MATH 218 Final, Spring 2010

MATHEMATICS 218 SPRING SEMESTER 2008-09 FINAL EXAMINATION-SAMPLE SOLUTION Time: 120 minutes Date: June 3, 2009 Name:—————————————————————————- ID Number:———————————————————————– Circle your section number in the table below: Instructors Ejeili Fuleihan Itani Karam El Khoury Lyzzaik Nassif Tannous Section 7 3 11 1 10 6 2 4 Section 5 12 8 9 QUESTION GRADE 1 /25 2 /20 3 /20 4 /20 5 /20 6 /20 7 /20 8 /25 9 /30 TOTAL GRADE /200 Answer the following nine sets of questions on the allocated pages; the back of pages may be used if needed 1 2 1. Let T : R4 → R3 the transformation defined by T(x, y, z, w) = (x + y − z + w, 2x + y + 4z + w, 3x + y + 9z). (a) Show that T is a linear transformation. (4 points) Solution. T is a linear transformation since x+y−z+w, 2x+y+4z+w, and 3x + y + 9z are linear functions in x. y, and z. (b) Find the standard matrix [T]. (5 points) Solution. [T] =   1 1 −1 1 2 1 4 1 3 1 9 0   . (c) Find bases for the kernel of and range of T. (12 points) Solution. We have: [T] =   1 1 −1 1 2 1 4 1 3 1 9 0   ∼   1 1 −1 1 0 −1 6 −1 0 −2 12 −3   ∼   1 1 −1 1 0 −1 6 −1 0 0 0 −1   ∼   1 0 5 0 0 1 −6 1 0 0 0 1   ∼   1 0 5 0 0 1 −6 0 0 0 0 1   Thus the kernel of T consists of all vectors (x, y, z, w), where x = −5s, y = 6s, z = s, w = 0, or, kernel(T) =            −5s 6s s 0   : s ∈ R          =           s −5 6 1 0   : s ∈ R          ; Hence a basis for kernel(T) is the singleton            −5 6 1 0            . On the other hand, a basis for the range of T consists of the set        1 0 0   ,   0 1 0   ,   0 0 1        . (d) Find the rank and nullity of T. (4 points) Solution. It follows from (c) that rank (T)= 3 nullity (T)= 1. 3 2. Let T : P2 → R3 be the function defined by the formula T(p(x)) =   p(1) p(2) p(3)   ; here P2 is the vector space of all real polynomials of degree at most 2. (a) Show that T is a linear transformation. (8 points) Solution. Since for any p, q ∈ P2 and k ∈ R, T(p(x)+q(x)) =   p(1) + q(1) p(2) + q(2) p(3) + 3(3)   =   p(1) p(2) p(3)  +   q(1) q(2) q(3)   = T(p(x))+T(q(x)) and T(kp(x)) =   kp(1) kp(2) kp(3)   = k   p(1) p(2) p(3)   = kT(kp(x)), T is a linear transformation. (b) Show that T is one-to-one. (8 points) Solution. We show that kerT = {0}. Note that the general form of a polynomial in P2 is ax2 + bx + c, where a, b, c ∈ R. Suppose T(ax2 + bx + c) =   0 0 0   , or   a + b + c a + 2b + 4c a + 3b + 4c   =   0 0 0   . This yields the linear homogeneous system of equations a + b + c = 0 a + 2b + 4c = 0 a + 3b + 9c = 0 Since det   1 1 1 1 2 4 1 3 9   = 2 ̸= 0 the system admits only the trivial solution a = b = c = 0; hence kerT = {0} and T is one-to-one. (c) Show that T is onto. (4 points) Solution. Since dimP2 = dimR3 = 3 and T is one-to-one linear trans- formation, T is also onto. 4 3. Let P3 be the set of all polynomials of degree at most 3, and let W = {ax3 + bx2 + cx + d : b + c + d = 0}. (a) Show that W is a subspace of P3. (6 points) Solution. Let a1x3 +b1x2 +c1x+d1, a2x3 +b2x2 +c2x+d2 ∈ W. Then b1+c1+d1 = b2+c2+d2 = 0; consequently, b1+b2+c1+c2+d1+d2 = 0. This implies that (a1x3 + b1x2 + c1x + d1) + (a2x3 + b2x2 + c2x + d2) = (a1 + a2)x3 + (b1 + b2)x2 + (c1 + c2)x + (d1 + d2) ∈ W. Also, for any real k, k(a1x3 + b1x2 + c1x + d1) = ka1x3 + kb1x2 + kc1x + kd1 ∈ W since kb1 + kc1 + kd1 = k(b1 + c1 + d1) = 0. (b) Find a basis S for W. (6 points) Solution. We may write W = {ax3 + bx2 + cx − b − c = ax3 + b(x2 − 1) + c(x − 1) : a, b, c ∈ R.} Hence, P3 = span{x3, x2 − 1, x − 1.