Chapter 4 notes - ISM Linear Algebra Section 4.1 Chapter 4...

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ISM: Linear Algebra Section 4.1 Chapter 4 4.1 1. Not a subspace since it does not contain the neutral element, that is, the function f ( t ) = 0, for all t . 2. This subset V is a subspace of P 2 : The neutral element f ( t ) = 0 (for all t ) is in V . If f and g are in V (so that f (2) = g (2) = 0), then ( f + g )(2) = f (2) + g (2) = 0 + 0 = 0, so that f + g is in V . If f is in V (so that f (2) = 0), and k is any constant, then ( kf )(2) = kf (2) = 0, so that kf is in V . A polynomial f ( t ) = a + bt + ct 2 is in V if f (2) = a + 2 b + 4 c = 0, or a = - 2 b - 4 c . The general element of V is of the form f ( t ) = ( - 2 b - 4 c ) + bt + ct 2 = b ( t - 2) + c ( t 2 - 4), so that t - 2, t 2 - 4 is a basis of V . 3. This subset V is a subspace of P 2 : The neutral element f ( t ) = 0 (for all t ) is in V since f (1) = f (2) = 0. If f and g are in V (so that f (1) = f (2) and g (1) = g (2)), then ( f + g ) (1) = ( f + g )(1) = f (1) + g (1) = f (2) + g (2) = ( f + g )(2), so that f + g is in V . If f is in V (so that f (1) = f (2)) and k is any constant, then ( kf ) (1) = ( kf )(1) = kf (1) = kf (2) = ( kf )(2), so that kf is in V . If f ( t ) = a + bt + ct 2 then f ( t ) = b + 2 ct , and f is in V if f (1) = b + 2 c = a + 2 b + 4 c = f (2), or a + b + 2 c = 0. The general element of V is of the form f ( t ) = ( - b - 2 c ) + bt + ct 2 = b ( t - 1) + c ( t 2 - 2), so that t - 1 , t 2 - 2 is a basis of V . 4. This subset V is a subspace of P 2 : The neutral element f ( t ) = 0 (for all t ) is in V since 1 0 0 dt = 0. If f and g are in V so that 1 0 f = 1 0 g = 0 then 1 0 ( f + g ) = 1 0 f + 1 0 g = 0, so that f + g is in V . 179
Chapter 4 ISM: Linear Algebra If f is in V so that 1 0 f = 0 and k is any constant, then 1 0 kf = k 1 0 f = 0, so that kf is in V . If f ( t ) = a + bt + ct 2 then 1 0 f ( t ) dt = at + b 2 t 2 + c 3 t 3 1 0 = a + b 2 + c 3 = 0 if a = - b 2 - c 3 . The general element of V is f ( t ) = - b 2 - c 3 + bt + ct 2 = b t - 1 2 + c t 2 - 1 3 , so that t - 1 2 , t 2 - 1 3 is a basis of V . 5. If p ( t ) = a + bt + ct 2 then p ( - t ) = a - bt + ct 2 and - p ( - t ) = - a + bt - ct 2 . Comparing coe ffi cients we see that p ( t ) = - p ( - t ) for all t if (and only if) a = c = 0. The general element of the subset is of the form p ( t ) = bt . These polynomials form a subspace of P 2 , with basis t . 6. Not a subspace, since I 3 and - I 3 are invertible, but their sum is not. 7. The set V of diagonal 3 × 3 matrices is a subspace of R 3 × 3 : a. The zero matrix 0 0 0 0 0 0 0 0 0 is in V , b. If A = a 0 0 0 b 0 0 0 c and B = p 0 0 0 q 0 0 0 r are in V , then so is their sum A + B = a + p 0 0 0 b + q 0 0 0 c + r . c. If A = a 0 0 0 b 0 0 0 c is in V , then so is kA = ka 0 0 0 kb 0 0 0 kc , for all constants k . 8. This is a subspace; the justification is analogous to Exercise 7. 180
ISM: Linear Algebra Section 4.1 9. Not a subspace; consider multiplication with a negative scalar. I 3 belongs to the set, but - I 3 doesn’t. 10. Let v = 1 2 3 . Let V be the set of all 3 × 3 matrices A such that Av = 0. Then V is a subspace of R 3 × 3 : a. The zero matrix 0 is in V , since 0 v = 0. b. If A and B are in V , then so is A + B , since ( A + B ) v = Av + Bv = 0 + 0 = 0. c. If A is in V , then so is kA for all scalars k , since ( kA ) v = k ( Av ) = k 0 = 0. 11. Not a subspace: I 3 is in rref, but the scalar multiple 2 I 3 isn’t.