MTH5102 Spring 2017 HW Assignment 3: Sec. 1.5, #2(e), 9, 15, 20; Sec. 1.6, #7, 13, 29 The due date for this assignment is 2/01/17.

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1 MTH5102 Spring 2017 HW Assignment 3: Sec. 1.5, #2(e), 9, 15, 20; Sec. 1.6, #7, 13, 29 The due date for this assignment is 2/01/17. Sec. 1.5, #2(e). Determine whether the following sets are linearly dependent or linearly independent. (e) S = {(1, 1, 2), (1, 2, 1), (1, 1, 4)} in R 3. Solution. Suppose there exists scalars a, b, c R such that Then (0, 0, 0) = a (1, 1, 2) + b (1, 2, 1) + c (1, 1, 4). 0 = a + b + c, 0 = a 2b + c, 0 = 2a + b + 4c. We wish to solve this linear system for the unknowns a, b, c. This is equivalent to solving the matrix equation a b = c 0 To solve this we replace row 2 with row 1 plus row 2 yielding the equivalent system a b = c 0 To solve this we replace row 3 with row 3 minus 2 times row 1 yielding the equivalent system a b = c 0 Thus, we find that b = 2c and a = (b + c) = 3c. Hence, we find that all solutions are given by 3c 2c : c R. c As there are nontrivial (i.e., nonzero) solutions this implies that S is linearly dependent. 1

2 Sec. 1.5, #9. Let u and v be distinct vectors in a vector space V. Show that {u, v} is linearly dependent if and only if u or v is a multiple of the other. Proof. Let u and v be distinct vectors in a vector space V. Suppose that {u, v} is linearly dependent. Then there exists scalars a, b not both zero such that au + bv = 0. Without loss of generality we may assume that a 0. Then u = ( a 1 b ) v and so u is a multiple of v. Conversely, suppose that u or v is a multiple of the other. Without loss of generality we can assume that u is a multiple of v. Then there exists a scalar c such that implying that u = cv 1u + ( c) v = 0 implying that the set {u, v} is linearly dependent. This completes the proof. Sec. 1.5, #15. Let S = {u 1, u 2,..., u n } be a finite set of vectors. Prove that S is linearly dependent if and only if u 1 = 0 or u k+1 span ({u 1, u 2,..., u k }) for some k (1 k < n). Proof. Let S = {u 1, u 2,..., u n } be a finite set of vectors in a vector space V over a field F with n a positive integer. Suppose that S is linearly dependent. We will prove now that u 1 = 0 or u k+1 span ({u 1, u 2,..., u k }) for some k (1 k < n). Since S is linearly dependent there exists scalars a 1, a 2,..., a n F not all zero such that 0 = a 1 u 1 + a 2 u a n u n. If u 1 = 0 we are done. So suppose that u 1 0. Let k be the smallest integer in {1, 2,..., n 1} such that a k+1 0 (that such a k exists follows from this argument: suppose such a k didn t exist then a 2 = = a n = 0, but a 1 0 and since 0 = a 1 u 1 then u 1 = 0, a contradiction). Then 0 = a 1 u 1 + a 2 u a n u n = a k+1 u k+1 + v where v span ({u 1, u 2,..., u k }) implying u k+1 = a 1 k+1 v span ({u 1, u 2,..., u k }). Therefore, we have proven if S is linearly dependent then u 1 = 0 or u k+1 span ({u 1, u 2,..., u k }) for some k (1 k < n). Now we will prove the converse. Suppose that u 1 = 0 or u k+1 span ({u 1, u 2,..., u k }) for some k (1 k < n). If u 1 = 0 then S is linearly dependent since 0 = u 1 = 1u 1 with 0 1 F. 2

