(a) Let F be a field, and E F a subfield. Then F can be regarded as a vector space over E (with + and the field operations).

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1 Sample True/False To demonstrate in class (a) If f(x) is an even function, then so i Presentation Math 4310problem 1. Determine whether Prelim the following I Solutions statements (10/09/2015) are (always) true or (at least Math 1110 Name: Instructor 1 sometimes) false, and circle your response. Please give a brief explanation (in complete sentences!) - a reason why it's true, or an example where it fails. Sample True/False Math 4310 (a) If f(x) is an even function, then so is 9(x) : 2' f (x - 4) + Z Prelim 1 To demonstrate in class (b) If f Tnun (x) and I Falss g(x) both one-to-one fun 5 October 2012 Time: 50 minutes Presentation problem 1. Determine whether the following statements are (always) true or (at least sometimes) false, and circle your response. Please give a brief explanation (in complete sentences!) Solutions - a reason why it's true, or an example where it fails. (a) True/False. If f(x) is an You even did function, not need then toso justify is 9(x) your : 2' answers, f (x- 4) but + ZI have included reasons. Tnun I Falss Math 1110 Sample True/Fals (a) Let F be a field, and E F a subfield. Then F can be regarded as a vector space over E (with + and the field operations). Presentation (b) If f (x) and problem g(x) both 1. Determine one-to-one whet fun sometimes) false, and circle your respons - a reason why it's true, or an example w This is true. The vector space axioms for F are strictly weaker than the field axioms for F: (a) in order to be a vector space, we use just the field axioms for E, together If f(x) with is an the even function, then so i axioms about how multiplication distributes, etc. The zero vector is 0, and the negative of a vector is the additive inverse in the field. (b) If f (x) and g(x) both one-to-one functions { defined on all of lr., then f o g is also one-to-one. } (b) For any subset S R, the set W S = f(x) F un(r, R) f(s) = 0 for eachtnus s SI Felss is a subspace of F un(r, R). This is true; there are three things to check. First, the zero function is zero on every input, (b) If f (x) and g(x) both one-to-one functions defined on all of lr., then f o g is also one-to-one. so it s zero on each element of S. Second, if both f and g are zero on elements of S, then f(s) + g(s) = = 0, so f + g is zero on any element of S. Tnus I Felss (c) The function T : Pol(F) Pol(F) defined by T (p(x)) = x+p(x) is a linear transformation. This is false. Let 0 denote the zero vector, in this case the zero polynomial. Then we can calculate T ( 0) = x, but linear transformations must take the zero vector (b) If tof (x) zero. and g(x) both one-to-one fun (d) Subsets of linearly dependent sets are linearly dependent. This is false. The set { 0} is always linearly dependent, but we defined that the empty set { 0} is always linearly independent. (e) An infinite-dimensional vector space can have both finite-dimensional subspaces and infinitedimensional subspaces. This is true: technically, { 0} is always a subspace of any vector space, and it s 0-dimensional. Also, V is a subspace of itself, and we re assuming that is finite-dimensional. A more explicit example is V = Pol(F). Then Pol k (F) is a (k + 1)-dimensional subspace, and the polynomials with zero constant term, {p(x) p(0) = 0}, is an infinite-dimensional subspace.

2 Math 4310 Prelim I Solutions (10/09/2015) 2 Question 1. Subspaces. (a) Give the definition of a subspace of a vector space V (over a field F). In class, we defined a subspace by Definition 1. A nonempty subset U of a vector space V (over a field F) is called a subspace of V if for each u 1, u 2 U, we also have u 1 + u 2 U; and for u U and α F, we have α u U. We also proved that it is equivalent to define a subspace by Definition 2. A nonempty subset U of a vector space V (over a field F) is called a subspace of V if U is itself a vector space with the operations + and inherited from V. We showed that to check whether a subset is a subspace, it is enough to check that 0 U, and that U is closed under scalar multiplication and vector addition. That is the characterization we will use in (b). (b) Let F be a field, M 2 2 (F) the vector space of 2 2 matrices with entries in F, and { [ ] } α β W ε = M γ δ 2 2 (F) α + δ = ε. Show that W ε is a subspace of M 2 2 (F) if and only if ε = 0. We have two directions to prove. { [ ] } α β ( =) If ε = 0, we are considering the set W 0 = M γ δ 2 2 (F) α + δ = 0. We [ ] 0 0 check our three conditions. First, satisfies = 0, so 0 W Second, if [ ] [ ] [ ] α1 β 1 α2 β and 2 α1 + α satisfy α γ 1 δ 1 γ 2 δ i +δ i = 0 for i = 1, 2, then their sum 2 β 1 + β 2 2 γ 1 + γ 2 δ 1 + δ 2 satisfies α 1 + α 2 + δ 1 + δ 2 = (α 1 + δ 1 ) + (α 2 + δ 2 ) = = 0, [ ] α β so W 0 is closed under vector addition. Third, if satisfies α + δ = 0, then γ δ [ ] [ ] α β κ α κ β κ = satisfies γ δ κ γ κ δ κ α + κ α = κ (α + δ) = κ 0 = 0, so W 0 is closed under scalar multiplication. We deduce that W 0 is a subspace of M 2 2 (F). (= ) To prove [ the other ] direction, we prove the contrapositive (the negation). Suppose that 0 0 ɛ 0. Then satisfies = 0 ε, so 0 W 0 0 ε. Hence W ε cannot be a subspace of M 2 2 (F).

