 v 1 ) (p 2 + t 2 v 2 ) = t 3 (v 1 v 2 ), leads to a method of solution, then do the math and find t 1, t 2, t 3, and the shortest distance between the two lines g 1 and g 2. (Note: there is a hard way and an easy way to solve for t 1, t 2, t 3. Hard way: three equations in three unknowns. Easy way: compute the inner product with v 1 to obtain 2

3 one equation in t 1, t 2, compute the inner product with v 2 to obtain a second equation in t 1, t 2. Use these to solve for t 1, t 2, then find t 3.) 4.1. I will prove that f is injective (f(v 1 ), f(v 2 ),..., f(v n )) is linearly independent. You must prove the converse. Assume that f is injective. To prove that (f(v 1 ), f(v 2 ),..., f(v n )) is linearly independent, suppose α 1 f(v 1 ) + α 2 f(v 2 ) + + α n f(v n ) = 0 W. Since f is a linear map, we must have f(α 1 v 1 + α 2 v α n v n ) = 0 W. Since f is an injective linear map, we must have α 1 v 1 + α 2 v α n v n = 0 V. Since (v 1,..., v n ) is a basis of V, we must have α 1 = α 2 = = α n = 0 F. Hence (f(v 1 ), f(v 2 ),..., f(v n )) is linearly independent We know that V i /Ker f i = Im fi. Therefore which rearranges to dim V i dim Ker f i = dim Im f i, dim V i = dim Ker f i + dim Im f i. Now replace each dim V i by dim Ker f i +dim Im f i in n i=0 ( 1)i dim V i and explain how this can be transformed into n i=0 ( 1)i dim H i (C). 4. We must show that φ 2 is injective and surjective. I will prove that φ 2 is injective by a process called diagram-chasing. You should show that φ 2 is surjective by a similar process. To show that φ 2 is injective it suffices to show that φ 2 (v 2 ) = 0 implies v 2 = 0. Suppose that φ 2 (v 2 ) = 0. Then g 2 (φ 2 (v 2 )) = 0, therefore φ 1 (f 2 (v 2 )) = 3

4 0. Since φ 1 is an isomorphism, this implies that f 2 (v 2 ) = 0. Therefore v 2 Ker f 2, therefore v 2 Im f 3, therefore v 2 = f 3 (v 3 ) for some v 3 V 3. Therefore φ 2 (f 3 (v 3 )) = 0, therefore g 3 (φ 3 (v 3 )) = 0, therefore φ 3 (v 3 ) Ker g 3, therefore φ 3 (v 3 ) = Im g 4, therefore φ 3 (v 3 ) = g 4 (w 4 ) for some w 4 W 4. Since φ 4 is surjective, we must have w 4 = φ 4 (v 4 ) for some v 4 V 4. Therefore φ 3 (v 3 ) = g 4 (φ 4 (v 4 )). Therefore φ 3 (v 3 ) = φ 3 (f 4 (v 4 )). Since φ 3 is an isomorphism, this implies that v 3 = f 4 (v 4 ). Therefore v 2 = f 3 (f 4 (v 4 )). The hypothesis that the top row of the diagram is exact implies that f 3 f 4 = 0, therefore v 2 = 0, as desired. Here is how to start the proof that φ 2 is surjective: Let w 2 W 2 be given. We must show there exists v 2 V 2 such that w 2 = φ 2. We have g 2 (w 2 ) W 1. Since φ 1 is an isomorphism, g 2 (w 2 ) = φ 2 (v 3 ) for some v 3 V 3. Since the bottom row of the diagram is exact, g 1 (g 2 (w 2 )) = 0. Therefore g 1 (φ 2 (v 3 )) = 0. Now proceed as above. 4.2P. (a) Prove that f(x), f(y) = x, y for all x, y V implies f(x) = x for all x V. (b) Prove that f(x) = x for all x V implies f(x), f(y) = x, y for all x, y V. Method: assuming that f(x) = x for all x V, we have f(x), f(x) = x, x for all x V. Use this fact and the equation 2 f(x), f(y) = f(x) + f(y), f(x) + f(y) f(x), f(x) f(y), f(y) and the fact that f is a linear map to prove that f(x), f(y) = x, y for all x, y V To prove that rk A+rk n rk AB, work through the following steps: (a) Let f : BV ABV be the linear map defined by f(bv) = ABv. Using the dimension formula for linear transformations, explain why rk B dim Ker f = rk AB. (b) Explain why dim Ker f nullity A. Once proved, this implies rk B nullity A rk AB. 4

