1.5 Linear Dependence and Linear Independence

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1 14. For v span(s 1 S ) we may write v = n i=1 a i x i with x i S 1 and x i S. So v is an element of both span(s 1 ) and span(s ) and hence an element of span(s 1 ) span(s ). For example we have if S 1 = S = (1, 0) then they are the same and if S 1 = (1, 0) and S = (0, 1) then we have the left hand side is the set of zero vector and the right hand side is the the plane R. 15. If we have both a 1 v 1 + a v + + a n v n = b 1 v 1 + b v + b n v n then we have (a 1 b 1 )v 1 + (a b )v + + (a n b n )v n = 0. By the property we can deduce that a i = b i for all i. 16. When W has finite element the statement holds. Otherwise W {v}, where v W will be a generating set of W. But there are infinitely many v W. 1.5 Linear Dependence and Linear Independence 1. (a) No. For example, take S = {(1, 0), (, 0), (0, 1)} and then (0, 1) is not a linear combination of the other two. (b) Yes. It s because 10 = 0. (c) No. It s independent by the remark after Definition of linearly independent. (d) No. For example, we have S = {(1, 0), (, 0), (0, 1)} but {(1, 0), (0, 1)} is linearly independent. (e) Yes. This is the contrapositive statement of Theorem 1.6. (f) Yes. This is the definition.. (a) Linearly dependent. We have = 6. So to 4 8 check the linearly dependency is to find the nontrivial solution of equation a 1 x 1 + a x + + a n x n = 0. And x 1 and x are the two matrices here. (b) Linearly independent. (c) Linearly independent. (d) Linearly dependent. (e) Linearly dependent. (f) Linearly independent. (g) Linearly dependent. (h) Linearly independent. (i) Linearly independent. (j) Linearly dependent. 3. Let M 1,M,...,M 5 be those matrices. We have M 1 +M +M 3 M 4 M 5 = 0. 15

2 4. If a 1 e 1 + a e + + a n e n = (a 1,a,...,a n ) = 0, then by comparing the i-th entry of the vector of both side we have a 1 = a = = a n = It s similar to exercise it s similar to exercise Let E ij be the matrix with the only nonzero ij-entry= 1. Then {E 11,E } is the generating set. 8. (a) The equation x 1 (1, 1, 0)+x (1, 0, 1)+x 3 (0, 1, 1) = 0 has only nontrivial solution when F = R. (b) When F has characteristic, we have = 0 and so (1, 1, 0) + (1, 0, 1) + (0, 1, 1) = (0, 0, 0). 9. It s su cient since if u = tv for some t F then we have u tv = 0. While it s also necessary since if au + bv = 0 for some a, b F with at least one of the two coe cients not zero then we may assume a 0 and u = b a v. 10. Pick v 1 = (1, 1, 0), v = (1, 0, 0), v 3 = (0, 1, 0). And we have that none of the three is a multiple of another and they are dependent since v 1 v v 3 = Vector in span(s) are linear combinations of S and they all have di erent representation by the remark after Definition of linear independent. So there are n representations and so n vectors. 1. Since S 1 is linearly dependent we have finite vectors x 1,x,...,x n in S 1 and so in S such that a 1 x 1 +a x ++a n x n = 0 is a nontrivial representation. But the nontrivial representation is also a nontrivial representation of S. And the Corollary is just the contrapositive statement of the Theorem (a) Su ciency: If {u+v,u v} is linearly independent we have a(u+v)+ b(u v) = 0 implies a = b = 0. Assuming that cu + dv = 0, we can deduce that c+d c d c+d (u + v) + (u v) = 0 and hence = c d = 0. This means c = d = 0 if the characteristc is not two. Necessity: If {u, v} is linearly independent we have au+ bv = 0 implies a = b = 0. Assuming that c(u + v) + d(u v) = 0, we can deduce that (c + d)u + (c d)v = 0 and hence c + d = c d = 0 and c = d = 0. This means c = d = 0ifthe characteristc is not two. (b) Su ciency: If au + bv + cw = 0wehave a+b c (u + v) + a b+c (u + w) + a+b+c (v + w) = 0 and hence a = b = c = 0. Necessity: If a(u + v) + b(u + w) + c(v + w) = 0wehave(a + b)u + (a + c)v + (b + c)w = 0 and hence a = b = c = Su ciency: It s natural that 0 is linearly dependent. If v is a linear combination of u 1,u,...,u n,sayv = a 1 u 1 +a u +a n u n,thenv a 1 u 1 a u a n u n = 0 implies S is linearly dependent. Necessity: If S is 16

