Mathematics Department Qualifying Exam: Algebra Fall 2012


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1 Mathematics Department Qualifying Exam: Algebra Fall 202 Part A. Solve five of the following eight problems:. Let {[ ] a b R = a, b Z} and S = {a + b 2 a, b Z} 2b a ([ ]) a b Define ϕ: R S by ϕ = a + b 2. Prove that ϕ is a ring isomorphism. 2b a You may assume that R and S are commutative rings and that ϕ is a welldefined function. 2. (a) Let G and G 2 be groups and consider their direct product G G 2. Prove that G G 2 is Abelian if and only if both G and G 2 are Abelian. (b) The dihedral group D n of order 2n (n 3) has a subgroup H of order n consisting of rotations, and several subgroups of order 2. Prove that D n cannot be isomorphic to the direct product of H and any of the groups of order Let G be a cyclic group of order 8. Prove that G has exactly one element of order 2. Is there a nonabelian group of order 8 with this property? 4. Let G be a group and a, b G. Assume that a = 2, b = 22 and a b {e}, prove that a 6 = b. 5. An (additive) Abelian group G is said to be divisible if for every element a G, and every k Z \ {0}, there is an element x G such that kx = a. Show that Q is divisible but that Z is not. 6. Recall that the center of a group G is the subset Z(G) = {a G ax = xa for all x G}. (a) Prove that Z(G) is a subgroup of G; (b) Prove that Z(G) is normal in G. 7. Let v be a fixed vector in R n. Show that the set of all matrices A M n (R) such that Av = 0 is a left ideal of M n (R). 8. (a) Find the characteristic of the ring Z 4 4Z (which, using a different notation, is Z 4 4Z). (b) Let R be a commutative ring of characteristic 2. Prove that the set of all idempotents in R is a subring of R. Remark: x R is an idempotent iff x 2 = x. Part B is on the back!!!
2 Part B. Solve five of the following eight problems :. Let T : R 2 R 2 be a linear transformation given by T (x, x 2 ) = (x + x 2, 3x 3x 2 ). (a) Determine whether or not T is invertible. (b) Determine whether or not the vector w = (2, 5) is in the range of T. (c) Find the dimension of the kernel of T. 2. Suppose that B = P AP (where A, B and P are square matrices of the same size, and P is invertible) and that x is an eigenvector of A corresponding to an eigenvalue λ. Show that P x is an eigenvector of B corresponding also to λ. 3. The following 3 3 matrix depends on c: A = c (a) For each c, find a basis for the column space of A. (b) For each c, find a basis for the null space of A. (c) For each c, find the complete solution to Ax = c. 0 [ ] 0 4. Given A =, find A Suppose that A is an invertible matrix and similar to B. Show that B is invertible, and that A is similar to B. 6. Justify your answer to each part. (a) Let A be a fixed 3 3 matrix, and let V be the set of all 3 3 matrices B such that AB = BA. Is V a subspace of M 3 3 (R)? (b) Let N be the set of all continuous functions on [, ] so that f(x) < 0 for x [, ]. Is N a subspace of C[, ]? (c) Let V and V be vector spaces, and let T : V V be a linear transformation. Is the kernel of T a subspace of V? 7. Let V be a vector space, and u, v, v 2,..., v n, w V. Suppose that u Span{v, v 2,..., v n, w}, but u / Span{v, v 2,..., v n }. Show that w Span{ u, v, v 2,..., v n }. 8. Let f(t) = 2, g(t) = 2t, and h(t) = 7t 2 2t + 7. Consider the inner product < p(t), q(t) >= p( )q( ) + p(0)q(0) + p()q() in the vector space P 2 (note that f(t), g(t), h(t) P 2 ). Use the GramSchmidt process to determine an orthogonal basis for the subspace of P 2 spanned by the polynomials f(t), g(t), and h(t).
3 Algebra Qualifying Exam, Solutions Fall 202 Part A. [ ] [ ] a b c d. Let A = and let B =. First we show that ϕ preserves the two operations. 2b a 2d c We want to show: ϕ(a + B) = ϕ(a) + ϕ(b) and ϕ(ab) = ϕ(a)ϕ(b). for all A, B R. We have ϕ(a + B) = ϕ ([ a + c ]) b + d 2(b + d) a + c = a + c + (b + d) 2 ( = a + b ) ( 2 + c + d ) 2 = ϕ(a) + ϕ(b) and ([ ]) ac + 2bd ad + bc ϕ(ab) = ϕ 2ad + 2bc ac + 2bd = ac + 2bd + (ad + bc) 2 = (a + b 2)(c + d 2) = ϕ(a)ϕ(b). [ ] a b To show ϕ is onetoone, note that ker(ϕ) = {A R ϕ(a) = 0}. So, if A = is such 2b a that ϕ(a) = 0, then a + b 2 = 0. Hence, a = b = 0 because 2 / Q. It follows that A = 0, for all A ker(ϕ), and thus ker(ϕ) = {0}. Finally, we show ϕ is onto. For y = a + b [ ] a b 2 S, we have y = ϕ(a) where A = R. 2b a Thus the image of ϕ is all of S. 2. (a) Let (a, a 2 ), (b, b 2 ) be arbitrary elements in G G 2. Thus a, b G and a 2, b 2 G 2. Then (a, a 2 )(b, b 2 ) = (a b, a 2 b 2 ) and (b, b 2 )(a, a 2 ) = (b a, b 2 a 2 ) It follows that (a, a 2 )(b, b 2 ) = (b, b 2 )(a, a 2 ) if and only if (a b, a 2 b 2 ) = (b a, b 2 a 2 ), which is true if and only if a b = b a and a 2 b 2 = b 2 a 2. (b) Let R be a rotation in 360/n degrees of a regular ngon. Then, the rotation subgroup of order n of D n is R. Let H denote a subgroup of order 2 (generated by any reflection). Since both subgroups are cyclic, they are also both abelian. Hence, using part (a) we get that R H is also abelian. Since R H = 2n and R H = {e} then R H = D n, but this contradicts the fact that D n is not abelian. 3. Every cyclic group of order 8 is isomorphic to Z 8. The orders of elements in Z 8 are 0 = = 8 2 = 4 3 = 8 4 = 2 5 = 8 6 = 4 7 = 8 It follows that there is exactly one element in Z 8 having order 2.
