Midterm 1 Advanced Linear Algebra Math 4377 / 638 (Spring 215) March 5, 215 2 points 1. Mark each statement True or False. Justify each answer. (If true, cite appropriate facts or theorems. If false, explain why or give a counterexample that shows why the statement is not true in every case). (1) If S is a linearly dependent set, then each vector in S is a linear combination of other vectors in S. (2) Any set containing the zero vector is linearly dependent. (3) Subsets of linearly dependent sets are linearly dependent. (4) Subsets of linearly independent sets are linearly independent. (5) Every vector space that is generated by a finite set has a basis. (6) Every vector space has a finite basis. (7) If a vector space has a finite basis, then the number of vectors in every basis is the same. (7 ) The dimension of M m n (F ) is m + n. (8) Suppose that V is a finite-dimensional vector space, that S 1 is a linearly independent subset of V, and that S 2 is a subset of V that generates V. Then S 1 cannot contain more vectors than S 2. (9) If V is a vector space having dimension n, and if S is a subset of V with n vectors, then S is linearly independent if and only if S spans V. (1) If T : V W is linear, then nullity(t ) + rank(t ) = dim(w ). 2 points 2. The first four Chebyshev polynomials are 1, x, 2x 2 1, and 4x 3 3x. These polynomials arise naturally in the study of certain important differential equations. Show that the first four Chebyshev polynomials form a basis of P 3 (R). 2 points 3. Let T : R 3 R 3 be given by T (a, b, c) = (3a, 2a + c, b). Prove that T is an isomorphism and find T 1. 2 points 4. Let T : R 3 R 2 be given by T (a, b, c) = (a b, 2c). Show that T is a linear transformation. (b) Find bases for the null space and the range of T. (c) Compute the nullity and rank of T, and verify the dimension theorem. 2 points 5. Let W 1 and W 2 be subspaces of a vector space V. Prove that W 1 + W 2 is a subspace of V that contains both W 1 and W 2. (b) Prove that any subspace of V that contains both W 1 and W 2 must also contain W 1 + W 2. Page 1 of 5 Please go to the next page...
Midterm 1 (cont.) Math 4377 (15549) / 638 (14674) (215 Spring) March 5, 215 (c) Suppose W 1 = span{u 1,, u p }, W 2 = span{v 1,, v q }. where u 1,, u p and v 1,, v q are vectors in V. Show that W 1 + W 2 = span{u 1,, u p, v 1,, v q }. 2 points 6. (BONUS PROBLEM) Let V and W be finite-dimensional vector spaces and T : V W be an isomorphism. Let V be a subspace of V. Prove that T (V ) is a subspace of W. (b) Prove that dim(v ) = dim(t (V )). Page 2 of 5 Please go to the next page...
Midterm 1 (cont.) Math 4377 (15549) / 638 (14674) (215 Spring) March 5, 215 Problem 1. (1) False. (2) True. (3), False. (4) True. (5) True. (6) False. (7) True. (7 ) False. (8) True. (9) True. (1) False. Problem 2. Let β = {1, x, 2x 2 1, 4x 3 3x} and let γ = {1, x, x 2, x 3 } be the standard ordered basis for P 3 (R). We have the coordinate vectors of β in γ as: 1 [1] γ =, [x] γ = 1, [ 1 + 2x2 ] γ = 1 2 [ 3x + 4x3 ] γ = Note that the matrix with the coordinate vectors as columns have four pivots 1 1 Q = 1 3 2 4 3 4 Then {[1] γ, [x] γ, [ 1 + 2x 2 ] γ, [ 3x + 4x 3 ] γ } is linearly independent. By Thorem 2.21, β is linearly independent. Combined with the fact that β = dim(p 3 (R)) = 4, β is a basis for P 3 (R). Page 3 of 5 Please go to the next page...
