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# Sect4 1

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### Sect4 1

1. 1. SECTION 4.1 THE VECTOR SPACE R3 Here the fundamental concepts of vectors, linear independence, and vector spaces are introduced in the context of the familiar 2-dimensional coordinate plane R2 and 3-space R3. The concept of a subspace of a vector space is illustrated, the proper nontrivial subspaces of R3 being simply lines and planes through the origin. 1. a − b = (2, 5, −4) − (1, −2, −3) = (1, 7, −1) = 51 2a + b = 2(2,5, −4) + (1, −2, −3) = (4,10, −8) + (1, −2, −3) = (5,8, −11) 3a − 4b = 3(2,5, −4) − 4(1, −2, −3) = (6,15, −12) − (4, −8, −12) = (2, 23, 0) 2. a − b = (−1, 0, 2) − (3, 4, −5) = (−4, −4, 7) = 81 = 9 2a + b = 2(−1, 0, 2) + (3, 4, −5) = (−2, 0, 4) + (3, 4, −5) = (1, 4, −1) 3a − 4b = 3(−1, 0, 2) − 4(3, 4, −5) = (−3, 0, 6) − (12,16, −20) = (−15, −16, 26) 3. a − b = (2i − 3j + 5k ) − (5i + 3 j − 7k ) = −3i − 6 j + 12k = 189 = 3 21 2a + b = 2(2i − 3j + 5k ) + (5i + 3 j − 7k ) = (4i − 6 j + 10k ) + (5i + 3j − 7k ) = 9i − 3j + 3k 3a − 4b = 3(2i − 3j + 5k ) − 4(5i + 3 j − 7k ) = (6i − 9 j + 15k ) − (20i + 12 j − 28k ) = − 14i − 21j + 43k 4. a − b = (2i − j) − ( j − 3k ) = 2i − 2 j + 3k = 17 2a + b = 2(2i − j) + ( j − 3k ) = (4i − 2 j) + ( j − 3k ) = 4i − j − 3k 3a − 4b = 3(2i − j) − 4( j − 3k ) = (6i − 3 j) − (4 j − 12k ) = 6i − 7 j + 12k 5. v = 3 2 u, so the vectors u and v are linearly dependent. 6. au + bv = a(0, 2) + b(3, 0) = (3b, 2a ) = 0 implies a = b = 0, so the vectors u and v are linearly independent. 7. au + bv = a(2, 2) + b(2, −2) = (2a + 2b, 2a − 2b) = 0 implies a = b = 0, so the vectors u and v are linearly independent. 8. v = − u, so the vectors u and v are linearly dependent. In each of Problems 9-14, we set up and solve (as in Example 2 of this section) the system
2. 2. u v1   a   w1  au + bv =  1  b  = w  = w u 2 v2     2 to find the coefficient values a and b such that w = au + bv,  1 −1  a  1  9.  −2 3   b  =  0  ⇒ a = 3, b = 2 so w = 3u + 2 v      3 2 a  0 10.  4 3   b  =  −1 ⇒ a = 2, b = −3 so w = 2u − 3v      5 2  a  1 11. 7 3   b  = 1 ⇒ a = 1, b = −2 so w = u − 2 v      4 −2   a  2 12. 1 −1  b  =  −2  ⇒ a = 3, b = 5 so w = 3u + 5 v      7 3   a  5 13.  5 4   b  =  −2  ⇒ a = 2, b = −2 so w = 2u − 3v       5 −6   a  5  14.  −2 4   b  =  6  ⇒ a = 7, b = 5 so w = 7u + 5 v      In Problems 15-18, we calculate the determinant u v w so as to determine (using Theorem 4) whether the three vectors u, v, and w are linearly dependent (det = 0) or linearly independent (det ≠ 0). 3 5 8 15. −1 4 3 = 0 so the three vectors are linearly dependent. 2 −6 −4 5 2 4 16. −2 −3 5 = 0 so the three vectors are linearly dependent. 4 5 −7 1 3 1 17. −1 0 −2 = − 5 ≠ 0 so the three vectors are linearly independent. 2 1 2
3. 3. 1 4 3 18. 1 3 −2 = 9 ≠ 0 so the three vectors are linearly independent. 0 1 −4 In Problems 19-24, we attempt to solve the homogeneous system Ax = 0 by reducing the coefficient matrix A = [u v w ] to echelon form E. If we find that the system has only the trivial solution a = b = c = 0, this means that the vectors u, v, and w are linearly independent. Otherwise, a nontrivial solution x = [a b c ] ≠ 0 provides us with a nontrivial linear T combination au + bv + cw ≠ 0 that shows the three vectors are linearly dependent.  2 −3 0   1 0 −3  19.  0 1 −2  →  0 1 −2  = E A =     1 −1 −1   0 0 0    The nontrivial solution a = 3, b = 2, c = 1 gives 3u + 2v + w = 0, so the three vectors are linearly dependent. 5 2 4 1 0 2  20.  5 3 1  → 0 1 −3 = E A =     4 1 5    0 0 0    The nontrivial solution a = –2, b = 3, c = 1 gives –2u + 3v + w = 0, so the three vectors are linearly dependent.  1 −2 3  1 0 11 21.  1 −1 7  → 0 1 4  = E A =      −2 6 2    0 0 0    The nontrivial solution a = 11, b = 4, c = –1 gives 11u + 4v – w = 0, so the three vectors are linearly dependent. 1 5 0  1 0 0  22. 1 1 1  → 0 1 0  = E A =     0 3 2    0 0 1    The system Ax = 0 has only the trivial solution a = b = c = 0, so the vectors u, v, and w are linearly independent. 2 5 2  1 0 0  23.  0 4 −1 →  0 1 0  = E A =      3 −2 1    0 0 1   