4 The nonabelian groups of order 8 are D 4 and Q 8. We know that D 4 contains three nontrivial reflections (all of them having order 2), hence this cannot be our example. Let us look at the orders of elements in Q 8 = i = i = j = j = k = k = 4 = 2 So, Q 8 has the desired property. 4. First we note that a b is a subgroup of the groups a and b, thus (Lagrange s Theorem) a b divides a = 2 and b = 22. So, a b divides gcd(2, 22) = 2. Since a b {e}, then a b, and so a b = 2. Hence, there is exactly one nonidentity element in a b, call it x, and x = 2. Working similarly to problem 3 above, or using that the number of elements of order 2 in a cyclic group equals φ(2) =, we get that the only element of order 2 in a must be x. Since a 6 = 2 (using a = 2) then a 6 = x. We get that b = x in the same way. 5. Let a Q, and k Z \ {0}. We want to find an element x Q such that kx = a. Just solving for x in kx = a we get x = a, which is a rational (recall that k 0). Done. k Now we want to find a Z, and k Z \ {0} such that there is no element x Z such that kx = a. The idea would be to force x to be a rational that is not an integer. So, let us take a = 3 and k = 2. This forces x = 3 2 / Z. 6. Solution. (a) First note that e Z(G) since e x = x e, for all x G. Now let m, n Z(G), and x G. Then, using associativity a few times, we get x(mn) = (xm)n = (mx)n = m(xn) = m(nx) = (mn)x therefore mn Z(G). Let n Z(G), and x G. Since x and n commute we have x = nxn and thus n x = xn, which means that n commutes with x. Therefore n Z(G). It follows that Z(G) G. (b) To show that Z(G) is normal in G, let m Z(G) and x G. Then xm = mx and thus xmx = m Z(G). Solution 2. For every x G consider the function φ x : G G defined by φ x (g) = xgx. This function is an automorphism of G, as (i) φ x (gh) = x(gh)x = xg(xx )hx = ( xgx ) ( xhx ) = φ x (g)φ x (h). (ii) Ker(φ x ) = {g G; xgx = e} = {g G; xg = x} = {g G; g = e} = {e}. (iii) For any given g G, φ x (x gx) = g. recall that Aut(G) is a group under composition. We define the function φ : G Aut(G) by φ(x) = φ x. This is a homomorphism, as φ xy (g) = (xy)g(xy) = xygy x = x ( ygy ) x = φ x ( ygy ) = φ x (φ y (g)) = (φ x φ y ) (g) for all g G. Finally, Ker(φ) = {x G; φ x = e} = {x G; φ x (g) = g, for all g G} = {x G; xgx = g, for all g G} = Z(G) Since Ker(φ) = Z(G) and the domain of φ is G then Z(G) G.
5 7. Let I = {A M n (R); Av = 0}. Clearly I M n (R), and O I (zero matrix, that is). Let A, B I, then (A B)v = Av Bv = 0 0 = 0 so, A B I. Now let a I and B M n (R), then and thus BA I. (BA)v = B(Av) = B0 = 0 8. (a) We know that char(z 4 ) = 4, since 4a 0 (mod 4), for all a Z 4 and that there is no positive integer k < 4 such that ka 0 (mod 4), for all a Z 4. On the other hand, char(4z) = 0 since there is no positive integer n such that n(4l) = 0 for all 4l 4Z. Now let (a, 4l) Z 4 4Z. Then n(a, 4l) = (na, 4nl) for any positive integer n. The observation above implies that there is no way to get a 0 in the second component of (na, 4nl), thus there is no positive integer n such that n(a, 4l) = (0, 0) Z 4 4Z. Hence, char(z 4 4Z) = 0. (b) Let R be a commutative ring with char(r) = 2. Then 2x = 0 for all x R (note that this is equivalent to x = x for all x R). Recall that x R is an idempotent if x 2 = x. Let S = {x R x 2 = x} be the set of all idempotents in R. Clearly S since 0 S (for 0 2 = 0). Let x, y S (thus x 2 = x and y 2 = y). Then (x y) 2 = (x y)(x y) = x 2 xy yx + y 2 = x 2 2xy + y 2 (since R is commutative). But this is equivalent to (x y) 2 = x 2 + y 2 = x + y = x y, since y = y. Hence x y S. Moreover, (xy) 2 = x 2 y 2 (since R is commutative), which is equivalent to (xy) 2 = xy. Therefore xy S and S is a subring of R, by the Subring Test. Note that if rings were considered to always have a one, then we would also have that S, as 2 =.
6 Part B.. (a) We want to find the standard matrix of T (that is, the matrix of T relative to the standard basis for R 2 ). For that, we compute T [ (, 0) = (, ] 3) and T (0, ) = (, 3), and therefore the standard matrix of T is A =. 3 3 Then det A = 0, so the matrix A is singular (NOT invertible), which implies that the linear transformation T is NOT invertible. (b) Recall that the range of a function T is the set of all images T (x). So, a vector w = (2, 5) is in the range of T if the equation T (x) = w has solutions. This equation is equivalent to the system (c) x + x 2 = 2 3x 3x 2 = 5 where we wrote x = (x, x 2 ), and used the formula for T. We can easily see that the above system is inconsistent (it has no solutions), thus w = (2, 5) is NOT the range of T. ker T = { (x, x 2 ) R 2 T (x, x 2 ) = (0, 0) } = { (x, x 2 ) R 2 (x + x 2, 3x 3x 2 ) = (0, 0) } = { (x, x 2 ) R 2 x + x 2 = 0 and 3x 3x 2 = 0 } = { (x, x 2 ) R 2 x + x 2 = 0 } = {(x, x ); x R} = Span({(, )}) Therefore dim(ker T ) =. 2. If x is an eigenvector of A corresponding to λ, then Ax = λx, and x 0. Since P is invertible it cannot be the zero matrix, thus P x 0. Then we have B(P x) = ((P AP )P )x = (P A)x = P (Ax) = P (λx) = λ(p x) which shows that P x is an eigenvector of B corresponding to λ. 3. (a) Call the columns of this matrix C, C 2, C 3, and C 4. We notice that 2C + C 3 = C 4, and thus Span{C, C 2, C 3, C 4 } = Span{C, C 2, C 3 }. Also, since C 3 / Span{C, C 2 } then C 3 will be in the basis of the column space we will find. The only thing left to check is whether C and C 2 are linearly independent. But this is easy, as their first component is the same. It follows that we have two cases: (i) When c = 3 we get that C = C 2 and thus {C, C 3 } is a basis for the column space. (ii) When c 3 we get that C and C 2 are linearly independent, and thus a basis is given by {C, C 2, C 3 }, or any three column vectors in R 3, as these three columns generate the whole space R 3 (column vectors). A different solution for part (a) follows: (a) Elimination gives c 3 4 4, which gives two cases: c = 3 and c 3. If c 3, then c 3 is a pivot. Hence, c ,
7 which implies that a basis for the column space C(A) is the first three columns of A: 2 3, c, If c = 3, then c 3 = 0, so c , so a basis for the column space is the first and third columns of A: 2 3, (b) Since the column space has dimension 2 or 3, depending on the value of c, we have to take cases: (i) If c 3, the column space has dimension 3, and thus the nullity must be equal to. Solving the homogeneous system gives the basis for the null space N(A) as 2 0. (ii) Here the nullity should be 2, as the dimension of the column space is 2 as well. Solving the homogeneous system gives the basis for the null space N(A) as 2 0, 0. 0 (c) Then again, given the two cases that naturally arise in this problem, we must consider two cases here as well. In either case we just look for a particular solution, to later attach to it those vectors in the null space... or, if you prefer, just solve the system that you get when you take the cases. What is interesting here is that the particular solution is common to both cases: 0 x p = 0 0 Hence, the complete solutions are: x 0 2 c 3 x 2 x 3 = 0 + t 0 for any scalar t. x 4 0 x 0 2 c = 3 x 2 x 3 = 0 + s 0 + t 0 for scalars s and t. 0 0 x 4 4. We need to first diagonalize A. The characteristic equation for A, gives 0 = A λi = λ 2 + 3λ + 2 = (λ + )(λ + 2),