Midterm 1 (cont.) Math 4377 (15549) / 638 (14674) (215 Spring) March 5, 215 Problem 3. Let β = {e 1, e 2, e 3 } be the standard ordered basis for R 3. The matrix representation of T in β is 3 [T ] β = 2 1 1 Note that the augmented matrix 3 1 1 1/3 [[T ] β I 3 ] = 2 1 1 1 1 = [ I 3 ([T ] β ) 1] 1 1 1 2/3 1 So we have 1/3 [T ] 1 β = 1 2/3 1 By Theorem 2.18, T is invertible and [T 1 ] β = ([T ] β ) 1. We have T 1 (a, b, c) = (a/3, c, 2a/3 + b). Problem 4. Note that, in the matrix and column vector notation, we have ( ) 1 1 x R 3 T (x) = Ax, A =. 2 Then, T is linear. (b) To find the null space of T, row reduce the augmented matrix corresponding to Ax = ( ) ( ) 1 1 1 1 2 1 We have Then x 1 x 2 1 x 2 = x 2 = x 2 1 x 3 1 the null space of T = span 1 The range of T is the column space of A and we have the range of T = span {pivot columns of A} = span (c) The nullity of T is 1 and the rank of T is 2 We have Then, the dimension theorem is verified. nullity(t ) + rank(t ) = 1 + 2 = 3 = dim(r 3 ). {( ) ( )} 1, 1 Page 4 of 5 Please go to the next page...
Midterm 1 (cont.) Math 4377 (15549) / 638 (14674) (215 Spring) March 5, 215 Problem 5. W 1 +W 2 is a subspace of W : Closed under vector addition, because if u, v W 1 +W 2, then there exist u 1, v 1 W 1 and u 2, v 2 W 2 such that u = u 1 + u 2 and v = v 1 + v 2, and then u + v = u 1 + u 2 + v 1 + v 2 = (u 1 + v 1 ) + (u 2 + v 2 ) W 1 + W 2. For scalar multiplication, au = a(u 1 + u 2 ) = au 1 + au 2 W 1 + W 2. Finally, W 1 + W 2 contains since both W 1, W 2 are subspaces and therefore contain. W 1 + W 2 contains both W 1 and W 2 : Every vector in W 1 + W 2 has the form x + y with x W 1, y W 2. Set y = to obtain all vectors in W 1 and x = to obtain all vectors in W 2. That is, any vector x W 1 or y W 2 is also present in W 1 + W 2. (b) A subspace W of V that contains both W 1 and W 2 must also contain all vectors of the form x + y with x W 1, y W 2, since it is closed under addition. Therefore it contains W 1 + W 2. (c) ( ) W 1 + W 2 span{u 1,, u p, v 1,, v q }: For any u W 1 = span{u 1,, u p } and v W 2 = span{v 1,, v q }, there exist c 1,, c p and d 1,, d q such that Then u = c 1 u 1 + + c p u p, v = d 1 v 1 + + d q v q. u + v = c 1 u 1 + + c p u p + d 1 v 1 + + d q v q span{u 1,, u p, v 1,, v q } ( ) span{u 1,, u p, v 1,, v q } W 1 +W 2 : For any w span{u 1,, u p, v 1,, v q }, there exist c 1,, c p and d 1,, d q such that w = (c 1 u 1 + + c p u p ) + (d 1 v 1 + + d q v q ) W 1 + W 2 Problem 6. (BONUS PROBLEM) (1) T (V ) contains W, since V V and T ( V ) = W. (2) Let u 1, u 2 T (V ), then there exist v 1, v 2 V such that T (v 1 ) = u 1 and T (v 2 ) = u 2. Then v 1 + v 2 V, and T (V ) T (v 1 + v 2 ) = T (v 1 ) + T (v 2 ) = u 1 + u 2. (3) Similarly for scalar multiplication, let u T (V ), then there exists v V such that T (v) = u. Then av V, and T (V ) T (av) = at (v) = au. Combining (1)-(3) shows that T (V ) is a subspace of W. (a ) Let β = {u 1,, u n } be a basis for V. T being linear, we have T (V ) = {T (v), v V } = {T (a 1 u 1 + + a n u n ), a 1,, a n F } = {a 1 T (u 1 ) + + a n T (u n ), a 1,, a n F } = span{t (u 1 ),, T (u n )} Then T (V ) is a subspace of W. (b) Let β = {u 1,, u n } be a basis for V. T (β) is then a basis for T (V ) (from (a )), since it spans T (V ) and its vectors are linearly independent: a 1 T (u 1 ) + + a n T (u n ) = T (a 1 u 1 + + a n u n ) = gives a 1 u 1 + + a n u n = since T is an isomorphism, and a 1 = = a n = since β is a basis for V. Thus, n = dim(v ) = dim(t (V )). When you finish this exam, you should go back and reexamine your work for any errors that you may have made. Page 5 of 5 End of exam.