5 (c) Using (b), explain why rk A + rk B n rk AB. To prove that rk AB min(rk A, rk B), work through the following steps: (d) The column space of AB is a subspace of the column space of A (prove this), therefore the rank of AB is the rank of A (explain). (e) Let b 1, b 2,..., b k be a basis for the column space of B. Then Ab 1, Ab 2,..., Ab k spans the column space of AB (prove this), therefore the dimension of the column space of AB is the rank of B (explain). Putting together (d) and (e), we are done You want to show α 1 w 1 + α 2 w 2 + α 3 w 3 = 0 V implies α 1 = α 2 = α 3 = 0. Follow these steps: (a) Explain why α 1 w 1 + α 2 w 2 + α 3 w 3 = 0 V α 2 α 3 = 0 α 1 + 2α 2 + α 3 = 0 α 1 α 2 + α 3 = 0 2α 1 α 2 + α 3 = 0 implies the system of equations (b) Find all solutions to α 1, α 2, α Some background material on matrix representations of endomorphisms: Let V be a vector space and let f : V V be an endomorphism (linear map from V to itself). If (v 1,..., v n ) is a basis for V and if f(v i ) = a 1i v 1 + a 2i v a ni v n = q a qi v q for each i n, then we say that A = (a ij ) is the matrix which represents f with respect to the basis (v 1, v 2,..., v n ). If g is a second endomorphism of V and B = (b ij ) is the matrix which represents g with respect to the same basis (v 1,..., v n ), then we have f(g(v i )) = f(b 1i v 1 + b 2i v b ni v n ) = p b pi f(v p ) = 5

6 ( b pi p q a qp v q ) = q ( ) a qp b pi v q, which implies that the matrix AB represents the endomorphism f g with respect to the basis (v 1,..., v n ). In particular, if f is invertible and g = f 1, then we must have AB = I, which implies B = A 1. Now suppose that A represents f with respect to the basis (v 1,..., v n ) and that B represents f with respect to the basis (w 1,..., w n ). How are the matrices A and B related? Let g : V V be the linear transformation defined by g(v i ) = w i for i n. Let C be the matrix which represents g with respect to the basis (v 1,..., v n ). We have p f(w i ) = p b pi w p. Therefore f(g(v i )) = p b pi w p. Therefore g 1 (f(g(v i ))) = p b pi v p. Therefore B represents g 1 f g with respect to the basis (v 1,..., v n ). On the other hand, so does C 1 AC. Therefore we have C 1 AC = B. (a) Let f : R 2 R 2 be rotation clockwise through 45. Let (v 1, v 2 ) = ( 1, 0, 0, 1 ), (w 1, w 2 ) = ( 1, 1, 1, 1 ). Let A, B, C, g be defined as in the paragraph above. Find the coefficients of A, B, and C, then verify that C 1 AC = B. Note that the latter statement is equivalent to AC = CB. We are ready to tackle 5. We are told that f has the same matrix representation with respect to all bases for V. Let A be the matrix representation 6

7 with respect to one arbitrary basis (v 1,..., v n ). Let (w 1,..., w n ) be any other basis. Let C be the matrix of the linear map g : V V defined by g(v i ) = w i for i n. Then we know that C 1 AC = A. So this relationship is true of every invertible matrix C. Use this fact to prove that A = λi for a some real number λ. This is equivalent to saying that f(v) = λv for all v V. I will do the 2 2 case for F = R. You will do the 2 2 case for F = Z 2 and the general n n case for F = R. We will find it easier to work with the relation AC = CA for all invertible matrices C. Let ( ) a11 a A = 12. a 21 a 22 Let C = ( ) x 0 0 y for two arbitrary field elements x, y, being careful not to form a singular matrix. We have ( ) ( ) ( ) ( ) a11 a 12 x 0 x 0 a11 a = 12, a 21 a 22 0 y 0 y a 21 a 22 ( ) ( ) a11 x a 12 y a11 x a = 12 x. a 21 x a 22 y a 21 y a 22 y Choosing x = 1 and y = 2, this forces a 12 = a 21 = 0. Now let ( ) x y C =. 0 z We have ( ) ( ) ( ) ( ) a11 0 x y x y a11 0 =, 0 a 22 0 z 0 z 0 a 22 ( ) ( ) a11 x a 11 y a11 x a = 22 y. 0 a 22 z 0 a 22 z Choosing x = y = z = 1 we have we have a 11 = a 22. Therefore A = a 11 I. (b) Do the argument for 2 2 matrices when F = Z 2. 7

