Name: Handed out: 14 November Math 4310 Take-home Prelim 2 Solutions OFFICIAL USE ONLY 1. / / / / / 16 6.
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1 Math 43 Take-home Prelim 2 Solutions Name: Handed out: 4 November 27 INSTRUCTIONS PLEASE READ THIS NOW Please answer all the questions on this exam to the extent that you can. Some are very challenging. If you don t complete a problem, please do include a partial solution. A partial solution should still be well written in careful mathematical language, and you can point out where and why you got stuck. SHOW YOUR WORK. To receive full credit, your answers must be neatly written and logically organized. You should include a complete logical justification, and your answers should be written in grammatically correct mathematical language. Please do your initial work on separate paper, and then transfer your final solutions to this exam booklet. Please carefully write all your final solutions on the page they are posed. If you need more space, please write on the back side of the preceding sheet, but be sure to label your work clearly. Look over your test packet as soon as possible. If you find any missing pages or problems please ask for another test booklet. You may use as much time during this week to complete this exam as you desire. This is a partially open book exam. You may consult any of the textbooks, or suggested texts for this course, your notes, the course website, posted solutions to homework, and handouts for the course. You may ask the professor and TAs questions during office hours (although we will be limited in what we can say), but you may not consult any other humans or websites. In particular, you may not talk to your peers about the exam, and you may not pose questions on homework-help websites or mathoverflow type sites. Academic integrity is expected of every Cornell University student at all times, whether in the presence or absence of a member of the faculty. Understanding this, I declare I shall not give, use, or receive unauthorized aid in this examination. I will not discuss this exam with other students until the end of class on 2 November, 27. Please sign below to indicate that you have read and agree to these instructions. OFFICIAL USE ONLY. / 6 2. / 6 3. / 6 4. / 2 5. / 6 6. / 6 Total: / Signature of Student
2 Math 43 Prelim 2 Solutions 2 Question. True/False. 6 points. Please circle TRUE if the statement is always true, or FALSE if it fails in at least one example. You should justify your answer in the space provided: give a proof of the statement or demonstrate that it must fail. (a) The set of invertible elements of an F-algebra is an F-vector space. TRUE FALSE FALSE. An algebra A is a vector space over F, its zero element O F is never invertible. However every vector subspace of A must contain F. Therefore the set of invertible elements of A cannot be a vector space. (b) Let V be a vector space over a field F. If a linear transformation T : V V is injective, then it must also be surjective. TRUE FALSE FALSE. While this is true for finite dimensional V, it doesn t need to be true more generally. As an example, consider V = F[x], where F is a field. Define T : V V to be T(f(x)) = xf(x). T is a linear transformation (T(f + g) = x(f + g) = xf + xg = T(f) + T(g), similarly for scalar multiplication), it has zero kernel (T(f) = implies that xf(x) =, therefore f(x) = ), yet it is not surjective (since image(t))
3 Math 43 Prelim 2 Solutions 3 Question continued. True/False. Please circle TRUE if the statement is always true, or FALSE if it fails in at least one example. You should justify your answer in the space provided: give a proof of the statement or demonstrate that it must fail. (c) Let F be a field, and T L (F n, F m ) and S L (F m, F n ). Then S T is invertible if and only if T S is invertible. TRUE FALSE FALSE. Almost any example works here, as long as m n: for example, ( ) let m = 2, n =, let T be the linear transformation with matrix (in standard bases) A =, and let S be the linear transformation with matrix B = ( ) ( ). Then the matrix of ST is BA =, which is clearly not invertible, and the matrix of TS is AB = ( ), which is invertible. (d) Let V and W be vector spaces over a field F, and S and T non-zero linear transformations in L (V, W). If S(V) T(V) = {}, then S and T are linearly independent in L (V, W). TRUE FALSE TRUE. Suppose that as + bt = L (V,W), for some a, b F not both zero. Note that since S and T are non-zero transformations, then if one of a, b is zero, the other must be as well. Let c = b/a, then for every v V, we have that, after dividing by a, and moving the T term to the other side, S(v) = ct(v) = T(cv). Since S is non-zero, there exists a vector v V such that S(v) W. But then S(v) S(V), and also S(v) = T(cv) T(V). But S(V) T(V) =, therefore S(v) =, a contradiction.