8 so the eigenvalues of A are  and 2. For λ = : Solve (A + I)v = 0 to obtain that a basis for the eigenspace is {v } = For λ = 2: Solve (A+2I)v = 0 to obtain that a basis for the eigenspace is {v 2 } = We can form the invertible matrix P = [ [ ] ] v v 2 =, and with D = 2 get A = P DP. It follows that [ ] [ ] [ ] A 2 = P D 2 P ( ) = 2 0 ( 2) 2 = {[ ]}. {[ 2]}. [ ] 0 we 0 2 [ ] If A and B are similar matrices, then B = P AP, for some invertible matrix P. Since A, P and P are invertible matrices, and since the product of invertible matrices is again an invertible matrix, we conclude that B is invertible. Moreover, we have B = (P AP ) = (P ) A P = P A P which implies that A and B are similar matrices. 6. (a) First of all, it is clear that the zero matrix is in V. Now take two matrices B, C in V, thus we know that AB = BA and AC = CA. Is B + C in V? A(B + C) = AB + AC = BA + CA = (B + C)A So, B + C V. Similarly, for α R we get So, V is a subspace of M 3 3 (R). A(αB) = αab = αba = (αb)a (b) Note that the zero function does not belong to this set, as it is not negative (it is always zero). So, N is not a subspace of C[, ] (c) Let 0 V and 0 V be the zero vectors for V and V, respectively. T is linear implies that T (0 V ) = 0 V ; hence, 0 V ker(t ). Therefore, ker(t ) is nonempty. Let u and v be elements of ker(t ). Then, by the linearity of T, T (u + v) = T (u) + T (v) = 0 V + 0 V = 0 V Thus, u + v ker(t ). Let u be an element of ker(t ) and let α be a scalar with the associated vector space V. Then, again by the linearity of T, T (αu) = αt (u) = α0 V = 0 V Thus, αu ker(t ). Since ker(t ) is nonempty and is closed under vector addition and scalar multiplication, ker(t ) is a subspace of V. 7. Since u Span{v, v 2,..., v n, w}, there exist constants c, c 2,..., c n, d such that u = c v + c 2 v c n v n + dw. Solving for dw gives dw = u (c v + c 2 v c n v n ). If d = 0, then this would imply that u = c v + c 2 v c n v n,
9 but this is impossible because u / Span{v, v 2,..., v n }. Therefore, we can write w = d u d c v d c 2 v 2 d c n v n. Hence, w can be written as a linear combination of u, v, v 2,..., v n, and thus w is an element in Span{u, v, v 2,..., v n }. 8. We need first to find a basis for the space Span{f(t), g(t), h(t)}. Since the equation α(2) + β( 2t ) + γ(7t 2 2t + 7) = 0 has only the trivial solution α = β = γ = 0, the set {f(t), g(t), h(t)} is linearly independent, and therefore forms a basis for Span{f(t), g(t), h(t)}. So, we will apply the GramSchmidt process to the basis B = {f(t), g(t), h(t)}. Let Then we have and thus p 2 (t) = 2t. Moreover, we have p (t) = f(t) = 2 p 2 (t) = g(t) < g, p > < p, p > p (t) p 3 (t) = h(t) < h, p > < p, p > p (t) < h, p 2 > < p 2, p 2 > p 2(t) < g, p > = ()(2) + ( )(2) + ( 3)(2) = 6 < p, p > = (2)(2) + (2)(2) + (2)(2) = 2 < h, p > = (6)(2) + (7)(2) + (2)(2) = 70 < h, p 2 > = (6)(2) + (7)(0) + (2)( 2) = 8 < p 2, p 2 > = (2)(2) + (0)(0) + ( 2)( 2) = 8 implying that p 3 (t) = 7t The set {p (t), p 2 (t), p 3 (t)} is an orthogonal basis for the space Span{f(t), g(t), h(t)}.
10 Mathematics Department Qualifying Exam : Algebra Spring 202 Part A. Solve five of the following eight problems:. Prove that the set {[ m n M = 2n m ] } ; m, n Z is isomorphic, as a ring, to the ring Z[2] = {m + n 2 m, n Z}. 2. (a) How many elements of order 7 does the group Z 2 Z 35 have? Justify your answer. (b) How many cyclic subgroups of order 7 does the group Z 2 Z 35 have? Justify your answer. Note: You may prefer to use the notation Z 2 Z 35 instead of Z 2 Z 35. You are free to do so. 3. (a) Find all group homomorphisms from Z 42 to Z 2 (give formulas for each). How many homomorphisms are there from Z 42 onto Z 2? (b) Are all the group homomorphisms in part (a) also ring homomorphisms? 4. Let G = GL(2, R) and H = {A G ; det(a) = 3 k, k Z}. Prove that H is a subgroup of G. Moreover, show that H G. 5. Let n N. Prove that all ideals in Z n are principal. 6. Let R be a commutative ring with one. Prove that R is a field if and only if {0} and R are the only (twosided) ideals in R. 7. (a) Suppose that H is a normal subgroup of G with [G : H] = 24 and H =. If x G and x = e, prove that x H. (b) Let H = {(), (2)(34)}. Prove or disprove: H is normal in A Let G be a group. Assume that G is isomorphic to all its nontrivial subgroups. Prove that G is isomorphic to either Z or Z p, for some prime p. Part B is on the back!!!
11 Part B. Solve five of the following eight problems : Notation: (a) M n (K) is the set of n n matrices with entries in K. (b) P n is the set of all polynomials (with real coefficients) with degree at most n (including the zero polynomial).. Let v be a fixed vector in R n, where n >. Assume that M M n (R) \ {0} is such that Mv = 0. Prove that there exists a matrix N M n (R) such that (MN)v Define T : M 2 (C) M 2 (C) by T (A) = A + A T. (a) Show that T is a linear transformation. (b) Show that the range of T is the set of all B in M 2 (C) with the property that B T = B. (c) Describe the kernel of T. 3. Recall that the trace of a square matrix is the sum of the entries on the main diagonal. Let W be the subset of M 2 (R) of matrices with trace equal to 0. (a) Show that W is a subspace of M 2 (R). (b) Let S = {[ 0 0 ], [ ], [ ]} W. Show that S is a basis for W. 4. Let V be the space C[0, ] of realvalued continuous functions defined on [0, ], and consider the inner product on V defined by f, g = 0 f(t)g(t) dt. Let W be the subspace spanned by the polynomials p (t) =, p 2 (t) = 2t, and p 3 (t) = 2t 2. Use the GramSchmidt process to find an orthogonal basis for W. [ ] p(0) 5. Define T : P 2 R 2 by T (p) =. p(2) (a) Show that T is a linear transformation. (b) Find the kernel of T, and a basis for it. (c) Find the matrix for T relative to the standard bases for P 2 and R 2 respectively. 6. Let A be a square matrix. Prove or disprove the following statements: (a) The matrices A and A T have the same eigenvalues, counting multiplicities. (b) If A is an n n diagonalizable matrix, then each vector in R n can be written as a linear combination of eigenvectors of A. (c) If A is diagonalizable, then the columns of A are linearly independent. 7. Let T : P P be a linear transformation such that T (a + bt) = ( 2a + b) + (a + 2b)t. Find the eigenvalues of T. Then choose one of the eigenvalues and find a basis for the corresponding eigenspace. 8. Show that B = {t, t + } forms a basis for P, and find the changeofcoordinates (i.e. change of basis) matrix from the standard basis C = {, t} to the basis B.
