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SUBSPACE TEST Strategy: We want to see if H is a subspace of V. 1.) Is the zero vector of V also in H? If no, then H is not a subspace of V. If yes, then move on to step 2. 2.) Identify c, u , v , and list any “facts”. 3.) Is u + v in H? If yes, then move on to step 4. If no, then give a specific example to show that u and v are in H but u + v is not in H. This shows that H is not a subspace of V. 4.) Is cu in H? If yes, then H is a subspace of V. If no, then give a specific example to show that u is in H but cu is not in H. This shows that H is not a subspace of V. Note: If H is defined with the variables u, v, or c, feel free to use different letters. For example, let your scalar be d instead of c or let your two vectors be g and h instead of u and v. There is nothing special about using c, u, and v. Just be sure to be consistent. Don't start using d as your scalar and end up using c as your scalar. 1 Example 1 a + b + c Let H = 2a + 3b a, b, c ∈ R . Is H a subspace of R 3 ? 4c 1.) Is the zero vector of R 3 also in H? a + b + c 0 We need to see if the equation 2a + 3b = 0 has a solution. A solution to this equation is 4c 0 a = b = c = 0 . So 0 is in H. 2.) Identify d, u , v , and list any “facts”. (I’m going to use d as my scalar because c has already been used.) u1 + u 2 + u 3 v1 + v 2 + v 3 d ∈ R , u = 2u1 + 3u 2 , and v = 2v 1 + 3v 2 4u 3 4v 3 3.) Is u + v in H? u1 + u 2 + u 3 + v 1 + v 2 + v 3 (u1 + v 1 ) + (u 2 + v 2 ) + (u 3 + v 3 ) u + v = 2u1 + 3u 2 + 2v 1 + 3v 2 = 2(u1 + v 1 ) + 3(u 2 + v 2 ) 4u 3 + 4v 3 4(u 3 + v 3 ) Here a = u1 + v 1 , b = u 2 + v 2 , and c = u 3 + v 3 . So u + v is in H. 4.) Is du in H? d(u1 + u 2 + u 3 ) du1 + du 2 + du 3 du = d(2u1 + 3u 2 ) = 2du1 + 3du 2 d(4u 3 ) 4du 3 Here a = du1 , b = du 2 , and c = du 3 . So du is in H. This means that H is a subspace of R 3 . 2 Example 2 a b Let W = : c = a + 2b & d = a − 3b . Is W a subspace of R 4 ? c d 1.) Is the zero vector of R 4 also in W? a a 0 b b 0 has a solution. We need to see if the equation = = c a + 2b 0 d a − 3b 0 A solution to this equation is a = b = 0 . So 0 is in W. 2.) Identify k, u , v , and list any “facts”. u1 v1 u2 v k ∈R , u = , v = 2 u3 v 3 u v 4 4 Our “facts” are u 3 = u1 + 2u 2 , u 4 = u1 − 3u 2 , v 3 = v 1 + 2v 2 , and v 4 = v 1 − 3 v 2 . 3.) Is u + v in W? u1 + v 1 u2 + v 2 u+v = u + v3 3 u + v 4 4 Here a = u1 + v 1 , b = u 2 + v 2 , c = u 3 + v 3 , and d = u 4 + v 4 . We need to know if c = a + 2b and d = a − 3b . c = u 3 + v 3 = (u1 + 2u 2 ) + (v 1 + 2v 2 ) = (u1 + v 1 ) + 2(u 2 + v 2 ) = a + 2b d = u 4 + v 4 = (u1 − 3u 2 ) + (v 1 − 3v 2 ) = (u1 + v 1 ) − 3(u 2 + v 2 ) = a − 3b So u + v is in W. 4.) Is ku in W? ku 1 ku 2 ku = ku 3 ku 4 Here a = ku1 , b = ku 2 , c = ku 3 , and d = ku 4 . We need to know if c = a + 2b and d = a − 3b . c = ku 3 = k (u1 + 2u 2 ) = ku1 + 2ku 2 = a + 2b and d = ku 4 = k (u1 − 3u 2 ) = ku1 − 3ku 2 = a − 3b So ku is in W. This means that W is a subspace of R 4 . 