3 linearly dependent and S {0} we have some nontrivial representation a 0 u 0 + a 1 u a n u n = 0 with at least one of the coe cients is zero, say a 0 = 0 without loss the generality. Then we can let v = u 0 = 1 a 0 (a 1 u 1 + a u + + a n u n ). 15. Su ciency: If u 1 = 0 then S is linearly independent. If u k+1 span({u 1,u,...,u k }) for some k, sayu k+1 = a 1 u 1 +a u ++a k u k,thenwehavea 1 u 1 +a u ++ a k u k u k+1 = 0 is a nontrivial representation. Necessary: If S is linearly dependent, there are some integer k such that there is some nontrivial representation a 1 u 1 + a u + + a k u k + a k+1 u k+1 = 0. Furthermore we may assume that a k+1 0 otherwise we may choose less k until that a k+1 = 0. Hence we have a k+1 = 1 a k+1 (a 1 u 1 + a u + + a k u k ) and so a k+1 span({u 1,u,...,u k }). 16. Su ciency: We can prove it by contrapositive statement. If S is linearly dependent we can find a 1 u 1 + a u + + a n u n = 0. But thus the finite set {u 1,u,...,u n } would be a finite subset of S and it s linearly dependent. Necessary: This is the Threorem Let C 1,C,...,C n be the columns of M. Let a 1 C 1 + a C + + a n C n = 0 then we have a n = 0 by comparing the n-th entry. And inductively we have a n 1 = 0,a n = 0,...,a 1 = It s similar to exercise We have a 1 A t 1 + a A t + + a k A t k = 0 implies a 1A 1 + a A + + a k A k = 0. Then we have a 1 = a = = a n = If {f,g} is linearly dependent, then we have f = kg. But this means 1 = f(0) = kg(0) = k 1 and hence k = 1. And e r = f(1) = kg(1) = e s means r = s. 1.6 Bases and Dimension 1. (a) No. The empty set is its basis. (b) Yes. This is the result of Replacement Theorem. (c) No. For example, the set of all polynomials has no finite basis. (d) No. R has {(1, 0), (1, 1)} and {(1, 0), (0, 1)} as bases. (e) Yes. This is the Corollary after Replacement Theorem. (f) No. It s n + 1. (g) No. It s m n. (h) Yes. This is the Replaceent Theorem. 17

4 (i) No. For S = 1,, a subset of R, then5= = (j) Yes. This is Theorem (k) Yes. It s {0} and V respectly. (l) Yes. This is the Corollary after Replacement Theorem.. It s enough to check there are 3 vectors and the set is linear independent. (a) Yes. (b) No. (c) Yes. (d) Yes. (e) No. 3. (a) No. (b) Yes. (c) Yes. (d) Yes. (e) No. 4. It s impossible since the dimension of P 3 (R) is four. 5. It s also impossible since the dimension of R 3 is three. 6. Let E ij be the matrix with the only nonzero ij-entry= 1. Then the sets {E 11,E 1,E 1,E }, {E 11 +E 1,E 1,E 1,E }, and {E 11 +E 1,E 1,E 1,E } are bases of the space. 7. We have first {u 1,u } is linearly independent. And since u 3 = 4u 1 and u 4 = 3u 1 +7u, we can check that {u 1,u,u 5 } is linearly independent and hence it s a basis. 8. To solve this kind of questions, we can write the vectors into a matrix as 18

5 below and do the Gaussian elimintaion. M = And the row with all entries 0 can be omitted 1.So{u 1,u 3,u 6,u 7 } would 1 Which row with all entries is important here. So actually the operation here is not the 19

