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Mathematics 2270 EXAM I-Solutions Fall 2004 1. a) Find the rank of the following matrices: 1 1 0 A = 2 3 4 1 0 1 Solution: By rowreduction you find that rank A = 3 1 0 5 6 B = 0 1 0 −1 0 0 0 0 Solution: B is already in row reduced form, then rank B = 2. b) Give the number of solutions for the following linear systems (you do not need to solve them! You can use part a)); explain: A~x = ~b Solution: As rank A = 3 it is invertible, which means that the system has always uniquesolution. 0 B~y = 1 0 Solution: The augmented matrix of the system is 1 0 5 6 |0 0 1 0 −1 | 1 0 0 0 0 |0 This system is consistent and has rank B = 2 it has two free variables, then it has infinite number of solutions. 1 1 2 2 √ 2. Let T : R → R be the projection on the line parallel to ~u = 5 . −2 a) Find the matrix representing T . Solution: We name the matrix A; using the formula we find 1/5 −2/5 A= −2/5 4/5 b) Show that T is not one-to-one. Solution: We prove that by showing that ker T 6= {0}. It is sufficient then to find a non-zero vector in ker T . Moreover as know that T (x) = (x.u)u we will choose we 2 . Then it perpendicular to u. So we take x = 1 1 x.u = √ (2 × 1 + (−2) × 1) = 0 5 and then T (x) = 0 which means that x ∈ ker T ; and as x 6= 0 then ker T 6= {0} so T is not one-to-one. 3. Let T : R2 → R3 and U : R3 → R2 be two linear maps represented by the matrices A and B respectively, where 1 −1 0 2 0 A= 2 0 and B = 1 −1 1 −1 2 a) Compute the matrix product C = BA. Solution: We compute the product: C = BA = 1 −1 0 2 0 4 0 2 0 = 1 −1 1 −2 1 −1 2 b) Deduce from a) that U ◦ T is a bijection (invertible). Solution: C = BA represents U ◦ T . Moreover C is invertible because det C = 4 + 2 = 6 6= 0 so U ◦ T is a bijection. c) Show that U is onto. Solution: Let z be a vector in R2 , we look for an antecedent of z by U (meaning we look for a vector y ∈ R3 such that U (y) = z). As U ◦ T is a bijection we know that it exists a (unique) vector x ∈ R3 such that U ◦ T (x) = z. If we define y = T (x) ∈ R3 we have z = U ◦ T (x) = U (T (x)) = U (y) Which proves that U is onto. 4. Define the following matrices: A= 1 −1 0 2 0 1 2 −1 and B = 2 −1 −4 2 Let T : R3 → R2 and U : R2 → R3 be the linear maps represented by A and B respectively. a) Compute the matrix product C = AB. Deduce the map T ◦ U . Solution: We compute 0 0 C = AB = 0 0 As C represents T ◦U , then T ◦U = 0 (the zero map, that is for all x ∈ R2 , T ◦U (x) = 0. b) Prove that the 2 vectors −1 2 v1 = 2 and v2 = −1 2 −4 are in the kernel of T . Solution: for i = 1, 2 we have vi = U (ei ), then T (vi ) = T (U (ei )) = T ◦ U (ei ) = 0 because T ◦ U = 0 Then v1 , v2 ∈ ker T . c) [Extra credit]: General case: prove that for 2 linear maps f : Rm → Rp and g : Rp → Rn such that g ◦ f = 0 (that is the zero map, for all ~x ∈ Rm , g ◦ f (~x) = ~0) we have Im f ⊂ ker g Let y ∈ Im f . Then it exists x ∈ Rm such that f (x) = y. Then we Solution: compute g(y) = g(f (x)) = g ◦ f (x) = 0 because g ◦ f is the zero map So that y ∈ ker g. So we proved Im f ⊂ ker g. 5. a) Show that ( x1 5x1 − x2 + 3x3 = 0 3 V = x2 ∈ R | x1 + 6x3 = 0 x3 is a linear subspace of R3 . Solution: We define the following linear map T : R 3 → R2 x1 5x − x + 3x 1 2 3 x2 7→ x1 + 6x3 x3 then we have ker T = V , which proves that V is a linear subspace. 2t + u b) Show that W = −t + 3u ∈ R3 | t, u ∈ R is a vector subspace of R3 . 5t − u Solution: We have 2 1 2t + u W = −t + 3u ∈ R3 | t, u ∈ R = t −1 + u 3 ∈ R3 | t, u ∈ R 5 −1 5t − u 1 2 −1 , 3 = span −1 5 which proves that W is alinear subspace. x1 ∈ R2 | x2 = x21 is not a linear subspace of R3 . c) Show that X = x2 Solution: We will prove that X is not closed under scaling. We remark that 1 1 2 ∈ / X because 2 6= 22 = 4. Then ∈ X because it satisfies 1 = 1 . But then 2 1 1 X is not closed under scaling, then X is not a linear subspace.