4 Math 43 Prelim 2 Solutions 4 Question 2. ( 6 points) Consider the set F = R { } with two operations defined by min(a, b) a a b = b for a, b R for a R and b = for a = and b R for a = b = a + b, and a b = for a, b R for a R and b = for a = and b R for a = b =. Which defining properties of a ring does (F,, ) satisfy, and which ones fail? Is it an R-algebra? Solution. The intent of the problem was that is addition, and is multiplication, and that a ring is (as we defined it in class) always a ring with identity. This is a bit of a messy problem, because we have so many cases to check. The ring axioms that we must check are for those of a ring with identity, that is: axioms: A (commutativity of addition), A2 (associativity of addition), A3 (existence of additive identity), A4 (existence of additive inverse), M2 (associativity of multiplication), M3 (existence of multiplicative identity), and M5 (distributivity). It will turn out that, with F := and F = R, that all of these axioms hold, except for one: A4: there is no additive inverse. Therefore, F is not a ring. Also, F cannot be an algebra over R, because it doesn t even satisfy the axioms for it to be a vector space over R, as it fails A4. Basically, every algebra is also a ring, so if it fails to be a ring, it certainly fails to be an algebra too. (A) and (A2) are simple, if one makes the simplifying definition that min(a, ) = a, for a R, or a =. (A) says that min(a, b) = min(b, a), and (A2) says that min(a, min(b, c)) = min(min(a, b), c), which holds since both are min(a, b, c), even if some or all of a, b, c are. (A3) a + F = min(a, ) = a, so F = is an additive identity. (A4) Fails! If a R, and b R, then min(a, b) is never, and so the equation a b = has no solution for b. The rest are straightforward to check, just a bit tedious. I m leaving them out because I want to get these solutions to you as quickly as possible. If you want to see details, ask in office hours.
5 Math 43 Prelim 2 Solutions 5 Question 3. (6 points) Let A = [a a 2... a 7 ] be a 4 7 matrix over the field F 7, whose row reduced echelon form is Let T : F 7 F 4 be the linear transformation defined by T(v) := Av, for a vector v F 7. Your answers for the problems below can involve the vectors a i, if needed. (a) Find a basis for ker(t). What is its dimension? Solution We use the results of Homework 7, problem (a). Specifically, this problem tells us that the number of pivots is the rank of A, therefore rank(a) = 3. By rank-nullity, we know that dim ker A = 7 rank(a) = 4. We also know that ker A = ker B, where B is the row reduced echelon form above. By the method used in class, or inspection, we can write down 4 vectors in the kernel of B and A: 3 2 w =, w 2 =, w 3 = 3, w 4 = 2 6 Note that each entry is an element of F 7, so that for example the last vector also equals: 6 w 4 = 5. Finally, it is clear that these are linearly independent, since w is the only vector with an element in the first row, similarly, w 2, w 3, w 4 are the only ones among these four with an element in rows 2, 4, 6 (respectively). Therefore, since they are LI, and the number matches the dimension, they form a basis. (b) Find a basis B for image(t). What is its dimension? Solution. In part (a), we have already seen that the dimension of image(t) is 3. By Homework 7, part(b), we know that a basis of the image is given by the columns of A which are pivot columns (of B): B = (a 2, a 4, a 6 ).
