Mah 35: Linear Algebra s o Assignmen 6 # Which of he following ses of vecors are bases for R 2? {2,, 3, }, {4,, 7, 8}, {,,, 3}, {3, 9, 4, 2}. Explain your answer. To generae he whole R 2, wo linearly independen vecors are needed. We go wo vecors in each of hese pairs. Hence, a given pair generaes R 2 if he vecors are linearly independen. Two non-zero vecors are linearly dependen if one of hem is a muliple of he oher, and independen oherwise. If a pair conains a zero vecor, he vecors of he pair are dependen. Therefore, only he firs wo pairs of vecors are independen. The hird pair conains a zero vecor. The vecors in he fourh pair are muliples of each oher. #2 Find a vecor ha can be added o he se {v, v 2 } o produce a basis for R 3. v =, 4, 2, v 2 = 3, 8, 4. Pu hose vecors as rows ino a marix and bring i o he row-echelon form. This resuls in 4 2 3 8 4 4 2.5 As you see, he resuling marix has wo leading ones in columns and 2. Adding o i a vecor wih a leading one in he hird column, e.g., vecor,, solves he problem.
#3 Which of he following ses of vecors are bases for P 2? a 3x + 2x 2, + x + 4x 2, 7x b + x + x 2, x + x 2, x 2. Explain your answer. Conver he polynomials ino he vecor represenaions and pu he obained vecors as rows ino a 3 x marix. This resuls in marices, respecively A = 3 2 4 7 and B = 2 To span P 2, hree vecors are needed. A se of 3 vecors is a basis of P 2 if he vecors are linearly independen. The vecors are linearly independen, if he corresponding marices have non-zero deerminans. Since dea = and deb = 2, only he firs se forms a basis of P 2. #4 Deermine he dimension of and a basis for he soluion space of he sysem a x + x 2 x 3 = 2x x 2 + 2x 3 = x + x 3 = b x 4x 2 + 3x 3 x 4 = 2x 8x 2 + 6x 3 2x 4 = The reduced row-echelon forms of he involved marices are, respecively 4 3 A = and B = The soluion se for he sysem A is given by =, so he soluion se is of dimension and,, is is basis. The soluion se for he sysem B is given by 4k 3l + k l = k 4 + l 3 + so he soluion se is of dimension 3 and he shown vecors form is basis. 2,
#5 Le {v, v 2, v 3 } be a basis for a vecor space V. Prove ha {u, u 2, u 3 } is also a basis, where u = v, u 2 = v + v 2, u 3 = v + v 2 + v 3. Since V = spanv, v 2, v 3, one has dimv = 3. Hence, he se of hree vecors {u, u 2, u 3 } is a basis of V iff hese vecors are linearly independen. Assume, for he sake of conradicion, ha hey are linearly dependen. So, here exiss consans λ, λ 2, λ 3, no all equal o, such ha λ u + λ 2 u 2 + λ 3 u 3 =. Using he represenaion of u, u 2, u 3 in erms of v, v 2, v 3, one derives which implies λ v + λ 2 v + v 2 + λ 3 v + v 2 + v 3 =, λ + λ 2 + λ 3 v + λ 2 + λ 3 v 2 + λ 3 v 3 =. Bu he vecors v, v 2, v 3 are linearly independen, so only heir rivial linear combinaion resuls in he zero vecor. This implies λ + λ 2 + λ 3 = λ 2 + λ 3 = λ 3 = which implies λ = λ 2 = λ 3 =, which conradics he fac ha no all λ, λ 2, λ 3 are and complees he proof. #6 Deermine if b is in he column space of A, and if so, express b as linear combinaion of column vecors of A. [ ] [ ] 3 2 A =, b =. 4 6 The reduced row-echelon form of he augmened marix of he sysem Ax = b is 3 2 4 6 which implies he soluion x, x 2 =,. Hence b = x c + x 2 c 2 = c c 2. 3
#7 Find a basis for he row space, column space and null space of A = Firs, compue he reduced row-echelon form of A: 6 9 3 5 4 4 7 6 2 The non-zero rows of his marix form he basis of he row-space. The columns of he original marix, corresponding o he leading ones ha is firs wo columns form a basis of he column space. Concerning he soluion se, i is given by 6, 9,, so he vecor 6, 9, forms is basis.. #8 In each par use he informaion in he able o find he dimension of he row space of A, he column space of A, he null-space of A, and he null-space of A T. a b c d e Size of A 3 3 3 3 3 3 5 9 6 2 Rank of A 3 2 2 2 a b c d e dimr.s. A 3 2 2 2 dimc.s. A 3 2 2 2 dimn.s. A 2 7 dimn.s. A T 2 3 4 #9 Deermine ranka as a funcion of for A = Subracing from he second and he hird rows he muliples and of he firs row resuls in 2 4.
If =, he second and he hird rows are zeros, so he rank is. If, he second and he hird row can be divided by, which gives + In his case he row-echelon form of his marix is 2 The hird row is zero for = 2, resuling in he rank value 2. Therefore,, if = ranka = 2, if = 2 3, oherwise # Le A and B be n n marices and deb Prove ha rankab = ranka. Since ranka = ranka T for any marix, is i equivalen o show ha rankab T = ranka T. Noe ha AB T = B T A T. Since B is inverible, so i is B T. In his case B T is a produc of he elemenary marices E E 2... E k for some k. Taking all his ino accoun, we have o show ha ranke E 2... E k A T = ranka T. Muliplicaion of A T by an elemenary marix on he lef is equivalen o applying o A T an elemenary row operaion. Hence, he marix E E 2... E k A T is obained from A T by applying o i a series of k elemenary row operaions. We know ha no elemenary row operaion changes he rank of a marix. This implies he proof. 5