12 Algebra Qualifying Exam, Solutions Spring 202 Part A.. Define f : Z[2] M by f(m + n [ m n 2) = 2n m ]. We have f((m + n 2) + (r + s 2)) = f((m + r) + (n + s) [ ] m + r n + s 2) = 2(n + s) m + r [ ] [ ] m n r s = + = f(m + n 2) + f(r + s 2) 2n m 2s r and f((m + n 2)(r + s 2)) = f((mr + 2ns) + (ms + nr) [ mr + 2ns ms + nr 2) = 2(ms + nr) mr + 2ns [ ] [ ] m n r s = = f(m + n 2)f(r + s 2). 2n m 2s r ] Hence, f is a ring homomorphism. [ ] [ ] m n 0 0 Since = iff m = n = 0, we have ker f = {0} and so f is onetoone. If 2n m 0 0 [ ] m n M then m, n Z, and then m + n 2 Z[2]. Moreover, 2n m f(m + n [ ] m n 2) = 2n m and thus f is onto. Hence f is an isomorphism of rings. 2. (a) Let (a, b) Z 2 Z 35. Then (a, b) = lcm( a, b ). So, when (a, b) = 7 we have two cases: Case : a = 7 and b = or 7. Since Z 2 has a unique cyclic group of order 7 and any cyclic group of order 7 has six generators, there are six choices for a. Similarly, there are seven choices for b (one for b = and six for b = 7). This gives 42 choices for (a, b). Case 2: a = and b = 7. Since Z 35 has a unique cyclic group of order 7 and any cyclic group of order 7 has six generators, there are six choices for b. There is only one choice for a. So, this case yields six more possibilities for (a, b). Thus Z 2 Z 35 has = 48 elements of order 7. (b) Because each cyclic group of order 7 has six elements of order 7 and no two of the cyclic subgroups can have an element of order 7 in common, there must be 48/6 = 8 cyclic subgroups of order (a) We know that a homomorphism f : Z 42 Z 2 is determined by its action on a generator of Z 42. Assume that f() = a, for some a Z 2. Then a divides Z 2 = 2 and Z 42 = 42 (why?), so a divides gcd(2, 42) = 6. Case : If a =, then a = 0. Case 2: If a = 2, then a = 6. Case 3: If a = 3, then a = 4 or a = 8. Case 4: If a = 6, then a = 2 or a = 0. Notice that if f() = a, then f(x) = f(x ) = xf() = xa, for all x Z, because f is operation preserving. Therefore, the homomorphisms from Z 42 to Z 2 are: f (x) = 0 f 2 (x) = 6x f 3 (x) = 4x f 4 (x) = 8x f 5 (x) = 2x f 6 (x) = 0x
13 None of the above functions are onto, since none of the possible values for a is a generator for Z 2. Thus there are no homomorphisms from Z 42 onto Z 2. (b) We want to check whether any of the homomorphisms between the (additive) groups Z 42 and Z 2 found in (a) are ring homomorphisms. If such an f is then Out of the six possible a s listed above we get a = f() = f( ) = f() f() = a = = = = = = 4 0 and thus the only candidates to be ring homomorphisms are f (x) = 0 and f 3 (x) = 4x. f is clearly a homomorphism of rings, and f 3 (xy) = 4(xy) = 4 2 (xy) = (4x)(4y) = f(x)f(y) which means that f 3 is also a homomorphism of rings. 4. Since det(a) = 3 k 0, for all A H then H G. Moreover, det(i) = = 3 0, and thus the identity matrix lives in H. Let A, B H then det(a) = 3 k and det(b) = 3 t, for some k, t Z. Then det(ab ) = det(a) det(b ) = det(a) det(b) = 3 k 3 t = 3 k t Since k t Z, then AB H. It follows that H G. In order to show that H G we let A H (with det(a) = 3 k ) and B G (with det(b) = n 0) and we want to show that BAB H. Note that det(bab ) = det(b) det(a) det(b ) = det(b) det(a) det(b) = n 3 k n = 3 k which implies that BAB H. 5. We know that the subgroups of (Z n, +) look like < [d] >, where d n. But < [d] >= {[d]k; k Z} = {[dk]; k Z} = {[dk]; [k] Z n } = {[d][k]; [k] Z n } = ([d])z n = ([d]) where ([d]) denotes the ideal generated by [d]. Note that we are using that Z n is a commutative ring with one to get that ([d])z n = ([d]). 6. (= ) If R is a field and I {0} is an ideal in R, then there exists a I, a 0. Since R is a field, every nonzero element is a unit, and therefore = a a I, since I is an ideal. Hence for all r R we have r = r I, and I = R. ( =) R is a commutative ring with one, so to show that R is a field we only need to show that all nonzero elements in R are units. Let 0 a R and consider I = (a), the principal ideal generated by a. Note that, since R is a commutative ring with one, then (a) = Ra. I {0} since 0 a I. If {0} and R are the only ideals in R, then we must have I = R. But R is a ring with one, hence I. In other words, there exists r R such that = ra. Hence r = a (using that R is a commutative ring), and thus a is a unit. 7. (a) First note that (xh) = x H = eh = H, by hypothesis. Hence xh divides. But since G/H is a group of order 24, we must also have that xh divides 24. Since gcd(, 24) =, it must be the case that xh =. That is, xh = H. It follows that x H, as required. (b) As (23)H(23) = (23)H(32) = {e, (4)(23)} H, then H is not normal in A If G is isomorphic to all its nontrivial cyclic subgroups then it must be cyclic, as it would be isomorphic to all its nontrivial cyclic subgroups. Hence,  If G is infinite we are done, as G = Z.  If G is finite then G = Z n, for some n. If n were not prime, then there would be a prime q dividing n. Consider H =< [q] >. This is a subgroup of Z n of order n/q. We get a contradiction because if G were isomorphic to H then n = n/q and that is false. So, n must be prime, and G is isomorphic to Z p, for some prime p.
