M A T H F A L L CORRECTION. Algebra I 1 4 / 1 0 / U N I V E R S I T Y O F T O R O N T O

Transcription

1 M A T H F A L L HOMEWORK ASSIGNMENT #4 CORRECTION Algebra I 1 4 / 1 0 / U N I V E R S I T Y O F T O R O N T O P r o f e s s o r : D r o r B a r - N a t a

2 Correctio Homework Assigmet #4 Exercise 8 page 41-42: Let S = {(1,1,0), (1,0,1), (0,1,1)} be a subset of the vector space F 3 (a) Prove that if F = R, the S is liearly idepedet Let S = {(1,1,0), (1,0,1), (0,1,1)} be a subset of the vector space R 3 Suppose a 1, a 2, a 3 R such that (0,0,0) = a 1 (1,1,0) + a 2 (1,0,1) + a 3 (0,1,1) We obtai the followig system of equatio: a 1 + a 2 = 0 a 1 + a 3 = 0 2a 1 = 0 a 1 + a 3 = 0 + a 1 = 0 a 1 + a 3 = 0 1/2 a 1 = 0 a 3 = 0 a 1 = 0 a 3 = 0 a 2 = 0 Therefore the oly solutio for this system is a 1 = a 2 = a 3 = 0. Si S is liearly idepedet. (b) Prove that if F has characteristic 2, the S is liearly depedet Let S = {(1,1,0), (1,0,1), (0,1,1)} be a subset of the vector space F 3 where F is a field with characteristic 2. Sice F is a field with characteristic 2 we have that: = 0. Therefore if we cosider the liear combiatio of the vectors of S where a 1, a 2, a 3 F ad a 1 = a 2 = a 3 = 1, we get: a 1 (1,1,0) + a 2 (1,0,1) + a 3 (0,1,1) = 1 (1,1,0) + 1 (1,0,1) + 1 (0,1,1) = (1,1,0) + (1,0,1) + (0,1,1) = (1 + 1,1 + 1,1 + 1) = (0,0,0) Therefore, i a field with characteristic 2, a 1 = a 2 = a 3 = 1 is a fiite liear combiatio of vectors of S whose coefficiets are ot all equals to 0. Hece, S is liearly depedet. Exercise 9 page 41: Let u ad v be distict vectors i a vector space V. Show that {u, v} is liearly depedet if ad oly if u or v is a multiple of the other. Let u ad v be distict vectors i a vector space V over a field F. We show first that u or v is a multiple of the other {u, v} liearly depedet. - Suppose u is a multiple of v. The there exists q F such that u = vq. From this, we have that: u vq = 0 V. Thus, there exists a o trivial represetatio of the zero vector as a liear combiatio of distict vectors of {u, v} sice we have a 1 = 1. - Suppose v is a multiple of u. The there exists q F such that v = uq. From this, we have that: v uq = 0 V. Thus, there exists a o trivial represetatio of the zero vector as a liear combiatio of distict vectors of {u, v} sice we have a 1 = 1. Hece u or v is a multiple of the other {u, v} liearly depedet. 2