4. 4. The system Ax = 0 has only the trivial solution a = b = c = 0, so the vectors u, v, and w are linearly independent. 1 4 −3 1 0 0 24.  4 2 3  → 0 1 0  = E A =      5 5 −1   0 0 1   The system Ax = 0 has only the trivial solution a = b = c = 0, so the vectors u, v, and w are linearly independent. In Problems 25-28, we solve the nonhomogeneous system Ax = t by reducing the augmented coefficient matrix A = [u v w t ] to echelon form E. The solution vector x = [a b c ] appears as the final column of E, and provides us with the desired linear T combination t = au + bv + cw. 1 3 1 2 1 0 0 2  25.  −2 0 −1 −7  → 0 1 0 −1 = E A =     2 1 2 9   0 0 1 3    Thus a = 2, b = –1, c = 3 so t = 2u – v + 3w. 5 1 5 5  1 0 0 1  26.  2 5 −3 30  → 0 1 0 5  = E A =      −2 −3 4 −21   0 0 1 −1   Thus a = 1, b = 5, c = –1 so t = u + 5v – w.  1 −1 4 0  1 0 0 2  27.  4 −2 4 0  → 0 1 0 6  = E A =      3 2 1 19    0 0 1 1    Thus a = 2, b = 6, c = 1 so t = 2u + 6v + w. 2 4 1 7  1 0 0 1 28.  5 1 1 7  → 0 1 0 1 = E A =      3 −1 5 7    0 0 1 1   Thus a = 1, b = 1, c = 1 so t = u + v + w. 29. Given vectors (0, y , z ) and (0, v, w) in V, we see that their sum (0, y + v, z + w) and the scalar multiple c(0, y, z ) = (0, cy, cz ) both have first component 0, and therefore are elements of V.
5. 5. 30. If ( x, y , z ) and (u , v, w) are in V, then ( x + v ) + ( y + u ) + ( z + w) = ( x + y + z ) + (u + v + w) = 0 + 0 = 0, so their sum ( x + u , y + v, z + w) is in V. Similarly, cx + cy + cz = c( x + y + x) = c(0) = 0, so the scalar multiple (cx, cy, cz ) is in V. 31. If ( x, y , z ) and (u , v, w) are in V, then 2( x + u ) = (2 x) + (2u ) = (3 y ) + (3v) = 3( y + v), so their sum ( x + u, y + v, z + w) is in V. Similarly, 2(cx) = c(2 x) = c(3 y ) = 3(cy ), so the scalar multiple (cx, cy, cz ) is in V. 32. If ( x, y , z ) and (u , v, w) are in V, then z + w = (2 x + 3 y ) + (2u + 3v) = 2( x + u ) + 3( y + v), so their sum ( x + u , y + v, z + w) is in V. Similarly, cz = c(2 x + 3 y ) = 2(cx) + 3(cy ), so the scalar multiple (cx, cy, cz ) is in V. 33. (0,1, 0) is in V but the sum (0,1, 0) + (0,1, 0) = (0, 2, 0) is not in V; thus V is not closed under addition. Alternatively, 2(0,1, 0) = (0, 2, 0) is not in V, so V is not closed under multiplication by scalars. 34. (1,1,1) is in V, but 2(1,1,1) = (1,1,1) + (1,1,1) = (2, 2, 2) is not, so V is closed neither under addition of vectors nor under multiplication by scalars.
6. 6. 35. Evidently V is closed under addition of vectors. However, (0, 0,1) is in V but (−1)(0, 0,1) = (0, 0, −1) is not, so V is not closed under multiplication by scalars. 36. (1,1,1) is in V, but 2(1,1,1) = (1,1,1) + (1,1,1) = (2, 2, 2) is not, so V is closed neither under addition of vectors nor under multiplication by scalars. 37. Pick a fixed element u in the (nonempty) vector space V. Then, with c = 0, the scalar multiple cu = 0u = 0 must be in V. Thus V necessarily contains the zero vector 0. 38. Suppose u and v are vectors in the subspace V of R3 and a and b are scalars. Then au and bv are in V because V is closed under multiplication by scalars. But then it follows that the linear combination au + bv is in V because V is closed under addition of vectors. 39. It suffices to show that every vector v in V is a scalar multiple of the given nonzero vector u in V. If u and v were linearly independent, then — as illustrated in Example 2 of this section — every vector in R2 could be expressed as a linear combination of u and v. In this case it would follow that V is all of R2 (since, by Problem 38, V is closed under taking linear combinations). But we are given that V is a proper subspace of R2, so we must conclude that u and v are linearly dependent vectors. Since u ≠ 0, it follows that the arbitrary vector v in V is a scalar multiple of u, and thus V is precisely the set of all scalar multiples of u. In geometric language, the subspace V is then the straight line through the origin determined by the nonzero vector u. 40. Since the vectors u, v, w are linearly dependent , there exist scalars p, q, r not all zero such that pu + qv + rw = 0. If r = 0, then p and q are scalars not both zero such that pu + qv = 0. But this contradicts the given fact that u and v are linearly independent. Hence r ≠ 0, so we can solve for p q w = − u − v = au + b v , r r thereby expressing w as a linear combination of u and v. 41. If the vectors u and v are in the intersection V of the subspaces V1 and V2, then their sum u + v is in V1 because both vectors are in V1, and u + v is in V2 because both are in V2. Therefore u + v is in V, and thus V is closed under addition of vectors. Similarly, the intersection V is closed under multiplication by scalars, and is therefore itself a subspace.