3 If u 1 0 then u k+1 span ({u 1, u 2,..., u k }) for some k (1 k < n) implying there exists scalars a 1, a 2,..., a k F such that so that a k+1 = a 1 u 1 + a 2 u a k u k 0 = a 1 u 1 + a 2 u a k u k + ( 1) a k+1 and as these coeffi cients are not all zero this implies S is linearly dependent. This completes the proof. Sec. 1.5, #20. Let f, g F(R, R) be the functions defined by f (t) = e rt and g (t) = e st. Prove these functions are linearly independent in F(R, R) if and only if r s. Proof. Suppose that a, b R and af + bg = 0. Then for all x R we have 0 = af (t) + bg (t) = ae rt + be st. Taking the derivative of this equality on both sides we find that 0 = are rt + bse st. This implies that ( ) ( ) e rt e st a re rt se st = b ( 0 0). This equation has a nontrivial solution, i.e., (a, b) (0, 0) if and only if ( ) e rt e 0 = det st re rt se st = se rt e st re rt e st if and only if 0 = s r. Therefore, f and g are linearly independent in F(R, R) if and only if r s. Sec. 1.6, #7. The vectors u 1 = (2, 3, 1), u 2 = (1, 4, 2), u 3 = ( 8, 12, 4), u 4 = (1, 37, 17), and u 5 = ( 3, 5, 8) generate R 3. Find a subset of the set {u 1, u 2, u 3, u 4, u 5 } that is a basis for R 3. Solution. To do this we follow the algorithm in the proof of Theorem 1.9. First, u 1 (0, 0, 0) (the zero vector of R 3 ) so that the set {u 1 } is linearly independent. Next, {u 1, u 2 } is linearly independent set since u 1 and u 2 are not scalar multiples of each other (see Exercise 9 in section 1.5). Next, as u 3 = 4u 1 we know that {u 1, u 2, u 3 } is linearly dependent, thus we check to see whether {u 1, u 2, u 4 } is linearly independent or linearly dependent. Suppose that there exists scalars a, b, c R such that (0, 0, 0) = au 1 + bu 2 + cu 4 = (2a + b + c, 3a + 4b + 37c, a 2b 17c), 3

4 that is, that is, 0 = 2a + b + c, 0 = 3a + 4b + 37c, 0 = a 2b 17c, a b = c Then this is equivalent to the system by replacing row 2 with row 2 plus 3 times row 3, that is, a b = c 0 This implies that implies Thus, for any c R we have b = 7c, 0 = 2a + b + c = 2a 6c a = 3c. (2a + b + c, 3a + 4b + 37c, a 2b 17c) = (2 (3c) + ( 7c) + c, 3 (3c) + 4 ( 7c) + 37c, (3c) 2 ( 7c) 17c) = (0, 0, 0). This implies that {u 1, u 2, u 4 } is linearly dependent. Finally, we conclude from the proof of Theorem 1.9 that we must have {u 1, u 2, u 5 } linearly independent (can also verify it). Therefore, by Corollary 2.(b) since dim ( R 3) = 3 and {u 1, u 2, u 5 } is a linearly independent subset of R 3 with three distinct elements then {u 1, u 2, u 5 } is a basis for R 3 which is a subset of {u 1, u 2, u 3, u 4, u 5 }, as desired. Sec. 1.6, #13. The set of solutions to the system of linear equations x 1 2x 2 + x 3 = 0 2x 1 3x 2 + x 3 = 0 is a subspace of R 3. Find a basis for this subspace. Solution. The system x 1 2x 2 + x 3 = 0 2x 1 3x 2 + x 3 = 0 is equivalent to the system (replace row 2 with row 2 minus 2 times row 1) x 1 2x 2 + x 3 = 0 x 2 x 3 = 0. 4

5 Thus, the set of solutions is {(c, c, c) : c R} = span {(1, 1, 1)}. As (1, 1, 1) (0, 0, 0) then {(1, 1, 1)} is a basis for this subspace of R 3. Sec. 1.6, #29. (a) Prove that if W 1 and W 2 are finite-dimensional subspaces of a vector space V, then the subspace W 1 +W 2 is finite-dimensional and dim (W 1 + W 2 ) = dim (W 1 ) + dim (W 2 ) dim (W 1 W 2 ). (b) Let W 1 and W 2 be finite-dimensional subspaces of a vector space V, and let V = W 1 + W 2. Deduce that V is the direct sum of W 1 and W 2 if and only if dim (V ) = dim (W 1 ) + dim (W 2 ). Proof. (a) First, by Exercise 23 of section 1.3, we know that if W 1 and W 2 are subspaces of a vector space V over a field F then W 1 + W 2 is a subspace of V containing both W 1 and W 2 and it is the smallest such subspace in V containing both W 1 and W 2. Hence, by Theorem 1.4 it follows that W 1 W 2 is a subspace of both V and W 1 + W 2. Thus, by Exercise 31 of section 1.3, the quotient space of W 1 +W 2 modulo W 1 W 2, i.e., (W 1 + W 2 ) / (W 1 W 2 ) is a vector space over F. Suppose now that W 1 and W 2 are finite-dimensional subspaces of V. Notice that if W 1 = {0} or W 2 = {0} then (a) is obviously true. Thus, suppose that W 1 {0} and W 2 {0}. Let β = {v 1,..., v n } and γ = {u 1,..., u m } be bases for W 1 and W 2, respectively. Then W 1, W 2 span ({v 1,..., v n } {u 1,..., u m }) W 1 + W 2. As span ({v 1,..., v n } {u 1,..., u m }) is a subspace of W 1 + W 2 containing W 1 and W 2 it follows from Exercise 23 of section 1.3 that span ({v 1,..., v n } {u 1,..., u m }) = W 1 + W 2. In particular, this implies S = {v 1,..., v n } {u 1,..., u m } generates W 1 + W 2 and so dim (W 1 + W 2 ) dim [span ({v 1,..., v n } {u 1,..., u m })] n + m = dim (W 1 ) + dim (W 2 ) so that dim (W 1 + W 2 ) is finite-dimensional. Next, it follows that W 1 / (W 1 W 2 ), W 2 / (W 1 W 2 ) are subspaces of (W 1 + W 2 ) / (W 1 W 2 ) with implying that [W 1 / (W 1 W 2 )] [W 2 / (W 1 W 2 )] = {W 1 W 2 }, (W 1 + W 2 ) / (W 1 W 2 ) = W 1 / (W 1 W 2 ) + W 2 / (W 1 W 2 ) (W 1 + W 2 ) / (W 1 W 2 ) = W 1 / (W 1 W 2 ) W 2 / (W 1 W 2 ) 5