14 Part B.. Let v R n and M M n (R) \ {0} be such that Mv = 0. Since M 0 then there is a vector w such that Mw 0. Consider N to be any matrix mapping v to w. Why does such a matrix exist? One way to think about it is to create two matrices: A that maps v to, let us say, (, 0, 0, 0) and B (invertible) mapping w to (, 0, 0, 0). Then N = B A would map v to w. It follows that (MN)v = M(Nv) = Mw (a) We need to show that T preserves addition and scalar multiplication of matrices. Let A, B M 2 (C) and c C. Then we have and T (A + B) = (A + B) + (A + B) T = (A + B) + ( A T + B T ) = (A + A T ) + ( B + B T ) = T (A) + T (B) T (ca) = (ca) + (ca) T = c(a) + c ( A T ) = c ( A + A T ) = ct (A). Therefore, T is a linear transformation. (b) Let B M 2 (C) such that B T = B (i.e. B is a symmetric matrix). Then consider A = B and observe that 2 A + A T = 2 B + ( 2 B ) T = 2 B + 2 BT = 2 B + 2 B = B and thus T (A) = B. These calculations show that the range of T contains all matrices in M 2 (C) such that B = B T. Moreover, notice that T (A) T = ( A + A T ) T = A T + ( A T ) T = A T + A = T (A) and thus the range of T is exactly the set of all matrices in M 2 (C) such that B = B T. (c) We have that ker T = {A M 2 (C) : T (A) = 0} = {A M 2 (C) : A + A T = 0} {[ ] [ ] [ ] [ ]} a b a b a c 0 0 = M c d 2 (C) : + = c d b d 0 0 {[ ] [ ] [ ]} a b 2a b + c 0 0 = M c d 2 (C) : =. c + b 2d 0 0 [ ] a b Thus if A = ker T, then 2a = b + c = 2d = 0, or equivalently, a = d = 0 and c d {[ ] } 0 b c = b, for some b R. Therefore, ker T = : b C. b 0 3. (a) Let A, B W and c R. Clearly, W M 2 (R) and 0 W. Now using that the trace is an additive function we get tr(a B) = tr(a) tr(b) = 0 0 = 0 tr(ca) = c tr(a) = c 0 = 0
15 implies that W is a subspace of M 2 (R) (b) Since M 2 (R) has dimension four and there are matrices in M 2 (R) \ W (for instance the identity matrix), then dim(w ) 3. Note that S contains three elements, all of them in W. Hence, if they were linearly independent then they would be forced to form a basis of W. But this is immediate, as [ yields x = y = z = 0. ] = x [ 0 0 ] [ 0 + y 0 0 ] [ z 0 ] [ ] x y = z x 4. Let q = p, and p 2, q = ˆ 0 (2t )() dt = (t 2 t) So, p 2 and q are already orthogonal. Hence, p 2 = q 2. Now we compute p 3, q = ˆ 0 q, q = p 3, q 2 = q 2, q 2 = ˆ 0 (2t 2 )() dt = (4t 3 ) ˆ 0 ˆ 0 ()() dt = t 0 0 = (2t 2 )(2t ) dt = 2 (2t 2 ) 2 dt = 6 (2t )3 Then, the projection of p 3 onto W 2 = span{q, q 2 } is given by Hence, 0 = 4 0 = 0 = 3 p 3, q q, q q + p 3, q 2 q 2, q 2 q 2 = 4 q + 2 /3 q 2 = 4q + 6q 2 q 3 = p 3 (4q + 6q 2 ) and thus q 3 (t) = 2t 2 4 6(2t ) = 2t 2 2t + 2. It follows that the orthogonal basis is {q, q 2, q 3 }. 5. (a) Let p, q P 2 and c R. Then [ ] (p + q)(0) T (p + q) = = (p + q)(2) T (cp) = [ (cp)(0) (cp)(2) ] = [ c(p(0)) c(p(2)) [ p(0) p(2) ] ] [ q(0) + q(2) [ p(0) = c p(2) ] = T (p) + T (q) ] = ct (p). Hence, T is a linear transformation. (b) Using the definition of the kernel of a linear transformation, we have: { [ ] [ ]} p(0) 0 Ker(T ) = p P 2 T (p) = = p(2) 0 { [ ] [ ]} = a + bt + ct 2 a 0 P 2 = a + 2b + 4c 0 = { a + bt + ct 2 P 2 a = 0, a + 2b + 4c = 0 }
16 If a = 0, the equation a+2b+4c =0 is equivalent to b + 2c = 0, or to b = 2c, where c is free. Therefore, we arrive at: Ker(T ) = {( 2c)t + ct 2 c R} = Span{ 2t + t 2 } which implies that { 2t + t 2 } forms a basis for Ker(T ). (c) The standard basis for P 2 is {, t, t 2 }, thus we need to compute the images under T of the polynomials, t and t 2. [ ] T () = [ ] 0 T (t) = 2 [ ] T (t 2 0 ) = 4 Therefore [ the matrix ] for T relative to the basis {, t, t 2 } for P 2 and the standard basis for 0 0 R 2 is (a) True. Matrices A and A T have the same characteristic polynomial, because det(a T λi) = det(a T (λi) T ) = det(a λi) T = det(a λi) and thus the same eigenvalues, counting multiplicities. (b) True. If A is an n n diagonalizable matrix, then A has n linearly independent eigenvectors v,..., v n in R n. Since dim(r n ) = n, the set {v,..., v n } forms a basis for R n, and therefore each vector in R n can be written as a linear combination of v,..., v n. (c) False. If A is a diagonal matrix with (at least one) 0 on the diagonal then the columns of A are not linearly independent (since a set of vectors containing the zero vector is linearly dependent). 7. We find first [T ] B, the matrix of T relative to the standard basis B = {, t} for P. Since T () = 2 + t and T (t) = + 2t, the Bmatrix of T has the following form: [ ] 2 [T ] B = 2 The eigenvalues of T are exactly the eigenvalues of the matrix A = [T ] B. The characteristic polynomial of A is χ A = λ 2 5, thus the eigenvalues of A (and hence the eigenvalues of T ) are λ = ± 5. For λ = 5: A [ ] 2 5 5I = 2 and the equation (A 5I)x = 0 amounts 5 to ( 2 5)x + x 2 = 0 x + (2 5)x 2 = 0 Observe that these two equations are equivalent. So, x 2 = (2 + 5)x, and x free. The general solution to the equation (A [ ] 5I)x = 0 is x = x 2 +, with x 5 R. This tells us that the eigenspace V λ= 5 is dimensional, and that if {p(t)} is a basis for [ ] V λ= 5, then the Bcoordinate vector of p(t) is [p(t)] B = Therefore, {p(t) = + (2 + 5)t} is a basis for the eigenspace V λ= 5 corresponding to the eigenvalue λ = 5. The computations for λ = 5 are similar.
17 8. We will denote the changeofcoordinates matrix from the standard basis C to the basis B by P B C. The set B contains 2 vectors, and dim P = 2, thus to show that B forms a basis for P it suffices to show that it is a linearly independent set. If a(t ) + b(t + ) = 0 = 0t +, for some scalars a and b, then a + b = 0 and a + b = 0. Then a = 0 = b, implying that B is linearly independent and, therefore, a basis for P. Recall that the changeofcoordinates matrix from the basis C = {, t} to the basis B = {t, t + } is given by P B C = [ [] B [t] B ], so we need to find the Bcoordinate vectors of the polynomials/vectors contained in the basis C. [ ] /2 Since = 2 (t ) + 2 (t + ) we have [] B =. /2 Similarly, since t = 2 (t ) + 2 (t + ), it implies that [t] B = Then P B C = [ [] B [t] B ] = [ /2 /2 /2 /2 [ /2 /2 ]. ].
