Name: Solutions 1. (5 + 5 + 15 points) Let F = F 7. Recall we defined for a positive integer n a vector space P n (F ) by P n (F ) = {f(x) F [x] : deg(f) n}. (a) What is the dimension of P n (F ) over F? How many elements are in P n (F )? Give a short justification for your answer. A basis for P n (F ) is given by {1, x, x 2,..., x n }. There are n + 1 elements in the basis, so P n (F ) has dimension n + 1 over F. Since elements of P n (F ) are polynomials of degree less than or equal to n, there are 7 choices for each coefficient. Thus, there are 7 n+1 elements in P n (F ). (b) Let V = P 3 (F ). Consider bases of V given by B = {1, x, x 2, x 3 } and C = {1, (x + 1), (x + 1) 2, (x + 1) 3 }. Define T Hom F (V, V ) by Determine the matrix [T ] C B. T (1) = 1 + (x + 1) + 3(x + 1) 2 + 2(x + 1) 3, T (x) = 4 + (x + 1) + 5(x + 1) 2 + 2(x + 1) 3, T (x 2 ) = 3(x + 1) + 6(x + 1) 2 + 6(x + 1) 3, T (x 3 ) = 1 + (x + 1) + 2(x + 1) 2 + 2(x + 1) 3. The matrix is given by 1 4 1 A = [T ] C B = 1 1 3 1 3 5 6 2. 2 2 6 2 (c) Give a basis for the kernel and image of T. We put [T ] C B in row echelon form. This is easily calculated to be 1 4 1 U = 1 1 1 1. We have the image of A is the same as the image of U, which has as a basis the 1 4 vectors w 1 =, w 2 = 1, and w 3 = 1 1. Translating this back to T we have w 1 corresponds to 1, w 2 corresponds to 4+(x 1), and w 3 corresponds to (x 1) (x 1) 2. Thus, a basis for the image of T is given by {1, 4 + (x 1), (x 1) + (x 1) 2 }.
2 The next step is to find a basis for the kernel of T. Note the kernel of A is the same as the kernel of U. Suppose v = a 1 a 2 a 3 a 4 is in the kernel of U. Then we have the equations a 1 + 4a 2 + a 4 =, a 2 a 3 =, and a 3 a 4 =. Set a 2 = t. Then the kernel is spanned 3 by the vector. Translating this back to T we have ker(t ) has 3 + x + x2 x 3 as a basis. 1 1 1 2. (8+8+9 points) Let V be a finite dimensional F -vector space. Let T Hom F (V, V ). (a) Define the minimal polynomial m T (x) of T. Given v V, define the annihilating polynomial m T,v (x). Prove that for any v V, m T,v (x) m T (x). The minimal polynomial is the monic polynomial m T (x) of minimal degree so that m T (T )(w) = for all w V. The annihilating polynomial of a specific element v V is the monic polynomial m T,v (x) of minimal degree so that m T,v (T )(v) =. We have that since m T (T )(v) = and m T,v (x) has minimal degree amongst such polynomials, we must have deg(m T,v (x)) deg(m T (x)). Using the division algorithm we can write m T (x) = m T,v (x)q(x) + r(x) for some polynomials q, r F [x] with r(x) = or deg(r(x)) < deg(m T,v (x)). This gives = m T (T )(v) = m T,v (T )(v) + r(t )(v) = r(t )(v). If r(x), then we can scale r so that it is a monic polynomial of degree smaller than m T,v (x) but r(t )(v) =. This contradicts the minimality of the degree of m T,v (x), so it must be that r(x) = and thus m T,v (x) mt (x). Recall that we say λ F is an eigenvalue of T if there exists a nonzero vector v V so that T (v) = λv. We call such a v an eigenvector. (b) Prove that λ is an eigenvalue of T if and only if λ is a root of the minimal polynomial m T (x). First, suppose that λ is an eigenvalue of T with nonzero eigenvector v. Then we have T (v) = λv, i.e., if we set f(x) = x λ, then f(t )(v) =. This gives that m T,v (x) f(x). Since v, deg(m T,v (x)) 1. Thus, m T,v (x) = x λ. Since m T,v (x) m T (x) and λ is a root of m T,v (x), it is a root of m T (x). Now let λ be a root of m T (x). We can write m T (x) = (x λ)q(x) for some q(x) F [x]. Since m T (x) is the minimal polynomial and deg(q(x)) < deg(m T (x)), there exists w V
