Solutions to Homework 5 - Math 3410

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1 Solutions to Homework 5 - Math 34 (Page 57: # 489) Determine whether the following vectors in R 4 are linearly dependent or independent: (a) (, 2, 3, ), (3, 7,, 2), (, 3, 7, 4) Solution From x(, 2, 3, ) + y(3, 7,, 2) + z(, 3, 7, 4) = (,,, ) we have x + 3y + z = 2x + 7y + 3z = 3x + y + 7z = x 2y 4z = By applying Gauss-Jordan elimination we can see that this system has infinitely many solutions, and for example x = 2, y =, z = is a solution of this system This shows that these three vectors are linearly dependent 2 (Page 57: # 49(b)) Show that the following functions f, g, h are linearly independent: f(t) = e t, g(t) = e 2t, h(t) = t Solution Suppose that for all t R c f(t) + c 2 g(t) + c 3 h(t) = where c, c 2, c 3 R In other words, c e t + c 2 e 2t + c 3 t = for all t R We now plug in the values t =,, 2 into this last equation to obtain c e + c 2 e + c 3 = c e + c 2 e 2 + c 3 = c e 2 + c 2 e 4 + c 3 2 = In matrix form this may be written as c e e 2 c 2 = e 2 e 4 2 c 3 We label this 3 3 matrix as A We now row reduce this matrix: R 2 R 2 /e, R 3 R 3 /e 2 e e 2 2e 2 R 2 R 2 R, R 3 R 3 R e e e 2 2e 2

2 R 2 R 2 /(e ), R 3 R 3 /(e 2 ) e e 2e 2 e 2 R 3 R 3 R 2 e e 2e 2 e 2 e e Clearly A has 3 pivots and thus it follows that c = c 2 = c 3 = is the only solution of the system of linear equations Thus p (t), p 2 (t), p 3 (t) are linearly independent 3 (Page 57: # 493) Suppose u, v, w are linearly independent vectors Investigate that whether S is linearly independent or not, where (a) S = { u + v 2 w, u v w, u + w}, (b) S = { u + v 3 w, u + 3 v w, v + w} Solution (a) From x( u + v 2 w) + y( u v w) + z( u + w) =, we have (x + y + z) u + (x y) v + ( 2x y + z) w = Since u, v, and w are linearly independent we have x + y + z = x y = 2x y + z = Solving this system will give x = y = z = So S is linearly independent (b) From x( u + v 3 w) + y( u + 3 v w) + z( v + w) =, we have (x + y) u + (x + 3y + z) v + ( 3x y + z) w = Since u, v, and w are linearly independent we have x + y = x + 3y + z = 3x y + z = This system has non-trivial solutions For example x =, y =, and z = 2 is a solution So S is linearly dependent 4 (Page 57: # 495) Let V be a vector space over a field K Suppose S = { v,, v n } is linearly independent Prove that S is linearly independent where (a) S = {a v,, a n v n } and a i for i n (b) S = { v,, v k, w, v k+,, v n } where w = n b i v i i= and b k (Note that in this problem S is obtained from S by replacing v k by w) Solution (a) Suppose that c (a v ) + c n (a n v n ) = 2

3 for some c,, c n K Thus (c a ) v + (c n a n ) v n = Since { v,, v n } are linearly independent it follows that c a =, c 2 a 2 =,, c n a n = Since a j for j n it follows that c j = (c j a j )a j = a j = for j n Thus {a v,, a n v n } are linearly independent (b) Suppose that c v + c k v k + c k w + c k+ v k+ + c n v n = or some c,, c n K Thus c v + c k v k + c k (b v + b n v n ) + c k+ v k+ + c n v n = Regrouping the terms we have (c +b c k ) v + (c k +b k c k ) v k +c k b k v k +(c k+ +b k+ c k ) v k+ + (c n +b n c k ) v n = Since { v,, v n } are linearly independent it follows that c + b c k = c k + b k c k = c k b k = c k+ + b k+ c k = c n + b n c k = Now looking at the k-th line we see that c k b k = and thus c k = which follows by multiplying both sides of the equation by b k as in part (a) Now we look at the other k equations, namely c j + b j c k = where j n and j k Since c k = it follows that and thus c j + b j = c j = for j n and j k It follows that c j = for j n and thus S is linearly independent 3