18 Mathematics Department Qualifying Exam : Algebra Fall 20 Part A. Solve five of the following eight problems:. Let G and H be groups. (a) Prove that S(G, H) = {f : G H; f is a function} is a group with the operation defined by (f g)(x) = f(x) g(x) for all x G, where is the operation in H. (b) Prove that S(G, H) is abelian if and only if H is abelian. 2. (a) Can there be a group homomorphism from Z 6 Z 6 onto Z 2? (b) Can there be a group homomorphism from Z 8 onto Z 3 Z 3? 3. (a) Let a, b and c be elements of a commutative ring with unity, and suppose that a is a unit. Prove that b divides c in this ring if and only if ab divides c in this ring. (b) Let a belong to a ring R. Let S = {x R ax = 0}. Show that S is a subring of R. 4. Let G, H be groups and φ : G H be a homomorphism. (a) Prove that φ(g i ) = φ(g) i, for all i Z. (b) Show that if φ is an isomorphism and G is cyclic, then H is cyclic. 5. Prove the following claims for permutations in S n. (a) (a a 2 a k ) = (a a k ) (a a 3 )(a a 2 ) = (a a 2 )(a 2 a 3 ) (a k a k ), where a, a 2,..., a k are distinct. Conclude that every permutation is the product of transpositions. (b) Any permutation can be written as a product of the transpositions (2), (3),, (n). 6. Let S = R\{ }. Define on S by a b = a + b + ab. Show that S is a group. 7. Show, in full detail, that I = {p(x) R[x]; p(0) = p() = 0} is a principal ideal of R[x]. 8. Let d be a positive integer. Prove that Q[ d] = {a + b d a, b Q} is a field. Part B is on the back!!!
19 Part B. Solve five of the following eight problems :. Let P 2 be the vector space of real polynomials of degree 2 or less. Consider the basis B = {, x, x 2 } and B = {, x +, x 2 + x + } for P 2. (a) Find the transition matrix from B to B. (b) Let p(x) = 3 x + 2x 2 P 2. Find the coordinates of p(x) relative to B. 2. Find an orthonormal basis for the subspace of R 4 spanned by the vectors u = (0, 0,, 0), u 2 = (, 3, 0, 0), and u 3 = (, 0,, ). 3. Find a square root of the matrix 3 3 A = Let V be an ndimensional vector space and W a subspace of V. Let B = {b, b 2,..., b m } be a basis for W. Prove that there exist vectors b m+, b m+,..., b n in V such that B = {b, b 2,..., b m, b m+,..., b n } is a basis for V. 5. Gaussian elimination changes Ax = b to a reduced Rx = d. The complete solution is x = s t (a) What is the reduced row echelon form matrix R? (b) What is d? Consider the linear transformation with matrix A = Find a basis for the kernel and a basis for the image of the transformation. 7. Let A be a nonzero n n skewsymmetrix real matrix. (a) Show that if n is odd, then A is not invertible. (b) What happens if n is even? Justify your answer. 8. (a) Show that if v is an eigenvector of the matrix product AB and Bv 0, then Bv is an eigenvector of BA. (b) Suppose that v is an eigenvector of A corresponding to an eigenvalue λ. Show that v is an eigenvector of 5I 3A + A 2, where I is the identity matrix of the same size as A. What is the corresponding eigenvalue?
20 Mathematics Department Qualifying Exam : Algebra Fall 20 Part A. Solve five of the following eight problems:. Let G and H be groups. (a) Prove that is a group with the operation defined by S(G, H) = {f : G H; f is a function} (f g)(x) = f(x) g(x) for all x G, where is the operation in H. (b) Prove that S(G, H) is abelian if and only if H is abelian. Solution. (a) Since this operation is weird, then we need to check everything about this group (to be). (i) Closure of in S(G, H): We take f, g S(G, H), we want to show that f g is an element of S(G, H). First of all, note that since f, g S(G, H) and x G then both f(x) and g(x) are in H. Moreover being closed in H forces f(x) g(x) H. Hence, (f g)(x) H for all x G. So, f g takes elements in G to elements in H, but is it a function? We need to check that images are unique! But this is immediate, as both f and g are functions, and thus f(x) and g(x) are uniquely defined... and also is their product. It follows that S(G, H) is closed under. (ii) Associativity: Let f, g, h S(G, H), then for any x G we get [f (g h)](x) = f(x) (g h)(x) = f(x) [g(x) h(x)] but now using that is associative in H we get f(x) [g(x) h(x)] = [f(x) g(x)] h(x) which is [(f g) h](x). Done. (iii) Identity? Let f S(G, H) and let e be the constant function e(x) = e H, where e H is the identity of H. Note that (f e)(x) = f(x) e(x) = f(x) e H = f(x) for al x G, and thus f e = f. Similarly, e f = f. Hence, e is the identity of S(G, H). (iv) Inverses? Let f S(G, H). We want to find g S(G, H) such that f g = e. Consider g(x) = f(x), where the inverse is taken in H. Then, and thus f g = e. Done. (f g)(x) = f(x) g(x) = f(x) f(x) = e H (b) This group is Abelian if H is Abelian, as (f g)(x) = f(x) g(x) = g(x) f(x) = (g f)(x) for all x G. If H is nonabelian, then there are elements a, b H such that a b b a. Then, we define the constant functions f(x) = a and g(x) = b, which are clearly in S(G, H) and do not commute. Hence, S(G, H) is Abelian if and only if H is Abelian. 2. (a) Can there be a group homomorphism from Z 6 Z 6 onto Z 2?
21 (b) Can there be a group homomorphism from Z 8 onto Z 3 Z 3? Solution. (a) No. The largest possible order of an element in Z 6 Z 6 is 6. But Z 2 has an element of order 2 (any generator would do). Since the order of ψ(x) divides the order of x, for all x Z 6 Z 6 then the largest order of an image under ψ is 6. Hence, the elements of order 2 is Z 2 are not in the range of ψ. (b) No. The homomorphic image of a cyclic group is cyclic. Having an onto homomorphism from Z 8 Z 3 Z 3 would imply Z 3 Z 3 is cyclic. A contradiction. 3. (a) Let a, b and c be elements of a commutative ring with unity, and suppose that a is a unit. Prove that b divides c in this ring if and only if ab divides c in this ring. (b) Let a belong to a ring R. Let S = {x R ax = 0}. Show that S is a subring of R. Solution. (a) Let R be a commutative ring with unity, and let a be a unit in R. Suppose that b c for some b, c R; that is, c = br for some r R. Then c = (a a)(br) = (ab)(a r) (we used here that R is commutative, that a exists in R and that multiplication is associative is a ring). But R is closed under multiplication, thus a r R, and then ab c, as desired. Conversely, suppose that ab c for some b, c R; that is c = (ab)s for some s R. Since R is commutative, it implies that c = b(as), thus b c (since as R). (b) First note that 0 S (since a0 = 0), thus S. Let x, y S. Then ax = 0 = ay, and we have: a(x y) = ax ay = 0 0 = 0 and a(xy) = (ax)y = 0y = 0 thus x y S and xy S. Then by the Subring Test, S is a subring of R. 4. Let G, H be groups and φ : G H be a homomorphism. (a) Prove that φ(g i ) = φ(g) i, for all i Z. (b) Show that if φ is an isomorphism and G is cyclic, then H is cyclic. Solution. (a) For i positive we get that φ(g i ) = φ(g i g) = φ(g i )φ(g). So, if we knew that φ(g i ) = φ(g) i then we would be done, as φ(g i ) = φ(g i )φ(g) = φ(g) i φ(g) = φ(g) i Note that we have just set an induction argument. That is, proving the icase by assuming the (i )case. Since the case i = 2 holds trivially, then the induction is done. For i negative, let us first look at φ(g ) = φ(g). Then putting this case together with the positive i case above will solve the case for all negative i s. Recall that φ(e) = e. Then, e = φ(e) = φ(gg ) = φ(g)φ(g ), which means that φ(g ) is the inverse of φ(g). Done. Now let us consider i for i positive. We need to show that φ(g i ) = φ(g) i : The case i = 0 is trivial. φ(g i ) = φ((g i ) ) = φ(g i ) = (φ(g) i ) = φ(g) i (b) Assume that G =< g >. We know, from part (a) that φ(g i ) = φ(g) i. Since every element in the domain of φ looks like g i, for some i Z, then every element in the image of G must look like φ(g) i. But since φ is onto then every element in H looks like φ(g) i. It follows that φ(g) generates H.
