SECTION 4.2

THE VECTOR SPACE Rn AND SUBSPACES
The main objective in this section is for the student to understand what ty...
5.   Suppose x = ( x1 , x2 , x3 , x4 ) and y = ( y1 , y2 , y3 , y4 ) are vectors in W, so

              x1 + 2 x2 + 3x3 +...
9.     The vector x = (1, 0) is in W, but its scalar multiple 2x = (2, 0) is not, because
       (2)2 + (0)2 = 4 ≠ 1. Henc...
Thus x3 = s and x4 = t are free variables. We solve for x1 = − s − 4t and x2 = −2t , so

                     x = ( x1 , x...
where u = (2, −1, −2,1, 0) and v = (3, −4, 5, 0,1).

           1 −3 −5 −6  1 0 1 0 
19.        2 1 4 −4  → 0 1 2 0...
where u = ( −6, 4, −3,1).

23.   Let u be a vector in W. Then 0u is also in W. But 0u = (0+0)u = 0u + 0u, so upon
      su...
31.   Let w1 and w2 be two vectors in the sum U + V. Then wi = ui + vi where ui is in U
      and vi is in V (i = 1, 2). T...
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Sect4 2

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Transcript of "Sect4 2"

  1. 1. SECTION 4.2 THE VECTOR SPACE Rn AND SUBSPACES The main objective in this section is for the student to understand what types of subsets of the vector space Rn of n-tuples of real numbers are subspaces — playing the role in Rn of lines and planes through the origin in R3. Our first reason for studying subspaces is the fact that the solution space of any homogeneous linear system Ax = 0 is a subspace of Rn. 1. If x = ( x1 , x2 , 0) and y = ( y1 , y2 , 0) are vectors in W, then their sum x + y = ( x1 , x2 , 0) + ( y1 , y2 , 0) = ( x1 + y1 , x2 + y2 , 0) and the scalar multiple cx = (cx1 , cx2 , 0) both have third coordinate zero, and therefore are also elements of W. Hence W is a subspace of R3. 2. Suppose x = ( x1 , x2 , x3 ) and y = ( y1 , y2 , y3 ) are vectors in W, so x1 = 5 x2 and y1 = 5 y2 . Then their sum s = x + y = ( x1 + y1 , x2 + y2 , x3 + y3 ) = ( s1 , s2 , s3 ) satisfies the same condition s1 = x1 + y1 = 5 x2 + 5 y2 = 5( x2 + y2 ) = 5s2 , and thus is an element of W. Similarly, the scalar multiple m = cx = (cx1 , cx2 , cx3 ) = (m1 , m2 , m3 ) satisfies the condition m1 = cx1 = c(5 x2 ) = 5(cx2 ) = 5m2 , and hence is also an element of W. Therefore W is a subspace of R3. 3. The typical vector in W is of the form x = ( x1 ,1, x3 ) with second coordinate 1. But the particular scalar multiple 2x = (2 x1 , 2, 2 x3 ) of such a vector has second coordinate 2 ≠ 1, and thus is not in W. Hence W is not closed under multiplication by scalars, and therefore is not a subspace of R3. (Since 2x = x + x, W is not closed under vector addition either.) 4. The typical vector x = ( x1 , x2 , x3 ) in W has coordinate sum x1 + x2 + x3 equal to 1. But then the particular scalar multiple 2x = (2 x1 , 2 x2 , 2 x3 ) of such a vector has coordinate sum 2 x1 + 2 x2 + 2 x3 = 2( x1 + x2 + x3 ) = 2(1) = 2 ≠ 1, and thus is not in W. Hence W is not closed under multiplication by scalars, and therefore is not a subspace of R3. (Since 2x = x + x, W is not closed under vector addition either.)
  2. 2. 5. Suppose x = ( x1 , x2 , x3 , x4 ) and y = ( y1 , y2 , y3 , y4 ) are vectors in W, so x1 + 2 x2 + 3x3 + 4 x4 = 0 and y1 + 2 y2 + 3 y3 + 4 y4 = 0. Then their sum s = x + y = ( x1 + y1 , x2 + y2 , x3 + y3 , x4 + y4 ) = ( s1 , s2 , s3 , s4 ) satisfies the same condition s1 + 2s2 + 3s3 + 4s4 = ( x1 + y1 ) + 2( x2 + y2 ) + 3( x3 + y3 ) + 4( x4 + y4 ) = ( x1 + 2 x2 + 3x3 + 4 x4 ) + ( y1 + 2 y2 + 3 y3 + 4 y4 ) = 0 + 0 = 0, and thus is an element of W. Similarly, the scalar multiple m = cx = (cx1 , cx2 , cx3 , cx4 ) = (m1 , m2 , m3 , m4 ) satisfies the condition m1 + 2m2 + 3m3 + 4m4 = cx1 + 2cx2 + 3cx3 + 4cx4 = c( x1 + 2 x2 + 3 x3 + 4 x4 ) = 0, and hence is also an element of W. Therefore W is a subspace of R3. 