3 so that q(t )(w). Set v = q(t )(w). Observe we have = m T (T )(w) = (T λ)q(t )(w) = (T λ)(v). Thus, T (v) = λv with v, so λ is an eigenvalue of T. 3. (25 points) Let F be a field. Prove that F [x] F F [y] = F [x, y]. The first step is to produce a well-defined linear map F [x] F F [y] F [x, y]. We use the universal property to do this. Define t : F [x] F [y] F [x, y] by t((f(x), g(y))) = f(x)g(y). This is clearly well-defined so we need to show this is bilinear. Let f 1 (x), f 2 (x) F [x], g(y) F [y], and c F. Then we have t((cf 1 (x) + f 2 (x), g(y))) = (cf 1 (x) + f 2 (x))g(y) = cf 1 (x)g(y) = f 2 (x)g(y) = ct((f 1 (x), g(y)) + t((f 2 (x), g(y))). Thus, t is linear in the first variable. The second variable follows in the same manner. Thus, t is bilinear. The universal property for tensor products gives a well-defined linear map T : F [x] F F [y] F [x, y] so that T (f(x) g(y)) = f(x)g(y). To see this is an isomorphism, we define an inverse linear map. Define S : F [x, y] F [x] F F [y] as follows. It is easy to see that to define a linear map on F [x, y], it is enough to define the map on x i y j for each i, j. Thus, we define S(x i y j ) = x i y j. It remains to show these are inverse maps. To do this, we first study the elements of F [x] F F [y] a bit. Let f(x) = m i=1 a ix i F [x] and g(y) = n j=1 b jy j. The we have f(x) g(y) = m n a i b j (x i y j ). i=1 j=1 Moreover, a general element of F [x] F F [y] is a sum of such elements. This makes it easy to see that it is enough to check S T is the identity on elements of the form x i y j and T S is the identity on elements of the form x i y j. However, this is obvious from the definition of S and T. Thus, we have the isomorphism.
4 4. (8 + 8 + 9 points) (a) Let T : R 3 R 3 be the linear map so that 2 [T ] E 3 E 3 = 1 1 1 3 2 where E 3 = {e 1, e 2, e 3 } is the standard basis of R 3. Compute the determinant of T by using the definition of the determinant given in this class. Recall given T Hom F (V, V ), for each κ 1 we have an induced map Λ k (T ) : Λ k (V ) Λ k (V ) given by Λ k (T )(v 1 v k ) = T (v 1 ) T (v k ). To compute the determinant, we evaluate Λ n (T ) if n = dim F V. We have that T (e 1 ) = e 2, T (e 2 ) = 2e 1 + e 2 + 3e 3, and T (e 3 ) = e 2 + 2e 3. We compute Thus, det T = 4. det(t )e 1 e 2 e 3 = Λ 3 (T )(e 1 e 2 e 3 ) = T (e 1 ) T (e 2 ) T (e 3 ) = (e 2 ) (2e 1 + e 2 + 3e 3 ) (e 2 + 2e 3 ) = (e 2 2e 1 + e 2 3e 3 ) (e 2 + 2e 3 ) = e 2 2e 1 2e 3 = 2e 1 e 2 2e 3 = 4e 1 e 2 e 3. (b) Give a basis for Λ 2 (R 3 ). Calculate the matrix of Λ 2 (T ) with respect to this basis. Elements of Λ 2 (R 3 ) are finite sums of elements of the form v 1 v 2 for v i R 3. We saw in class a basis is given by e 1 e 2, e 1 e 3, and e 2 e 3. We have Similarly, we compute Λ 2 (T )(e 1 e 2 ) = T (e 1 ) T (e 2 ) Λ 2 (T )(e 1 e 3 ) = 2e 2 e 3 = e 2 (2e 1 + e 2 + 3e 3 ) = 2e 1 e 2 + 3e 2 e 3. Λ 2 (T )(e 2 e 3 ) = 2e 1 e 2 + 4e 1 e 3 e 2 e 3. Thus, if we set B = {e 1 e 2, e 1 e 3, e 2 e 3 } then we have 2 2 [Λ 2 (T )] B B = 4. 3 2 1
5 (c) Now consider finite dimensional F -vector spaces V and W. Let T Hom F (V, W ). Prove that if T is a surjection, then Λ k (T ) is a surjection for all k 1. Prove that if T is an isomorphism, then Λ k (T ) is an isomorphism for all k 1. We know that the map Λ k (T ) is a linear map, so to see it is injective it is enough to show that every basis element of Λ k (W ) is in the image of Λ k (T ). Let C = {w 1,..., w n } be a basis of W. Since T is surjective, for each 1 i n there exists v i V so that T (v i ) = w i. Now a basis for Λ k (W ) is given by {w i1 w ik } where i 1 < i 2 < i k. Consider the element v i1 v ik Λ k (V ). By definition we have Λ k (T )(v i1 v ik ) = T (v i1 ) T (v ik ) = w i1 w ik. Thus, if T is surjective then Λ k (T ) is surjective for each k 1. If T is an isomorphism then dim F V = dim F W. This gives dim F Λ k (V ) = dim F Λ k (W ). Since Λ k (T ) is a surjection between finite dimensional vector spaces of the same dimension, it must be an isomorphism.