Math 030 Assignment 5 Solutions Question 1: Which of the following sets of vectors are linearly independent? If the set is linear dependent, find a linear dependence relation for the vectors (a) {(1, 0, ), (3, 1, 0), (0, 0, 1)} (b) {(0, 1,, 4, 1), (, 1, 7, 1, 5), (0, 0,, 1, 1), (3, 1,, 4, 5), (6, 6, 7, 1, 1), ( 1, 1, 0, 1, 1)} (c) {(1, 0, 1, 1), (0, 1,, 1), (3,, 7, 1)} Solutions: Recall the method from class to conclude whether or not a set of vectors in linearly independent is to write the relation c 1 v 1 + c v + + c n v n = 0 as the homogeneous linear system A x = 0 where A has columns v 1,, v n Once we ve done this we use Theorems, 3 and 4 from section 51 to get our answers If the homogeneous system has a unique solution (the trivial one x = 0) then the set is linearly independent If it has infinitely many solution then the set is linearly dependent (a) Writing the vectors as columns we get the matrix 1 3 0 0 1 0 0 1 Computing the determinant, we see that det(a) = 1, thus A x = 0 has only the trivial solution Thus by Theorem 3, the set {(1, 0, ), (3, 1, 0), (0, 0, 1)} is linearly independent Equivalent, you could row reduce A to the identity matrix and use Theorem 4 to conclude that the set is linearly independent (b) We can see that the vectors in the set are from R 5 Since there are 6 vectors in the set, we immediately know by Theorem that the set must be linearly dependent (c) We have 1 0 3 0 1 1 7 1 1 1 1
Row reducing A to row reduced echelon form gives 1 0 3 B = 0 1 0 0 0 0 0 0 By Theorem 5, the set of vectors is linearly dependent since there are columns of the row reduced echelon form of A which are not elementary columns A linear dependence relation amongst the vectors is (3,, 7, 1) = 3(1, 0, 1, 1) + (0, 1,, 1) Question : Let { v 1, v, v 3 } be a set of non-zero mutually orthogonal vectors in R 3 Note that mutually orthogonal means v 1 v = v 1 v 3 = v v 3 = 0 Show that this set is linearly independent Solution: Suppose that c 1 v 1 + c v + c 3 v 3 = 0 has a non trivial solution, that is, a solution where not all c 1, c, c 3 are zero We want to show that this can not occur Consider the following v 1 (c 1 v 1 + c v + c 3 v 3 ) = 0 This can be rewritten as c 1 ( v 1 v 1 ) + c ( v 1 v ) + c 3 ( v 1 v 3 ) = 0 Since v 1, v, v 3 are all orthogonal, we know v 1 v = 0 and v 1 v 3 = 0 Thus the above simplifies to c 1 ( v 1 v 1 ) = c 1 ( v 1 ) = 0 For this to be zero, either c 1 or v 1, must be zero Since v 1 is non zero, we know its norm must be v 1 > 0 Therefore c 1 = 0 The same calculation on v (c 1 v 1 + c v + c 3 v 3 ) = 0 and v 3 (c 1 v 1 + c v + c 3 v 3 ) = 0, shows the c = c 3 = 0 as well Thus, c 1 = c = c 3 = 0 Thus the set { v 1, v, v 3 } is linearly independent by Corollary 1 in Section 51 Question 3: Are the following subsets of R 4 subspaces? Prove or disprove (a) S = {(a, a, a 3, a 4 ) : a R}
(b) S = {(a, b, b, a + b) : a, b R} Solutions (a) We first note that S is non-empty Let u = (a, a, a 3, a 4 ), v = (b, b, b 3, b 4 ) S If S is a subspace then we need u + v S u + v = (a, a, a 3, a 4 ) + (b, b, b 3, b 4 ) = (a + b, a + b, a 3 + b 3, a 4 + b 4 ) If this vector is in S, then it needs to be the case that a + b = (a + b) But we know that (a + b) = a + ab + b Thus u + v is not in S So S is not a subspace of R 4 (b) We first note that S is non-empty Let u = (a, b, b, a + b) and v = (c, d, d, c + d) If S is a subspace then we need u + v S u = v = (a, b, b, a+b)+(c, d, d, c+d) = (a+c, b+d, b+d, a+b+c+d) = (k, l, l, k+l) where k = a + c and l = b + d Thus u + v S Secondly, for any c R, we need to show c u S c u = c(a, b, b, a + b) = (ca, cb, cb, ca + cb) = (k, l, l, k + l) where k = ca and l = cb Thus c u S Therefore, S must be a subspace Question 4: Find a set of vectors which spans the solution set of A x = 0 where ( ) 4 1 6 1 1 3 Solution: We first need to find the solution set by Gauss Jordan Elimination on the augmented system ( 4 1 6 0 1 1 3 0 ) The row reduced echelon form is 3