22 5. Prove the following claims for permutations in S n. (a) (a a 2 a k ) = (a a k ) (a a 3 )(a a 2 ) = (a a 2 )(a 2 a 3 ) (a k a k ). Conclude that every permutation is the product of transpositions. (b) Any permutation can be written as a product of the transpositions (2), (3),, (n). Solution. (a) Consider the functions σ = (a a 2 a k ) and τ = (a a k ) (a a 3 )(a a 2 ). We want to show σ = τ, so we need to show σ(x) = τ(x), for all x {, 2, 3,, n}. If x = a i for some i, then σ(a i ) = a i+ mod k. On the other hand, we have three cases: (i) If x = a then τ(a ) = a 2, as a 2 only appears in the farmost right transposition only. This is consistent with having σ(a ) = a 2. (ii) If x = a k then τ(a k ) = a, as a k only appears in the farmost left transposition only. This is consistent with having σ(a k ) = a. (iii) If x a, a k then a i appears exactly once in the transpositions of τ. Hence, a i will be fixed by whatever is to the right of (a a k ) (a a i+ )(a a i ) (in the product of transpositions of τ). Now it is easy to see that a i will be first moved to a and then to a i+. Since a i+ does not appear anymore in this product then τ(a i ) = a i+, which is consistent wit what σ does. If x a i, for all i then both σ and τ fix this element. Now consider the functions σ = (a a 2 a k ) and τ = (a a 2 )(a 2 a 3 ) (a k a k ). We want to show σ = τ, so we need to show σ(x) = τ(x), for all x {, 2, 3,, n}. If x = a i for some i, then σ(a i ) = a i+ mod k. On the other hand, we notice that every a i a, a k appears in exactly two cycles in τ. Hence, we need to take cases again: (I) If x = a then τ(a ) = a 2, as only the cycle (a a 2 ) affects a. We are done because σ(a ) = a 2. (II) If x = a k, then the first transposition on the right sends a k to a k, but then the next one sends a k to a k 2, and so on, until reaching a. It follows that τ(a k ) = a, which is exactly what σ does to a k. (III) If x a, a k then the first cycle (on the right) that moves a i is (a i a i+ ), which sends a i to a i+. At this point there are no a i+ s on the cycles to the left of (a i a i+ ), and thus τ(a i ) = a i+. Just like σ. Since every permutation is a product of cycles, and each cycle is a product of transpositions, then it follows that each permutation is a product of transpositions. (b) If we can get all transpositions out of products of (2), (3),, (n) then, by part (a), we would be done. We already have all transposition of the form (b), so let us consider (ab), where a and b are both different from and, of course a b. But that is easy, as (a)(b)(a) = (ab) 6. Let S = R\{ }. Define on S by a b = a + b + ab. Show that S is a group. Solution. See problem 6 in Spring 08 exam and problem in spring exam. 7. Show, in full detail, that I = {p(x) R[x]; p(0) = p() = 0} is a principal ideal of R[x]. Solution. Let p(x) R[x]. The factor theorem says that if p(α) = 0, for some α R then p(x) = (x α)q(x), for some q(x) R[x]. Hence, using this theorem twice we can rewrite I as I = {p(x) R[x]; p(x) = x(x )r(x), for some r(x) R[x]} which is the set of all multiples (in R[x]) of the polynomial t(x) = x(x ). It follows that I is the principal ideal of R[x] generated by t(x) = x(x ), which is most of the times denoted t(x)r[x].
23 8. Let d be a positive integer. Prove that Q[ d] = {a + b d a, b Q} is a field. Solution. If d Q, then Q[ d] Q (by closure of Q). On the other hand, if considering the elements of Q[ d] with b = 0 we get Q Q[ d]. It follows that, in this case, Q = Q[ d]. For d / Q. It suffices to show that Q[ d] is a subfield of R. First we observe that 0 = d Q[ d], so Q[ d]. Let a + b d, a2 + b 2 d Q[ d], thus a, a 2, b, b 2 Q. Then (a + b d) (a2 + b 2 d) = (a a 2 ) + (b b 2 ) d Q[ d]. Assume that a 2 + b 2 d 0, i.e., a2 0 and b 2 0. Then we observe that (a 2 + b 2 d) = = a 2 b 2 d a 2 a 2 + b 2 d a 2 2 = db2 2 a b 2 d Q[ d] db2 2 a 2 2 db2 2 since a 2 2 db (here using that d is not a square) and Then we have (a + b d)(a2 + b 2 d) ( a2 = (a + b d) a 2 2 db2 2 = a a 2 b b 2 d a 2 2 db2 2 Then, by the subfield test, Q[ d] is a subfield of R. Done. a 2 b 2 a 2 2, db2 2 a 2 2 Q. db2 2 + a b 2 + b a 2 a 2 2 db2 2 b 2 d ) a 2 2 db2 2 d Q[ d]. Part B is on the back!!!
24 Part B. Solve five of the following eight problems :. Let P 2 be the vector space of real polynomials of degree 2 or less. Consider the basis B = {, x, x 2 } and B = {, x +, x 2 + x + } for P 2. (a) Find the transition matrix from B to B. (b) Let p(x) = 3 x + 2x 2 P 2. Find the coordinates of p(x) relative to B. Solution. (a) To find the first column vector of the transition matrix, we must find scalars a, a 2 and a 3 such that a () + a 2 (x + ) + a 3 (x 2 + x + ) =. By inspection we see that the solution is a =, a 2 = 0, and a 3 = 0. Therefore, [] B = The second and third column vectors of the transition matrix can be found by solving the equations b () + b 2 (x + ) + b 3 (x 2 + x + ) = x and 0 0. c () + c 2 (x + ) + c 3 (x 2 + x + ) = x 2, respectively. The solutions are given by b =, b 2 =, and b 3 = 0, and c = 0, c 2 =, and c 3 =. Hence, the transition matrix is P B B = Remark: This can also be done by finding the inverse of P B B = 0, 0 0 which is obtained by hanging the coordinates of the vectors {, + x, + x + x 2 } in the standard basis. (b) The coordinate vector of p(x) = 3 x + 2x 2 relative to B is given by 3 [p(x)] B =, 2 and therefore [p(x)] B = P B B[p(x)] B = is the coordinate vector of p(x) = 3 x + 2x 2 relative to B. Notice also that 3 x + 2x 2 = 4() 3(x + ) + 2(x 2 + x + ). = Find an orthonormal basis for the subspace of R 4 spanned by the vectors u = (0, 0,, 0), u 2 = (, 3, 0, 0), and u 3 = (, 0,, ). Solution. Using the GramSchmidt process, we have that v, v 2, v 3 is an orthogonal basis, where v = u = (0, 0,, 0), v 2 = u 2 = (, 3, 0, 0) (since u u 2 = 0), and v 3 = (, 0,, ) (0, 0,, 0) 0 (, 3, 0, 0) = ( 9 0, 3 0, 0, ). We have v =, v 2 = 0, and v 3 = Therefore an orthonormal basis is {(0, 0,, 0), 0 0 (, 3, 0, 0), 90 ( 9 0, 3 0, 0, )}.