8 (c) Do the argument for 3 3 matrices when F = R. (d) Do the argument for general n n matrices when F = R. 5.3P. Think in terms of the determinant and use a trigonometric identity. 6. To orient R n is to choose a basis (v 1,..., v n ) for it. Given another basis (v 1,..., v n), one says that it is positively oriented or negatively oriented depending on whether the endomorphism f : R n R n defined by f(v i ) = v i has a positive or a negative determinant. Now let 1 k < n. To simultaneously orient all k-dimensional subspaces of R n is to choose a basis for each and to define orientation for each in the same way. Say that for each k-dimensional subspace W we choose the orientation (w 1, w 2,..., w k ). The problem is to show that there exist continuous functions v 1, v 2,..., v k from [0, 1] to R n such that the vectors f(t) = (v 1 (t),..., v k (t)) are linearly independent for each t [0, 1], the vectors (v 1 (0),..., v k (0)) are positively oriented, but the vectors (v 1 (1),..., v k (1)) are negatively oriented. Complete the following steps: (a) Let (a 1,..., a k ) be a positively oriented basis for a k-dimensional subspace of R n. Prove that there exists a vector b R n which is not in the span of (a 1,..., a k ). (b) Let v 1 (t) = a 1, v 2 (t) = a 2,..., v k 1 (t) = a k 1, v k (t) = (1 t)a k + tb. Prove that the list of vectors (v 1 (t), v 2 (t),... v k (t)) is linearly independent for 0 t 1. (c) If the vectors in the list (v 1 (1),..., v k (1)) are negatively oriented then we are done. But if they are positively oriented, prove that we can use the vector b instead of the vector b to get the job done You know that if x F n is a solution to Ax = b, where A M(n n, F), then the solution set to the matrix equation Ay = b is x+u, where U = {u F n : Au = 0}. If you can find A so that U is the subspace in question, you can set b = Ax. Then the system of equations in question is j a ijy j = b i, 1 i n. So the theorem is equivalent to the following statement: for any subspace U F n there exists a matrix A M(n n, F) such that U = {u F n : Au = 0}. Prove this statement. I will give an example for n = 5 adapt it to provide a general solution for an arbitrary U F n, proving any statements that require a proof. 8

9 Say that U is a 3-dimensional subspace of F 5. Then it has a basis {u 1, u 2, u 3 }. Expand this to a basis {u 1, u 2, u 3, u 4, u 5 } of F 5. Define f : F 5 F 5 by f(a 1 u 1 + a 2 u 2 + a 3 u 3 + a 4 u 4 + a 5 u 5 ) = a 4 u 4 + a 5 u 5. Then f is a linear map with kernel U. If A is the matrix representation of f with respect to the basis {e 1, e 2, e 3, e 4, e 5 } then U = {u F 5 : Au = 0}. 7. Complete the following steps. (a) Argue that if F is an infinite field then any matrix equation Ax = b has either no solutions or an infinite number of solutions, hence can never have exactly three solutions. Hence F must be finite. (b) Given that F is a finite field with a elements, any k-dimensional subspace of F must have a k elements. (c) Given that F is a finite field with a elements, if Ax = b has three solutions then a = 3. (d) Every finite field with three elements is isomorphic to Z P. The system of equations is equivalent to x 1 a 1 + x 2 a x n a n = b. Take the inner product of both sides with b and derive a contradiction using the properties of inner products (see page 31) Observe that a b 2 = a b, a b = (a c) + (c b), (a b) + (b c). Expand this out using the properties of the inner product and use the fact that a c is orthogonal to b c to simplify the expression Verify, carefully, that O(2) satisfies axioms (1), (2), and (3), but not axiom (4) of the group definition (page 145) Assume that such an inner product exists and derive a contradiction. Let e 1 through e n be the standard basis for R n. Define a ij = e i, e j. (a) Given x = (x 1,..., x n ), show that x, x = i,j x ix j a ij. (b) Given (a), prove that a ii = 1 for all i. Hint: compute e i 2 using the inner product and (a). 9