6 be the basis for W (the answer here will not be unique). 9. If a 1 u 1 + a u + a 3 u 3 + a 4 u 4 = (a 1,a 1 + a,a 1 + a + a 3,a 1 + a + a 3 + a 4 ) = 0 we have a 1 = 0 by comparing the first entry and then a = a 3 = a 4 = 0. For the second question we can solve (a 1,a,a 3,a 4 ) = a 1 u 1 +(a a 1 )u +(a 3 a )u 3 + (a 4 a 3 )u The polynomials found by Lagrange interpolation formula would be the answer. It would have the smallest degree since the set of those polynomials of Lagrange interpolation formula is a basis. (a) 4x x + 8. (b) 3x + 1. (c) x 3 + x + 4x 5. (d) x 3 x 6x If {u, v} is a basis then the dimension of V would be two. So it s enough to check both {u+v,au} and {au, bv} are linearly independent. Assuming s(u + v) + tau = (s + ta)u + sv = 0wehaves + ta = s = 0 and hence s = t = 0. Assuming sau + tbv = 0wehavesa = tb = 0 and hence s = t = If {u, v, w} is a basis then the dimension of V would be three. So it s enough to check {u + v + w,v + w,w} is lineaerly independent. Assuming a(u + v + w) + b(v + w) + cw = au + (a + b)v + (a + b + c)w = 0wehave a = a + b = a + b + c = 0 and hence a = b = c = We can substract the second equation by the two times of the first equation. And then we have x 1 x + x 3 = 0 x x 3 = 0 Let x 3 = s and hence x = s and x 1 = s. We have the solution would be {(s, s, s) = s(1, 1, 1) s R}. And the basis would be {(1, 1, 1)}. 14. For W 1 we can observe that by setting a = p, a 3 = q, a 4 = s, and a 5 = t we can solve a 1 = q+s. SoW 1 = {(q+s, p, q, s, t) = p(0, 1, 0, 0, 0)+q(1, 0, 1, 0, 0)+ s(1, 0, 0, 1, 0) + t(0, 0, 0, 0, 1) p, q, s, t F 5 }. And {(0, 1, 0, 0, 0), (1, 0, 1, 0, 0), (1, 0, 0, 1, 0), (0, 0, 0, 0, 1)} is the basis. The dimension is four. And similarly for W we may set a 4 = s, a 5 = t. And then we have a 1 = t, a = a 3 = a 4 = s and. And hence W = {( t, s, s, s, t) = s(0, 1, 1, 1, 0) + t( 1, 0, 0, 0, 1) s, t F 5 } {(0, 1, 1, 1, 0), ( 1, 0, 0, 0, 1)} is the basis of W. The dimension is two. standard Gaussian elimination since we can not change the order of two row here. 0

7 15. Just solve A 11 +A ++A nn = 0 and hence {E ij } i j {E ii E nn } i=1,,...n 1 } is the basis, where {E ij } is the standard basis. And the dimension would be n We have A ij = 0 for all i > j. Hence the basis could be {E ij } i j } and the dimension is n(n+1). 17. We have A ii = 0 and A ij = A ji. Hence the basis could be {E ij E ji } i<j and the dimension is n(n 1). 18. Let e i be the sequence with the only nonzero i-th term= 1. Then we have {e i } i 0 is a basis. To prove it, we have that every sequence is linear combination of the basis since we only discuss the sequence with finite nonzero entries. Furthermore we have for every finite subset of {e i } i 0 is linearly independent. 19. If every vector has a unique representation as linear combination of the set, this means every vector is a linear combination of. Furthermore, if there are nontrivial representation of 0, then we have there are two representations, say the trivial and nontrivial one, of 0. This is a contradiction. 0. (a) If S = or S = {0}, thenwehavev = {0} and the empty set can generate V. Otherwise we can choose a nonzero vector u 1 in S, and continuing pick u k+1 such that u k+1 span({u 1,u,...,u k }). The process would teminate before k > n otherwise we can find linearly independent set with size more than n. If it terminates at k = n, then we knoew the set is the desired basis. If it terminates at k < n, then this means we cannot find any vector to be the vector u k+1.so any vectors in S is a linear combination of = {u 1,u,...,u k } and hence can generate V since S can. But by Replacement Theorem we have n k. This is impossible. (b) If S has less than n vectors, the process must terminate at k < n. It s impossible. 1. Su ciency: If the vector space V is finite-dimensional, say dim= n, and it contains an infinite linearly independent subset, then we can pick an independent subset of such that the size of is n + 1. Pick a basis with size n. Since is a basis, it can generate V. By Replacement Theorem we have n n+1. It s a contradiction. Necessity: To find the infinite linearly independent subset, we can let S be the infinite-dimensional vector space and do the process in exercise 1.6.0(a). It cannot terminate at any k otherwise we find a linearly independent set generating the space and hence we find a finite basis.. The condition would be that W 1 W. Let and be the basis of W 1 W and W 1.SinceW 1 and W finite-dimensional, we have and are bases with finite size. First if W 1 is not a subset of W, we have some vector v W 1 W. But this means that v span( ) and hence {v} would be a 1