6 Math 43 Prelim 2 Solutions 6 (c) Note that we are not given the matrix A itself, only its row reduced echelon form. However, there is a choice of bases for F 7 and F 4 such that the matrix of T with respect to these new bases can be determined. Find these bases, and find the resulting matrix. Note: the bases might involve the a i, e j, or other vectors, but the resulting matrix should have only numbers in it. Solution. Notice that as we know the kernel of A, that tells us what A is in terms of the vectors a 2, a 4, a 6 : from part (a), we know that a 3 = 3a 2, a 5 = 2a 2 + 3a 4, and that a 7 = a 2 + 2a 4 + 6a 6, therefore ( A = a2 3a 2 a 4 2a 2 + 3a 4 a 6 ) a 2 + 2a 4 + 6a 6 Using the standard basis for F 7 (i.e. don t change that basis), but setting B = (a 2, a 4, a 6, v), where v is any vector not in im(t), then the matrix of T is Another possible solution is to use this basis for F 4, but use the basis (w,..., w 4, e 2, e 4, e 6 ) of F 7 : one checks that these seven explicitly given vectors are LI, therefore form a basis of F 7. With respect to these bases, the matrix is:
7 Math 43 Prelim 2 Solutions 7 Question 4. (2 points) If f F[x] is a monic polynomial of positive degree over a field F, let V f := F[x]/ f(x). V f is a vector space over F (you may assume this). (Note that we are not assuming that f is irreducible). Let g(x), h(x) F[x] be two monic polynomials of degrees m > and n >. Assume that gcd(g, h) =. Define a function T : V gh V g V h, which sends the equivalence class [α(x)] g(x)h(x) to the pair of equivalence classes ([α(x)] g(x), [α(x)] h(x) ). (a) If deg(f) >, find a basis for V f, and show that dim V f = deg(f). Find the dimensions of the domain and codomain of T. Claim: If deg(f) = n, then B = {[], [x], [x 2 ],..., [x n ]} is a basis for V f Proof: -Let [g] V f. Because g F[x], there are q, r F[x] such that g = qf + r and deg r < deg f. Thus [g] = [r] = [r + r x + + r n x n ] = r [] + + r n [x n ], and B is spanning. -Now assume r [] + + r n [x n ] = [r + + r n x n ] = []. Then by definition r + + r n x n = f(x)g(x). But since deg(f) = n, we must have g(x) = so r i = for all i. Thus B linearly independent and forms a basis. Hence dim V f = deg(f). Then the domain of T, V gh, has dimension deg(gh) = deg(g) + deg(h). The codomain of T, V g V h, has dimension deg(g) + deg(h) (see hw 6 glossary). (b) Show that T is well defined, and that it is a linear transformation. -If [α] gh = [β] gh then by definition α β = (gh)d. So by definition [β] g = [α] g and [β] h = [α] h. Hence T([α]) = T([β]), showing T well defined. -Further, T(r[α] gh ) = T([rα] gh ) = ([rα] g, [rα] h ) = rt([α] gh ). Similarly, T([α] gh + [β] gh ) = T([α + β] gh ) = ([α + β] g, [α + β] h ) = T([α] gh ) + T([β] gh ). Hence T linear. (c) Show that ker(t) =. Deduce that T is an isomorphism of vector spaces. -Since T linear, {[] gh } ker T. Now let [α] gh ker T. Then ([α] g, [α] h ) = ([] g, [] h ). So α = gg and α = hh. But then gg = hh so g divides hh. By Lemma 3.6 (p. 95), g and h relatively prime implies g divides h. Thus α = hh = h(gf) implies [α] gh = [] gh. -Thus ker T = {} showing T injective. By rank nullity, dim(im(t)) = dim(v gh ) = dim(v g V h ) showing T a surjection. Hence T is an isomorphism. (d) Show that if β(x) and γ(x) are polynomials, then there exists a polynomial α(x) such that α(x) β(x) (mod g(x)) and α(x) γ(x) (mod h(x)). (Suggestion: Use the fact that there are polynomials u(x), v(x) F[x] such that ug + vh = ). By definition, h h 2 (mod f) if and only if h h 2 is divisible by f. -We seek α such that [α] g = [β] g and [α] h = [γ] h. By Bezout s, there are u, v F[x] so that ug + vh =. Working in V g and V h, this shows [ug + vh] g = [vh] g = [] g and [ug + vh] h = [ug] h = [] h. So [βvh] g = [β] g and [γug] h = [γ] h. -So let α = βvh + γug. Then [α] g = [βvh] g = [β] g and [α] h = [γug] h = [γ] h. (e) Use the previous part to exhibit the inverse of the linear transformation T. Given ([β] g, [γ] h ), we need to determine [α] gh s.t. T([α] gh ) = ([α] g, [α] h ) = ([β] g, [γ] h ). So from part d we can take α = βvh + γug.