3 Correctio Homework Assigmet #4 Now let us show that {u, v} liearly depedet u or v is a multiple of the other Suppose {u, v} liearly depedet. The there exists a, b F such that a, b are ot both equal to zero such that au + bv = 0 V - Suppose a 0. The au + bv = 0 V au = bv. Moreover, sice a 0, a 1 F such that a a 1 = 1. Thus, we get: au = bv a 1 au = a 1 ( bv) 1 u = a 1 ( bv) u = ( ba 1 )v. Therefore u is a multiple of v. - Suppose b 0. The au + bv = 0 V bv = au. Moreover, sice b 0, b 1 F such that b b 1 = 1. Thus, we get: bv = au b 1 bv = b 1 ( au) 1 v = b 1 ( au) v = ( ab 1 )u. Therefore v is a multiple of u. Thus {u, v} liearly depedet u or v is a multiple of the other. Hece u or v is a multiple of the other {u, v} liearly depedet. Exercis1 page 42: Let S = {u 1, u 2,, u } be a liearly idepedet subset of a vector space V over a field Z 2. How may vectors are there i Spa(S)? Justify your aswer. Let S = {u 1, u 2,, u } be a liearly idepedet subset of a vector space V over the field Z 2. Let deote by the umber of elemets of S which we deote by S. If = 0, the S = (possible sice the empty set is liearly idepedet) the Spa(S) = {0}. Therefore, Spa(S) = 1 Suppose ow that 1. We have S = {(1)} because {(0)} is liearly depedet ad ay oempty subset of a vector space that cotai 0 is liearly depedet. Moreover we have (1) = 1 (1) ad (0) = 0 (1). Thus Spa(S) = Z 2 ad it has 2 elemets. So the umber of elemets of the spa of a liearly idepedet subset of a vector space V over a field Z 2 is 1 if the umber of elemets the subset is 0 ad 2 is the umber of elemet of the subset is greater tha 1. Geeral case for a liearly idepedet subset of a vector space V over a field Z 2 k with k a positive iteger Let S = {u 1, u 2,, u } be a liearly idepedet subset of a vector space V over the field Z 2 k. We claim that the umber of vectors i Spa(S) = 2. We will show it by iductio over S =. If = 0, the S = (possible sice the empty set is liearly idepedet) the Spa(S) = {0 V }. Therefore, Spa(S) = 1 = 2 0 Base case: We ow suppose that = 1. So S = {u 1 } is o empty. If v spa(s), there exists a Z 2 such that v = au 1 Sice Z 2 has oly two elemets we have that a = 1 or a = 0. Therefore, spa(s) = {u 1, 0 V } ad Spa(S) = 2 = 2 1 Suppose that Spa(S) = 2 for ay liearly idepedet subset of a vector space V over Z 2 k such that S = 1. Let us show that it is true for + 1. Suppose that S {u +1 } is liearly idepedet. Moreover S S {u +1 } Spa(S) Spa(S {u +1 }) Spa(S) = 2 Spa(S {u +1 }) sice S is fiite (ad thus so is 3

4 Correctio Homework Assigmet #4 its spa). We wat to kow how much ew vectors does u +1 brigs to the spa of this ew set. Suppose that v Spa(S u +1 ). The v = a i u i + a +1 u +1 For every ew vector of Spa(S u +1 ) we must have that a +1 = 1 otherwise v was already i Spa(S). So therefore we have that: v = a i u i + u +1 We claim that ay vector of this form is ot already i Spa(S). I fact if it were i the Spa(S) we would have that: v = b i u i This implies that (v + v = 0 becaus + 1 = = 0): v + v = 0 = b i u i + a i u i + u +1 = (a i + b i )u i + u +1 This would implies that there exists a o trivial liear combiatio of 0 with distict vectors of S u +1 sice a +1 = 1. Thus it cotradicts the fact that S u +1 is liearly idepedet. Thus every ew vectors i S u +1 is of the form: v = a i u i + u +1 Sice there ar vectors of the form a i u i sice Spa(S) = 2 by our iductio hypothesis. Therefore, there ar vectors of the form a i u i + u +1. Sice these vectors does ot belogs to Spa(S) we have that: Spa(S u +1 ) = = 2 +1 Hece, we proved that Spa(S) = 2 for ay liearly idepedet subset of a vector space V over Z 2 k such that S = 1. Actually as we proved it before, this is also true for = 0. Exercise 4 page 54: Do the polyomials x 3 2x 2 + 1,4x 2 x + 3, ad 3x 2 geerate P 3 (R)? Justify your aswer. Let us cosider the subset S = {x 3 2x 2 + 1,4x 2 x + 3, 3x 2} of the vector space P 3 (R). Moreover, we have that dim(p 3 (R)) = 4. By the corollary 2 of the replacemet theorem, we have that ay fiite geeratig set of a vector space of fiite dimesio should at least cotais elemets. Here we have that S = 3 < dim(p 3 (R)) = 4. Therefore, S does ot geerates P 3 (R). Exercise 5 page 54: Is {(1,4, 6), (1,5,8), (2,1,1), (0,1,0)} a liearly idepedet subset of R 3? Justify your aswer. Let us cosider the subset S = {(1,4, 6), (1,5,8), (2,1,1), (0,1,0)} of the vector space R 3. 4