6 so that [we appeal to the first part of part (b) here, which we prove below, which didn t require the statement (a) to prove it] ( ) dim [(W 1 + W 2 ) / (W 1 W 2 )] = dim [W 1 / (W 1 W 2 )]+dim [W 2 / (W 1 W 2 )]. Next, we will prove (*): if W is a finite-dimensional vector space with a subspace U that dim (W/U) = dim (W ) dim (U). To prove this let α be a basis for U then by Corollary 2.(c) we can extend α to a basis α for W and it can be easily shown that α \α + U = {x + U : x α \α} is a basis for W/U which implies dim (W/U) = α \α = α α = dim (W ) dim (U), where denotes here the number of distinct elements in a set. Thus, (*) is proven. Finally, it follows from (*) that ( ) dim [(W 1 + W 2 ) / (W 1 W 2 )] = dim (W 1 + W 2 ) dim (W 1 W 2 ), dim [W 1 / (W 1 W 2 )] = dim (W 1 ) dim (W 1 W 2 ), dim [W 2 / (W 1 W 2 )] = dim (W 2 ) dim (W 1 W 2 ) and therefore, by (**) and (***) we have dim (W 1 + W 2 ) = dim [(W 1 + W 2 ) / (W 1 W 2 )] + dim (W 1 W 2 ) [by (***)] = dim [W 1 / (W 1 W 2 )] + dim [W 2 / (W 1 W 2 )] + dim (W 1 W 2 ) [by (**)] = dim (W 1 ) dim (W 1 W 2 ) + dim (W 2 ) dim (W 1 W 2 ) [by (***)] + dim (W 1 W 2 ) = dim (W 1 ) + dim (W 2 ) dim (W 1 W 2 ). This proves (a). We now will prove (b). Let W 1 and W 2 be finite-dimensional subspaces of a vector space V, and let V = W 1 + W 2. Suppose that V is the direct sum of W 1 and W 2, i.e., W 1 W 2 = {0} and V = W 1 + W 2. Let β = {v 1,..., v n } and γ = {u 1,..., u m } be bases for W 1 and W 2, respectively. We claim that β γ is a basis for V and β γ =. First, as we proved above span (β γ) = W 1 + W 2. Hence β γ generates V. Next, if 0 = n i=1 a iv i + m for some scalars a 1,..., a n, b 1,..., b m then n i=1 a iv i = m i=1 b iu i i=1 b iu i W 1 W 2 = {0} implying 0 = n a iv i, 0 = m b iu i i=1 i=1 implying a 1 = = a n = b 1 = = b m = 0 implying β γ is linearly independent and in particular it follows that β γ =. Therefore, β γ is a basis for W 1 + W 2 = V so that dim (V ) = β γ = β\γ + γ = β + γ = dim (W 1 ) + dim (W 2 ). 6

7 Conversely, if dim (V ) = dim (W 1 ) + dim (W 2 ), where V = W 1 + W 2 then by part (a) we have implying implying dim (W 1 + W 2 ) = dim (W 1 ) + dim (W 2 ) dim (W 1 W 2 ) dim (W 1 W 2 ) = 0 W 1 W 2 = {0} implying V is the direct sum of W 1 and W 2. This completes the proof of part (b). Therefore, the statements (a) and (b) are true. 7