6. Suppose x = ( x1 , x2 , x3 , x4 ) and y = ( y1 , y2 , y3 , y4 ) are vectors in W, so x1 = 3x3 , x2 = 4 x4 and y1 = 3 y3 , y2 = 4 y4 . Then their sum s = x + y = ( x1 + y1 , x2 + y2 , x3 + y3 , x4 + y4 ) = ( s1 , s2 , s3 , s4 ) satisfies the same conditions s1 = x1 + y1 = 3x3 + 3 y3 = 3( x3 + y3 ) = 3s3 , s2 = x2 + y2 = 4 x4 + 4 y4 = 4( x4 + y4 ) = 4s4 , and thus is an element of W. Similarly, the scalar multiple m = cx = (cx1 , cx2 , cx3 , cx4 ) = (m1 , m2 , m3 , m4 ) satisfies the conditions m1 = cx1 = c(3 x3 ) = 3(cx3 ) = 3m3 , m2 = cx2 = c(4 x4 ) = 4(cx4 ) = 4m4 , and hence is also an element of W. Therefore W is a subspace of R3. 7. The vectors x = (1,1) and y = (1, −1) are in W, but their sum x + y = (2, 0) is not, because 2 ≠ 0 . Hence W is not a subspace of R2. 8. W is simply the zero subspace {0} of R2.
  3. 3. 9. The vector x = (1, 0) is in W, but its scalar multiple 2x = (2, 0) is not, because (2)2 + (0)2 = 4 ≠ 1. Hence W is not a subspace of R2. 10. The vectors x = (1, 0) and y = (0,1) are in W, but their sum s = x + y = (1,1) is not, because 1 + 1 = 2 ≠ 1. Hence W is not a subspace of R2. 11. Suppose x = ( x1 , x2 , x3 , x4 ) and y = ( y1 , y2 , y3 , y4 ) are vectors in W, so x1 + x2 = x3 + x4 and y1 + y2 = y3 + y4 . Then their sum s = x + y = ( x1 + y1 , x2 + y2 , x3 + y3 , x4 + y4 ) = ( s1 , s2 , s3 , s4 ) satisfies the same condition s1 + s2 = ( x1 + y1 ) + ( x2 + y2 ) = ( x1 + x2 ) + ( y1 + y2 ) = ( x3 + x4 ) + ( y3 + y4 ) = ( x3 + y3 ) + ( x4 + y4 ) = s3 + s4 and thus is an element of W. Similarly, the scalar multiple m = cx = (cx1 , cx2 , cx3 , cx4 ) = (m1 , m2 , m3 , m4 ) satisfies the condition m1 + m2 = cx1 + cx2 = c( x1 + x2 ) = c( x3 + x4 ) = cx3 + cx4 = m3 + m4 , and hence is also an element of W. Therefore W is a subspace of R3. 12. The vectors x = (1, 0,1, 0) and y = (0, 2, 0,3) are in W (because both products are 0 in each case) but their sum s = x + y = (1, 2,1,3) is not, because s1 s2 = 2 but s3 s4 = 3. Hence W is not a subspace of R2. 13. The vectors x = (1, 0,1, 0) and y = (0,1, 0,1) are in W (because the product of the 4 components is 0 in each case) but their sum s = x + y = (1,1,1,1) is not, because s1 s2 s3 s4 = 1 ≠ 0. Hence W is not a subspace of R2. 14. The vector x = (1,1,1,1) is in W (because all 4 components are nonzero) but the multiple 0x = (0, 0, 0, 0) is not. Hence W is not a subspace of R2. In Problems 15-22, we first reduce the coefficient matrix A to echelon form E in order to solve the given homogeneous system Ax = 0 . 1 −4 1 −4  1 0 1 4  15. 1 2 1 8  → 0 1 0 2  = E A =     1 1 1 6    0 0 0 0   
  4. 4. Thus x3 = s and x4 = t are free variables. We solve for x1 = − s − 4t and x2 = −2t , so x = ( x1 , x2 , x3 , x4 ) = ( − s − 4t , −2t , s, t ) = (− s, 0, s, 0) + (−4t , −2t , 0, t ) = s u + t v where u = ( −1, 0,1, 0) and v = ( −4, −2, 0,1).  1 −4 −3 −7  1 0 1 5  16.  2 −1 1 7  → 0 1 1 3  = E A =     1 2 3 11    0 0 0 0   Thus x3 = s and x4 = t are free variables. We solve for x1 = − s − 5t and x2 = − s − 3t , so x = ( x1 , x2 , x3 , x4 ) = ( − s − 5t , − s − 3t , s, t ) = (− s, − s, s, 0) + (−5t , −3t , 0, t ) = s u + t v where u = ( −1, −1,1, 0) and v = (−5, −3, 0,1). 1 3 8 −1  1 0 −1 2  17. 1 −3 −10 5  → 0 1 3 −1 = E A =     1 4 11 −2    0 0 0 0    Thus x3 = s and x4 = t are free variables. We solve for x1 = s − 2t and x2 = −3s + t , so x = ( x1 , x2 , x3 , x4 ) = ( − s − 5t , − s − 3t , s, t ) = ( s, −3s, s, 0) + (−2t , t ,0, t ) = s u + t v where u = (1, −3,1, 0) and v = ( −2,1, 0,1).  1 3 2 5 −1   1 0 0 −2 −3  18.  2 7 4 11 2  → 0 1 0 1 4  = E A =      2 6 5 12 −7     0 0 1 2 −5    Thus x4 = s and x5 = t are free variables. We solve for x1 = 2 s + 3t , x2 = − s − 4t , and x3 = − 2s + 5t , so x = ( x1 , x2 , x3 , x4 , x5 ) = (2 s + 3t , − s − 4t , −2s + 5t , s, t ) = (2s, − s, −2s, s, 0) + (3t , −4t ,5t , 0, t ) = s u + t v