4 5 (Page 58: # 497) Find a subset of u, u 2, u 3, u 4 that gives a basis for W = span( u i ) of R 5 where: (c) u = (,,,, ), u 2 = (,, 2,, ), u 3 = (2,, 3,, ), u 4 = (, 2,,, ) Solution We note that system of linear equations from x(,,,, )+y(,, 2,, )+ z(2,, 3,, ) + w(, 2,,, ) = (,,,, ) has non-trivial solution, and so these four vectors are linearly dependent Now we drop one of the vectors, for example u 3, and we for the equation x(,,,, ) + y(,, 2,, ) + z(, 2,,, ) = (,,,, ) Solving this system results in x = y = z =, so u, u 2, and u 4 are linearly independent and therefore they form a basis for W = span( u i ) 6 (Page 58: # 498) Consider the subspaces U = {(a, b, c, d) : b 2c + d = } and W = {(a, b, c, d) : a = d, b = 2c} of R 4 Find a basis and the dimension of (a) U, (b) W, (c) U W Solution (a) An element (a, b, c, d) U can be written as (a, 2c d, c, d) = a(,,, ) + c(, 2,, ) + d(,,, ) So {(,,, ), (, 2,, ), (,,, )} forms a basis of U, since these vectors are linearly independent Thus the dimension is 3 (b) An element (a, b, c, d) W can be written as (a, 2c, c, a) = a(,,, ) + c(, 2,, ) So {(,,, ), (, 2,, )} is a basis of W since these vectors are linearly independent Thus the dimension is 2 (c) By Gauss-Jordan elimination we solve this system of linear equations b 2c + d = a d = b 2c = We find out that the solutions are in the form (, 2c, c, ) So {(, 2,, )} is a basis of U W Thus the dimension is 7 (Page 58: # 499) Find a basis and the dimension of the solution space W of each of the following homogeneous systems: x + 2y 2z + 2s t = (a) x + 2y z + 3s 2t = 2x + 4y 7z + s + t = Solution The reduced echelon form of the coefficient matrix is in the form So the reduced system is x = 2y 4s+3t, z = s+t So,, is a basis of W and dim W = 3 8 (Page 58: # 42(a)) Find a basis and dimension of the subspace W of P(t) spanned by p (t) = t 3 + 2t 2 2t +, p 2 (t) = t 3 + 3t 2 3t + 4, p 3 (t) = 2t 3 + t 2 7t 7 Solution Let W = span(p (t), p 2 (t), p 3 (t)) = {d p (t) + d 2 p 2 (t) + d 3 p 3 (t) d, d 2, d 3 R} 3 4

5 We first check whether p (t), p 2 (t), p 3 (t) are linearly independent are not Suppose that c p (t) + c 2 p 2 (t) + c 3 p 3 (t) = for some c, c 2, c 3 R This reads or c (t 3 + 2t 2 2t + ) + c 2 (t 3 + 3t 2 3t + 4) + c 3 (2t 3 + t 2 7t 7) = (c + c 2 + 2c 3 )t 3 + (2c + 3c 2 + c 3 )t 2 + ( 2c 3c 2 7c 3 )t + (c + 4c 2 7c 3 ) = This equals zero for all t R only if each coefficient equals zero That is c + c 2 + 2c 3 = 2c + 3c 2 + c 3 = 2c 3c 2 7c 3 = c + 4c 2 7c 3 = We now write this as a matrix equation A x = where x = (c c 2 c 3 ) T and 2 A = R 3 R 3 + R R 2 R 2 2R, R 4 R 4 R R 4 R 4 3R R 3 R 3 / 6, R 2 R 2 + 3R 3, R R 2R 3, R R R 2 Thus we see that the solutions to our initial system of linear equations is c c 2 = c 3 which clearly are c = c 2 = c 3 = It follows that p (t), p 2 (t), p 3 (t) are linearly independent In addition, we have that p (t), p 2 (t), p 3 (t) span W (by definition) Thus p (t), p 2 (t), p 3 (t) are a basis of W and the dimension is 3 5

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