10 (c) Let x(i, j, α) denote the vector with 1 in position i and α in position j, where i j. Assuming that α < 1, prove that α(a ij +a ji ) = 1 by computing x(i, j, α) 2 using the inner product. (d) Explain why (c) implies that a ij + a ji = 0 whenever i j. (e) Explain why (b) and (d) imply that x, x = x x 2 n whenever x = (x 1,..., x n ). (f) Using (e), explain why x 2 = 1 results in a contradiction using x = (1, 1,..., 1) Let U be the set of subsequences in which only a finite number of terms are non-zero. Prove that U is a proper subspace of V and that if v, u = 0 for all u U then v = (0, 0,... ) Show that if M has at least 3 elements then there are two noncommuting elements in (Bij(M), ). 8.3P. The dot notation in physics represents the derivative with respect to a time variable t. To make this problem a little more concrete and manageable, make the following assumptions: V = R 3, φ(t) = (φ 1 (t), φ 2 (t), φ 3 (t)), φ (t) = (φ 1(t), φ 2(t), φ 3(t)), ψ(t) = (ψ 1 (t), ψ 2 (t), ψ 3 (t)), ψ (t) = (ψ 1(t), ψ 2(t), ψ 3(t)), and the inner product in V is given by the formula in problem P: To solve this problem you must show that A(Aw λw) = λ(aw λw) for all w R 2. You will be able to do [ this] provided you know what form a b the matrix A takes. Assume that A =. The assumption that there is c d exactly one eigenvalue tells you that P A (x) = (x λ) 2. Work out the determinant of A( x), compare its coefficients to the coefficients to the coefficients in (x λ) 2, and see what this tells you about a, b, c, d. 9.3: Eigenvalues of this map are real numbers λ such that f (x) = λf(x) for a non-zero function f(x). So λ appears as a constant in a second-order differential equation. Find the solutions to this equation and find the permissible values of λ. 10

11 10.2: By Theorem 10.3 of these notes, the existence of an inner product with respect to which f is self-adjoint implies that f is diagonalizable. To prove the converse, assume f is diagonalizable and let {v 1,..., v n } be a basis of eigenvectors. Assume that f(v i ) = λ i v i for each i. Define the inner product as α i v i, β i v i = α i β i. i i i (a) First, prove carefully that this construction has all the required properties of an inner product. (b) Second, prove that f is self-adjoint with respect to this inner product. 10.3: Choose an orthonormal basis {u 1,..., u k } for U. Expand this to a basis {u 1,..., u k, v k+1,..., v n } for V. The orthogonal projection map P U : V U is defined by k P U (v) = v, u i u i. (a) Prove that P U is self-adjoint. i=1 (b) It is harmless to regard P U as a mapping from V into V. Choose the basis described above for both domain and codomain, then compute the matrix representation A. (c) Compute the eigenvalues of P U by finding the roots of det A( x). (d) Find a basis for each eigenspace of P U. 10 : Let V be an n-dimensional Euclidean vector space. (a) Assume that f and g are self-adjoint linear maps from V to V which can be simultaneously diagonalized. This means that V has a basis {v 1,..., v n } with respect to which the matrix representation for f is A = diag(α 1,..., α n ) and the matrix representation for g is B = diag(β 1,..., β n ), using a notation for diagonal matrices which lists the diagonal elements only. Prove that f g(v i ) = g f(v i ) for 1 i n. This implies that f g(v) = g f(v) for all v V, which implies f g = g f. Now assume that f and g are self-adjoint linear maps from V to V and assume that f g = g f. There are two cases to consider: all the eigenvalues of f are equal; f has at least two different eigenvalues. 11

12 (b) Consider the case where all the eigenvalues of f are equal to λ. Explain why this implies that f(v) = λv for every v V. Explain why this implies that every matrix representation of f is diagonal. Explain why this implies f and g are simultaneously diagonalizable. (c) Now consider the case where f has at least two different eigenvalues. Then a basis of eigenvectors of f can be chosen for V in such a way that the matrix representation of f is of the form A = diag(λ 1,..., λ 1, λ 2,..., λ 2,..., λ k,..., λ k ), where λ 1 through λ k are the distinct eigenvalues of f. A has block matrix form diag(λ 1 I 1, λ 2 I 2,..., λ k I k ), where each I i is an identity matrix with n λi rows and columns. Let B the matrix representation of g with respect to this basis. Then B has the block matrix form B = (B ij ), where B ij is a matrix with n λi rows and n λj columns. Since f g = g f, we must have AB = BA. Explain why this implies B ij = 0 for i j. Hence B = diag(b 11, B 22,..., B kk ). (d) Fix an index i in the range {1, 2,..., k}. Let {v (i) 1, v (i) 2,..., v n (i) λi } be the corresponding basis vectors used to construct the matrix representation of f and g. Let V i = L R (v (i) 1, v (i) 2,..., v (i) n λi ). Explain why f(v i ) V i and g(v i ) V i. Hence f and g are commuting self-adjoint linear operators on V i and all the eigenvalues of f (restricted to V i ) are equal to λ i. Hence by part (c) above, f and g can be simultaneously diagonalized on V i, meaning V i has a basis consisting of vectors which are eigenvectors for both f and g. The union of all these bases forms a basis for V of eigenvectors for both f and g. 10.3P: Interpret Sym(n, R) to mean the set of symmetric n n matrices with entries in R. First prove that this is a subspace of all the real n n matrices, then compute its dimension by finding an appropriate basis of matrices. 12