3 Math 4310 Prelim I Solutions (10/09/2015) 3 Question 2. Linear dependence. (a) Give the definition of a linearly dependent set of vectors in a vector space V. Definition 3. A set of vectors S V is linearly dependent if there are vectors v 1,..., v k S and scalars α 1,..., α k F not all zero so that α 1 v α k v k = 0. (b) Consider the vector space V = F un(r, R) of real-valued functions over the field F = R. Determine whether the functions f(x) = sin(x) and g(x) = sin(2x) are linearly dependent. Two non-zero vectors are linearly dependent if and only if one is a (non-zero) scalar multiple of the other. In particular, in the case of functions, scalar multiples of non-zero functions must achieve zero on exactly the same subset of the domain. Neither of the two functions is the zero function, since f( π 2 ) = 1 and g( π 4 ) = 1. On the other hand, f( π 2 ) = 1 and g( π 2 ) = 0. So f and g are not scalar multiples of each other, and hence they are not linearly dependent.

4 Math 4310 Prelim I Solutions (10/09/2015) 4 Question 3. Dimension. (a) Give the definition of the dimension of a vector space V over a field F. Definition 4. Let V be a vector space over a field F. If V = { 0}, we define dim(v ) = 0. If V has a finite basis v 1,..., v n V, we define dim(v ) = n to be the number of elements in a basis. In all other cases, we define dim(v ) =. (b) Let U, V and W be finite-dimensional vector spaces over F, and S : U V and T : V W linear transformations. Show that dim(im(t S)) dim(im(t )). We recall that and Im(T S) = {w W u U so that T S(u) = w } Im(T ) = {w W v V so that T (v) = w }. Thus, both Im(T S) and Im(T ) are subspaces of W (we know from class that they are subspaces; we have just explained that they are subsets). Moreover, as subspaces of the finite-dimensional vector space W, each is finite-dimensional. Next, we notice that if w Im(T S), then w = T S(u). But then if we set v = S(u), we have that T (v) = T (S(u)) = T S(u) = w. We conclude that w Im(T ). Hence, Im(T S) Im(T ). But Im(T S) is a vector space, so in fact it must be a subspace of Im(T ). We proved in class (Theorem in the book) that the dimension of a subspace U of a vector space V is less than or equal to the dimension of V. Thus, because Im(T S) is a subspace of Im(T ), we conclude that as desired. dim(im(t S)) dim(im(t )),

5 Math 4310 Prelim I Solutions (10/09/2015) 5 Question 4. Kernel. (a) Give the definition of the kernel of a linear transformation. Definition 5. Let V and W be vector spaces over F, and T : V W a linear transformation. The kernel of T is the subset { } ker(t ) = v V T (v) = 0 of vectors which T maps to 0 W. (b) Suppose that T : V V is linear. Prove that T 2 = T T is the zero transformation if and only if Im(T ) ker(t ). We have two directions to prove. (= ) First we suppose that T T is the zero transformation. That is, for any v V, T (T (v)) = 0. Suppose that some vector w V is in the image of T. That means that there is a u V so that T (u) = w. But then, T (w) = T (T (u)) = T T (u) = 0. Thus, w ker(t ). We have thus shown that Im(T ) ker(t ). ( =) Now suppose that Im(T ) ker(t ). Let v V be any vector. Then T (v) Im(T ) ker(t ), and so T (T (v)) = 0. But T (T (v)) = T T (v). Thus, T T (v) = 0 for any vector v V. This precisely means that the transformation T T is the zero transformation, as desired.

6 Math 4310 Prelim I Solutions (10/09/2015) 6 Question 5. Equivalence relations. (a) Give the definition of an equivalence relation on a set S. Definition 6. An equivalence relation on a set S is a relation that satisfies (a) For each a S, a a ( is reflexive); (b) whenever a b, we also have b a ( is symmetric); and (c) whenever a b and b c, we also have a c ( is transitive). (b) Fix a positive integer n. Prove that the relation, congruence modulo n is an equivalence relation on the set of integers Z. We fix a positive integer n. We say that two integers a, b Z are congruent modulo n if when we divide a and b by n, they have the same remainder. We write a b mod n. Congruence modulo n is equivalent to n being a divisor of the difference (a b); that is, n (a b). This latter divisibility is easier to work with for this problem. We now show that congruence modulo n is an equivalence relation. (reflexive) For any a Z, the difference a a = 0 = 0 n, so n (a a),and hence a a. (symmetric) Suppose that a b, i.e. a b mod n. That means that n (a b), so (a b) = q n. But then (b a) = (a b) = q n, and so we have n (b a), and so b a mod n, or b a. (transitive) Finally, we assume that a b and b c. This means (a b) = q 1 n and (b c) = q 2 n. But then a c = a + ( b + b) c = (a b) + (b c) = q 1 n + q 2 n = (q 1 + q 2 ) n, so n (a c) and a c, as desired. We have checked the three defining properties, so we conclude that is an equivalence relation.