25 3. Find a square root of the matrix 3 3 A = Solution. Since A is upper triangular, its eigenvalues are its diagonal entries, and A can thus 0 0 be diagonalized (since it has distinct eigenvalues). Thus, we will have S AS = 0 4 0, where S is a matrix whose columns are eigenvectors of A for the respective eigenvalues. 0 0 The matrix B = S S will then be the square root of A. Carrying out the computations, one obtains S = 0, S = 0 0, giving B = Remark: Another approach to this problem would be to assume that a square root of A, call it B, is upper triangular (not so crazy to assume, as these matrices form a subring of M 3 (C). Since det(a) = 36 then det(b) = 6. It follows that a b c B = 0 d e 0 0 6/(ad) Easy computations show that a = and d = 2. After that, all other values follow easily. 4. Let V be an ndimensional vector space and W a subspace of V. Let B = {b, b 2,..., b m } be a basis for W. Prove that there exist vectors b m+, b m+,..., b n in V such that B = {b, b 2,..., b m, b m+,..., b n } is a basis for V. Solution. Since V is ndimensional, there exist a linearly independent set {v, v 2,..., v n } that spans V. We wish to add some of the v, v 2,..., v n to B to form a basis of V. If v span(b ), then let S = B ; otherwise, set S = {b, b 2,..., b m, v }. Do this for each of the vectors v 2, v 3,..., v n (i.e., if v k S, then leave S unchanged; otherwise, add v k to S). After each step, the set S is still linearly independent, since v k was only added to the set if v k was not in the span of the previous set of vectors. After n steps, v k span(s) for all k =, 2,..., n. Since {v, v 2,..., v n } spanned V, S spans V, and thus, S is a basis for V. So, letting b m, b m+,..., b n be equal to the vectors from {v, v 2,..., v n } that were added to S and defining B = S, we are done. 5. Gaussian elimination changes Ax = b to a reduced Rx = d. The complete solution is x = 0 + s + t (a) What is the reduced row echelon form matrix R? (b) What is d? Solution. Since R is in reduced row echelon form, we must have d =
26 The other two vectors provide solutions to the homogeneous system, and show that R has rank. Since R is in reduced row echelon form, the bottom two rows must be all 0, and the top row must be orthogonal to the vectors A few computations show that this vector could be [ 2 5] T. Hence, 2 5 R = Consider the linear transformation with matrix A = Find a basis for the kernel and a basis for the image of the transformation. 3z Solution. Set Ax = 0. Then the solution space is 4z z z R. Therefore it is of 0 3 dimension, and a basis is 4. 0 Since A is 3 4, the image is a subspace of R 3. Moreover, the dimension of the image is 3 since the dimension of the kernel is. A 3dimensional subspace of R 3 must be R 3 itself, so we may use the standard basis { [00] T, [00] T, [00] T }. 7. Let A be a nonzero n n skewsymmetrix matrix. (a) Show that if n is odd, then A is not invertible. (b) What happens if n is even? Justify your answer. Solution. (a) Since the matrix A is obtained from A by multiplying each of the rows of A by , and there are an odd number of rows, det( A) = ( ) n det(a) = det(a). Since A is skewsymmetrix, det(a) = det(a T ) = det( A) = det(a). Thus, det(a) = 0, and A is not invertible. (b) If n is even, then it could go either way. For example, the matrix [ ] 0 A = 0 is a 2 2 skewsymmetric matrix that is invertible. However, the matrix 0 0 A = is skewsymmetrix, but is clearly not invertible.
27 8. (a) Show that if v is an eigenvector of the matrix product AB and Bv 0, then Bv is an eigenvector of BA. (b) Suppose that v is an eigenvector of A corresponding to an eigenvalue λ. Show that v is an eigenvector of 5I 3A + A 2, where I is the identity matrix of the same size as A. What is the corresponding eigenvalue? Solution. (a) Suppose that Bv 0 and ABv = λv for some λ R. Then A(Bv) = λv. Leftmultiply each side by B, and obtain BA(Bv) = B(λv) = λ(bv). This equation says that Bv is an eigenvector of BA, because Bv 0. (b) Suppose that Av = λv, with v 0. Then (5I 3A + A 2 )v = 5v 3Av + A(Av) = 5v 3λv + λ 2 v = (5 3λ + λ 2 )v. Hence v is an eigenvector of 5I 3A + A 2 with eigenvalue 5 3λ + λ 2.
28 Mathematics Department Qualifying Exam : Algebra Spring 20 Part A. Solve five of the following eight problems:. Let G = {x R x }. For x, y G let x y = x + y + xy. Prove that is a binary operation on G and that (G, ) is a group. 2. Consider the group S n of permutations on n elements (n 3), and let A n denote the set of even permutations in S n. Prove that A n is a normal subgroup of S n. 3. Prove that there is no ring homomorphism from Z 4 Z 2 onto Z 8 Z (a) Let G be a group, and let a G. Prove that the function φ a : G G, defined by φ a (x) = axa for all x G is an automorphism of G (it is called the inner automorphism of G induced by a). (b) Let G be a group and Inn(G) = {φ a a G} be the set of all inner automorphisms of G. Prove that Inn(G) is a group under the operation of function composition. 5. (a) Let R be a commutative ring with unity and let I be an ideal of R. Prove that r + I is a unit (invertible) in R/I if and only if there is an element s in R such that rs I. (b) Show that the ring Z Z has infinitely many zerodivisors. (c) Find all units in the ring Z Z. 6. Let R be a ring. Recall that a R is called a nilpotent element if a n = 0 for some n Z + and it is called an idempotent element if a 2 = a. (a) Show that if a is an idempotent, then a is also an idempotent. (b) If f : R S is a ring homomorphism and a R is nilpotent, prove that f carries the element a to a nilpotent element in the ring S. (c) If R is an integral domain, prove that the only idempotents in R are 0 and. (d) Show that 0 a R is a zerodivisor if and only if aba = 0 for some b Let G be a group and let Z(G) = {g G gx = xg, x G}. (a) Show that if G/Z(G) is cyclic then G is Abelian. (b) Show that if G is nonabelian with G = p 3, where p is a prime, and Z(G) {e}, then Z(G) = p. 8. Suppose G is a cyclic group with at least 3 elements. Without using the Euler φ function, prove that G has an even number of generators. Part B is on the back!!!
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