3 Example 3 2a + b Let H = 2b + 4c − 1 a, b, c ∈ R . Is H a subspace of R a + b + c 1.) Is the zero vector of R 3 3 ? also in H? 2a + b 0 Does the equation 2b + 4c − 1 = 0 have a solution? No, this equation can be written as the a + b + c 0 2a + b = 0 system 2b + 4c = 1 . Looking at the augmented matrix a+b+c =0 2 1 0 0 1 0 − 10 0 2 4 1 ~ 0 1 2 0 we can see 1 1 1 0 0 0 0 1 that this system has no solution. This means that 0 is not in H. So H is not a subspace of R 3 . 4 Example 4 abc Let H = 2a + 3b a, b, c ∈ R . Is H a subspace of R 3 ? 4c 1.) Is the zero vector of R 3 also in H? abc 0 We need to see if the equation 2a + 3b = 0 has a solution. A solution to this equation is 4c 0 a = b = c = 0 . So 0 is in H. 2.) Identify d, u , v , and list any “facts”. (I’m going to use d as my scalar because c has already been used.) u1 * u 2 * u 3 v1 * v 2 * v 3 d ∈ R , u = 2u1 + 3u 2 , and v = 2v 1 + 3v 2 4u 3 4v 3 3.) Is u + v in H? u1 * u 2 * u 3 + v 1 * v 2 * v 3 u 1 * u 2 * u 3 + v 1 * v 2 * v 3 u + v = 2u1 + 3u 2 + 2v 1 + 3v 2 = 2(u1 + v 1 ) + 3(u 2 + v 2 ) 4u 3 + 4v 3 4(u 3 + v 3 ) Looking at the last two entries we see that a = u1 + v 1 , b = u 2 + v 2 , and c = u 3 + v 3 . However, abc = (u1 + v 1 )(u 2 + v 2 )(u 3 + v 3 ) ≠ u1 * u 2 * u 3 + v 1 * v 2 * v 3 , in general. 8 27 For example, let u1 = u 2 = u 3 = 2 , v 2 = 3 , and v 1 = v 3 = −3 . So u = 10 , v = 3 , and 8 − 12 35 abc 13 − 3b 13 − 3b u + v = 13 = 2a + 3b . Now we have that c = −1 , a = , and * b * (− 1) = 35 2 2 − 4 4c or 3b 2 − 13b − 70 = 0 . Using the quadratic formula, we get b = 13 ± 169 − 840 ∉ R . So, in 6 general, u + v is not in H. This means that H is not a subspace of R 3 . 5 EXERCISES n n For problems 1-5, determine if H is a subspace of R . If so, write it as the span of a set of vectors in R and the column space of a matrix. 1.) a + b + 2c H = ab + c a, b, c ∈ R 2.) a + 2b H = a + b + c 0 3.) x + 2y − z H = 4y + 6z 7 x − 3 y + 14z 4.) 2x + 3 y + 4z + 6 w y−z H = 1 x + y − z − w 5.) a − b + 7c 3a − 4b + 8c − 9d H = d−c 6a − 17d a, b, c ∈ R x, y, z ∈ R x, y, z, w ∈ R a, b, c, d ∈ R 4 6 1 2 3 6.) Let A = − 1 5 − 3 2 . Is the solution set of A x = 0 as subspace of R ? Is the solution set of A x = 0 as 1 2 3 4 4 subspace of R ? 7.) 4 6 1 2 1 3 Let A = − 1 5 − 3 2 and b = 2 . Is the solution set of A x = b as subspace of R ? Is the solution set 1 2 4 3 4 4 of A x = b as subspace of R ? n For problems 8-14, determine if W is a subspace of R . 8.) a W = b c abc = 0 9.) x W = y z x + y + z = 0 x 10.) W = y sin x = π 2 r 11.) W = s t r + s ≤ t x y 12.) W = z w 2x + 3 y = z + w a b 13.) W = c d a + b = 3c and b + c = 4d 14.) Let T : R n → R m be a linear transformation. Determine if W = x n subspace of R . () x ∈ R n and T x = 0 is a 6 ANSWERS TO THE ODD EXERCISES 2 1.) H is not a subspace of R . 3.) 1 H is a subspace of R where H = span 0 , 7 2 4 , − 3 2 − 1 1 − 1 A = 0 4 6 . 6 = ColA with 14 7 − 3 14 1 3 4 5.) H is a subspace of R where H = span , 0 6 − 1 − 4 0 , 0 7 8 − 1, 0 3 3 0 0 1 −1 7 8 − 9 − 9 3 − 4 = ColA with . A = 0 1 0 −1 1 − 17 6 0 0 − 17 4 7.) The solution set of A x = b is not a subspace of R or R . 9.) W is a subspace of R . 3 3 11.) W is not a subspace of R . 4 13.) W is a subspace of R . 7