8 Math 43 Prelim 2 Solutions 8 Question 5. (6 points) Let D : M at n n (F) F be a function (not necessarily n linear, or a determinant function). Suppose that D(AB) = D(A)D(B), for all matrices A and B in M at n n (F), and also suppose that D is not a constant function, i.e. there is no constant c F such that D(A) = c, for all A. Let P ij be the permutation matrix, which interchanges columns i and j. Suppose that D(P ij ), for all i j. (a) Show that D(I) =. D is not a constant function, so there is some matrix P such that D(P). Then D(P) = D(IP) = D(I)D(P) so D(P) F = D(P)D(P) = D(I)D(P)D(P) = = D(I). (b) Show that if A is invertible, then D(A) is non-zero. What is D(A )? By (a), D(AA ) = D(A)D(A ) =. So D(A), and we have D(A ) = D(A). (c) Show that D() =, i.e. D, applied to the zero matrix, gives. D is not a constant function, so there is P such that D(P). So D(P) = D(I) and D()D(P) = D()D(I). Thus D() =, else D() D()D(P) = D(P) = D() D()D(I) =. (d) Show that if A is obtained from A by interchanging two rows (or two columns), then D(A ) = D(A). -For any i, j, P ij P ij = I so D(P ij ) 2 = = (D(P ij ) )(D(P ij ) + ) = = D(P ij ) = ±. For i j, D(P ij ) by assumption, so D(P ij ) =. -Recall that left multiplying by P ij interchanges rows i and j, while right multiplying interchanges columns i and j. So A = P ij A or A = AP ij for i j, and D(A ) = D(A) (e) Show that D(A) = whenever A is not invertible. -If A not invertible, then A is not full rank so the nth row of rref (A) = E... E m A is zero. -Let E be the elementary matrix that adds row to row n, so E(rrefA) = A has row n equal to row. Hence P n A = A = D(P n )D(A ) = D(A ) = D(A ). -Now notice that since D(P n ) =, ( + ). So ( + )D(A ) = = D(A ) =. -E i and E invertible, so by part b D(A ) = D(E)D(E )... D(E m )D(A) = = D(A) =.
9 Math 43 Prelim 2 Solutions 9 Question 6. (6 points) Let V = F n, where F is a field. Recall that V = L (F n, F), let S := (e,..., e n) be the basis of V dual to the standard basis S of V. Recall that e i : F n F is the linear transformation which sends e j to, if j i, and sends e i to. (a) Find the matrix [e i ] S S, where S, S are the standard bases of F n and F. -Note that the standard basis S = (e,..., e n ) and the standard basis of S = {}. -So by definition we have [e i ] S S = [[e i (e )] S [e i (e 2)] S... [e i (e n)] S ] = [e i (e ) e i (e 2)... e i (e n)]. -Thus [e i ] S S is the n matrix with in the ith component and elsewhere. (b) Let A be an n n matrix, and let T : V V be the linear transformation T(v) := Av. From the first prelim, we know that T : V V is also a linear transformation. Find the matrix of T with respect to the basis S. That is, find [T ] S S. Your answer should involve the (entries of the) matrix A = (a ij ). (Recall that for f : V F, T (f) is the linear transformation V F which sends v to f(t(v)). -We will denote a j as the jth column of A, and a j as the jth row of A. -For each i, T (e i ) is the linear transformation that sends v Fn to e i (T(v)) = e i (Av) -Thus, T (e i )(e j) = e i (Ae j) = e i (a j). Using part a, this implies T (e i )(e j) = a ij. -Hence we have that T (e i ) = a ie + + a ine n (as these functions agree on the basis S). -By definition, [T ] S S = [[T (e )] S... [T (e n)] S ] -Thus the ith column of [T ] S S is a i, the ith row of A. So we see that [T ] S S = AT.
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