5 Correctio Homework Assigmet #4 Moreover, we have that dim(r 3 ) = 3. Let be the dimesio of a fiite dimesioal vector space. By the corollary 2 of replacemet theorem, we kow that if B is a basis the vector space the B =. Moreover a basis geerates the vector space by defiitio. Besides, by the replacemet theorem, we kow that if L is a liearly idepedet subset of the vector space that cotais m elemets ad G is a geeratig subset of the vector space, the L G. So i particular for the basis we have L = B. I this case, S = 4 > dim(r 3 ) = 3 so S is ot liearly idepedet. Exercise 9 page 55: The vectors u 1 = (1,1,1,1), u 2 = (0,1,1,1), u 3 = (0,0,1,1) ad u 4 = (0,0,0,1) for a basis for F 4. Fid the uique represetatio of a arbitrary vector (a 1, a 2, a 3, a 4 ) i F 4 as a liear combiatio of u 1, u 2, u 3 ad u 4. Sice B = {u 1, u 2, u 3, u 4 } is a basis o F 4 ad that (a 1, a 2, a 3, a 4 ) F 4, the! b 1, b 2, b 3, b 4 such that : (a 1, a 2, a 3, a 4 ) = b 1 u 1 + b 2 u 2 + b 3 u 3 + b 4 u 4 This implies the followig system of equatio: b 1 + b 2 = a 2 b 1 + b 2 + b 3 = a 3 b 1 + b 2 + b 3 + b 4 = a 4 e 4 b 1 + b 2 = a 2 b 1 + b 2 + b 3 = a 3 b 4 = a 4 a 3 e 4 b 1 + b 2 = a 2 b 3 = a 3 a 2 b 4 = a 4 a 3 e 4 b 2 = a 2 a 1 b 3 = a 3 a 2 b 4 = a 4 a 3 e 4 Therefore, the uique represetatio of a arbitrary vector (a 1, a 2, a 3, a 4 ) F 4 is (a 1, a 2 a 1, a 3 a 2, a 4 a 3 ) as a liear combiatio of u 1, u 2, u 3 ad u 4. Exercis2 page 55: Let u, v ad w be distict vectors of a vector space V. Show that if {u, v, w} is a basis for V the {u + v + w, v + w, w} is also a basis for V Let u, v ad w be distict vectors of a vector space V. Suppose that B = {u, v, w} is a basis for V. First otice that u + v + w, v + w, w are distict vectors sice: - u + v + w = v + w u = 0 B = {0, v, w} B is liearly depedet. This cotradicts the fact that B is a basis. u + v + w v + w - v + w = w v = 0 B = {u, 0, w} B is liearly depedet. This cotradicts the fact that B is a basis v + w w - u + v + w = w u + v = 0 There exists a o trivial liear combiatio of 0 with vectors of B B is liearly depedet. This cotradicts the fact that B is a basis. u + v + w w 5

6 Correctio Homework Assigmet #4 Therefore u + v + w, v + w, w ar distict elemets ad B = {u + v + w, v + w, w} has 3 elemets. Moreover let a, b, c F such that: 0 = a(u + v + w) + b(v + w) + cw = au + (a + b)v + (a + b + c)w Sice B is a basis we have that the oly liear combiatio of vectors of B that gives the zero vector is the liear combiatio where all the coefficiets equal zero. So by idetificatio we have that a = 0 a + b = 0 a + b + c = 0 a = 0 b = 0 c = 0 Therefore, the oly combiatio of vectors of B that gives the zero vector is the combiatio where all coefficiet are equal to 0. Therefore B is liearly idepedet. Sice B is liearly ad has exactly 3 elemets, B is a basis of V by corollary 2 of the replacemet theorem. 6