  5. 5. where u = (2, −1, −2,1, 0) and v = (3, −4, 5, 0,1).  1 −3 −5 −6  1 0 1 0  19.  2 1 4 −4  → 0 1 2 0  = E A =     1 3 7 1    0 0 0 1    Thus x3 = t is a free variable and x4 = 0. . We solve for x1 = −t and x2 = −2 t , so x = ( x1 , x2 , x3 , x4 ) = ( −t , −2t , t , 0) = t u where u = (−1, −2,1, 0).  1 5 1 −8  1 0 0 5  20.  2 5 0 −5  → 0 1 0 −3 = E A =      2 7 1 −9    0 0 1 2    Thus x4 = t is a free variable. We solve for x1 = −5t , x2 = 3 t , and x4 = −2t. so x = ( x1 , x2 , x3 , x4 ) = ( −5t ,3t , −2t , t ) = t u where u = (−5, 3, −2,1).  1 7 2 −3  1 0 0 3  21.  2 7 1 −4  → 0 1 0 −2  = E A =      3 5 −1 −5    0 0 1 4    Thus x4 = t is a free variable. We solve for x1 = −3t , x2 = 2 t , and x4 = −4t. so x = ( x1 , x2 , x3 , x4 ) = ( −3t , 2t , −4t , t ) = t u where u = ( −3, 2, −4,1). 1 3 3 3  1 0 0 6  22.  2 7 5 −1 → 0 1 0 −4  = E A =      2 7 4 −4    0 0 1 3    Thus x4 = t is a free variable. We solve for x1 = −6t , x2 = 4 t , and x4 = −3t. so x = ( x1 , x2 , x3 , x4 ) = ( −6t , 4t , −3t , t ) = t u
  6. 6. where u = ( −6, 4, −3,1). 23. Let u be a vector in W. Then 0u is also in W. But 0u = (0+0)u = 0u + 0u, so upon subtracting 0u from each side, we see that 0u = 0, the zero vector. 24. (a) Problem 23 shows that 0u = 0 for every vector u. (b) The fact that c0 = c(0 + 0) = c0 + c0 implies (upon adding –c0 to each side) that c0 = 0. (c) The fact that u + (–1)u = (1 + (–1))u = 0u = 0 means that (–1)u = –u. 25. If W is a subspace, then it contains the scalar multiples au and bv, and hence contains their sum au + bv. Conversely, if the subset W is closed under taking linear combinations of pairs of vectors, then it contains (1)u + (1)v = u + v and (c)u + (0)v = cu, and hence is a subspace. 26. The sum of any two scalar multiples of u is a scalar multiple of u, as is any scalar multiple of a scalar multiple of u. 27. Let a1u + b1v and a2u + b2v be two vectors in W = {au + bv}. Then the sum (a1u + b1v) + (a2u + b2v) = (a1 + a2 )u + (b1 + b2 ) v and the scalar multiple c( a1u + b1 v ) = (ca1 )u + (cb1 ) v are again scalar multiples of u and v, and hence are themselves elements of W. Hence W is a subspace. 28. If u and v are vectors in W, then Au = ku and Av = kv. It follows that A(au+bv) = a(Au) + b(Av) = a(ku) + b(kv) = k(au + bv), so the linear combination au + bv of u and v is also in W. Hence W is a subspace. 29. If Ax0 = b and y = x – x0, then Ay = A(x – x0) = Ax – Ax0 = Ax – b. Hence it is clear that Ay = 0 if and only if Ax = b. 30. Let W denote the intersection of the subspaces U and V. If u and v are vectors in W, then these two vectors are both in U and in V. Hence the linear combination au + bv is both in U and in V, and hence is in the intersection W, which therefore is a subspace. If U and V are non-coincident planes through the origin if R3, then their intersection W is a line through the origin.
  7. 7. 31. Let w1 and w2 be two vectors in the sum U + V. Then wi = ui + vi where ui is in U and vi is in V (i = 1, 2). Then the linear combination aw1 + bw2 = a(u1 + v1) + b(u2 + v2) = (au1 + bu2) + (av1 + bv2) is the sum of the vectors au1 + bu2 in U and av1 + bv2 in U, and therefore is an element of U + V. Thus U + V is a subspace. If U and V are noncollinear lines through the origin in R3, then U + V is a plane through the origin.

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