( 1 1 3 0 0 0 0 ) Writing the solution set we get x 1 = x + 1 x 3 3x 4 So x, x 3, x 4 are free Set x = s, x 3 = t, x 4 = u, withs, t, u R Thus the solution set can be written as x = s(, 1, 0, 0) + t( 1, 0, 1, 0) + u( 3, 0, 0, 1) Thus a set of vectors which spans the solution set of A x = 0 is {(, 1, 0, 0), ( 1, 0, 1, 0), ( 3, 0, 0, 1)} Question 5: Do the following sets of vectors form a basis for the indicated subspaces If not, either expand or reduce the set to find the appropriate basis (a) {(1, 0, 1), (1, 1, 1), (, 1, 0)} for span { (1, 0, 1), (1, 1, 1), (, 1, 0) } (b) {( 1, 1, 0, 0), (1, 1, 1, 0), (0, 0, 1, )} for R 4 (c) The spanning set in Question 4 for the subspace in Question 4 Solution: Recall that the basis of a subspace is a collection of vectors which is a linearly independent spanning set for the subspace So it is necessary to check that the given sets are spanning and linearly independent for the indicated subspaces (a) Since the subspace is S = span { (1, 0, 1), (1, 1, 1), (, 1, 0) }, we automatically know T = {(1, 0, 1), (1, 1, 1), (, 1, 0)} is a spanning set for S by definition All that remains is to determine if T is linearly independent We set up the matrix 1 1 0 1 1 1 1 0 Computing the row reduced echelon form we get 1 0 1 B = 0 1 1 0 0 0 4
Since we do not have 3 distinct elementary columns, we know that T is not linearly independent by Theorem 5 Also by Theorem 5, we know (, 1, 0) is a linearly combination of (1, 0, 1) and (1, 1, 1) Removing (, 1, 0) from T gives us a new set T = {(1, 0, 1), (1, 1, 1)} We know by Theorem in 5, that T is still a spanning set for S Since (1, 0, 1) and (1, 1, 1) correspond to distinct elementary columns in B we know they are linearly independent Therefore, T is a basis for S (b) Since the set T = {( 1, 1, 0, 0), (1, 1, 1, 0), (0, 0, 1, )} has only 3 vectors we know that it can not be a basis for R 4 This is because R 4 has dimension 4 so any basis for R 4 must have 4 vectors Thus we must extend T to a basis We use the procedure in Theorem Section 53 to extend T to a basis This procedure says that we place the vectors in T is a matrix along with the vectors from some spanning set for R 4 We know that the standard basis {e 1, e, e 3, e 4 } is a spanning set for R 4 so we will use this The row reduced echelon form of A is B = 1 1 0 1 0 0 0 1 1 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 1 0 0 Since columns 1,,3 and 4 are distinct elementary columns in B, the columns 1,,3 and 4 of A correspond to a basis for R 4 A basis for R 4 is {( 1, 1, 0, 0), (1, 1, 1, 0), (0, 0, 1, ), (1, 0, 0, 0)} (c) By Theorem 4 in Section 53 we know that {(, 1, 0, 0), ( 1, 0, 1, 0), ( 3, 0, 0, 1)} of the solution space is automatically a basis This is because Theorem 4 says that the dimension of the solution space is equal to n r which in this cause is 3 since n = 4 and r = 1 for the matrix A is question 4 Thus by Theorem 5 iii) Section 53 this set is automatically a basis for A x = 0 Question 6: Suppose that A is a matrix (a) Prove that if A is not square then either the rows or the columns of A are linearly dependent (b) Prove that if A is square, then the rows of A are linearly independent if and only if the columns are linearly independent 5
Solution: (a) Suppose A is a m n matrix with n > m Then we have n columns in A which we can think of as being vectors in R m By Theorem 5 part i), the columns of A can not be linearly independent since R m has dimension m, so any collection of vectors from R m with more than m vectors must be linearly dependent Similarly, if n < m, then you can think of the rows of A as a collection of m vectors in R n By the same argument as above, since R n has dimension n, a collection of vectors in R n with more than n vectors must be linearly dependent by Theorem 5 part i) in section 53 (b) We first assume that the rows of A are linearly independent Since the rows are LI by Theorem 3 in section 51, det(a T ) 0 Since det(a T ) = det(a), we can also conclude that det(a) 0 Thus by Theorem 3 in 51, we can conclude that the columns of A are also LI The same argument shows the other direction Assume that the columns of A are linearly independent Then by Theorem 3 in 51 we know det(a) 0 Since det(a) = det(a T ), we can also conclude that det(a T ) 0 Thus by Theorem 3 in 51, we can conclude that the rows of A are LI 6