} But {x3, x2 − 1, x − 1} is linearly independent since ax3 + b(x2 − 1) + c(x − 1) ≡ 0 means ax3 + bx2 + cx − b − c ≡ 0 and consequently a = b = c = 0. Hence S = {x3, x2 − 1, x − 1}. (c) Give one vector in P3 but not in W. (2 points) Solution. Any polynomial ax3 + bx2 + cx + d, where b + c + d ̸= 0 is not in W; for instance the polynomial x3 + x2 + x + 1. (d) Complete S to a basis for P3. (6 points) Solution. We show that B = {x3, x2−1, x−1, 1} is a basis for P3. Since P3 has dimension 4, it suffices to show that B is linearly independent. This holds since ax3 + b(x2 − 1) + c(x − 1) + d.1 ≡ 0 means ax3 + bx2 + cx − b − c + d ≡ 0, and consequently a = b = c = d = 0. 5 4. Let A =   1 2 −1 2 3 5 0 4 1 1 2 0   (a) Find a basis for the row space of A and a basis for its orthogonal complement. (16 points) Solution: (a) Since A =   1 3 5 7 2 0 4 2 3 2 8 7   ≡   1 3 5 7 0 1 1 2 0 0 0 0   , a basis for the row space is {(1, 3, 5, 7), (0, 1, 1, 2)}. Since the orthogonal complement of the row space of A is the nullspace of A, the latter consists of all the vectors (x1, x2, x3, x4) where x1 = −2s − t, x2 = −s − 2t, x3 = s, x4 = t, s, t ∈ R. Thus the nullspace of A, or, the orthogonal complement of the row space of A is the set of all vectors            −2s − t −s − 2t s t   : s, t ∈ R          =          s      −2 −1 1 0      + t      −1 −2 0 1      : s, t ∈ R          = span            −2 −1 1 0   ,   −1 −2 0 1            . (b) Find a subset of the column vectors of A that forms a basis for the column space of A. (4 points) Solution. The desired set is        1 2 3   ,   3 0 2   .      6 5. Let P2 be the set of all polynomials of degree at most 2, and let p(x) = a0 + a1x + a2x2 and q(x) = b0 + b1x + b2x2 be in P2. Define on P2 the operation < p, q >= a0b0 + 2a1b1 + 3a2b2. (a) Show that < ., . > is an inner product on P2. (10 points) Solution. Let r(x) = c0 +c1x+c2x2. Then < ., . > is an inner product on P2 since the following hold: < p, q >= a0b0 + 2a1b1 + 3a2b2 = b0a0 + 2b1a1 + 3b2a2 =< q, p >, < p + q, r > = (a0 + b0)c0 + 2(a1 + b1)c1 + 3(a2 + b2)c2 = a0c0 + b0c0 + 2a1c1 + 2b1c1 + 3a2c2 + 3b2c2 = (a0c0 + 2a1c1 + 3a2c2) + (b0c0 + 2b1c1 + 3b2c2) = < p, r > + < q, r >, < kp, q >= ka0b0+2ka1b1+3ka2b2 = k(a0b0+2a1b1+3a2b2) = k < p, q > for every k ∈ R, and < p, p >= a2 0 + 2a2 1 + 3a2 2 ≥ 0 with equality if and only if a0 = a1 = a2 = 0, or p = 0. (b) Show that (a0b0 + 2a1b1 + 3a2b2)2 ≤ (a2 0 + 2a2 1 + 3a2 2)(b2 0 + 2b2 1 + 3b2 2). (5 points) Solution. By the Cauchy-Schwarz inequality we have | < p, q > |2 ≤ ∥p∥2∥q∥2, or the desired inequality. (c) Determine the cosine of the angle between the polynomials 1 − x + x2 and 1 + x − x2. (5 points) Solution. Since < 1 − x + x2, 1 + x − x2 >= (1)(1) + 2(−1)(1) + 3(1)(−1) = −4, ∥1−x+x2∥ = [1+2(−1)2+3(1)2]1/2 = 61/2 and ∥1+x−x2∥ = [1+2(1)2+3(−1)2]1/2 = 61/2, we have, cos(1−x+x2, 1+x−x2) = < 1 − x + x2, 1 + x − x2 > ∥1 − x + x2∥∥1 + x − x2∥ = −4 61/261/2 = −2 3 . 7 6. Let A =   1 0 2 0 1 1 1 2 0   . (a) Use the Gram-Schmidt process to transform the column vectors of A to an orthonormal basis of R3. (16 points) Solution: Let u1 = (1, 0, 1), u2 = (0, 1, 2), u3 = (2, 1, 0). Take v1 = u1. Define v2 = u2 − < u2, v1 > ∥v1∥2 v1 = (0, 1, 2) − (1, 0, 1) = (−1, 1, 1). Also define, v3 = u3 − �< u3, v1 > ∥v1∥2 v1 + < u3, v2 > ∥v12∥2 v2 � = (2, 1, 0) − {(1, 0, 1) + 1 3(1, −1, 1)} = (2/3, 4/3, −2/3). Thus an orthonormal system resulting from {u1, u2, u3} consists of q1 = v1/∥v1∥ = (1/ √ 2, 0, 1/ √ 2), q2 = v2/∥v2∥ = (−1/ √ 3, 1/ √ 3, 1/ √ 3), q3 = v3/∥v3∥ = (1/ √ 6, 2/ √ 6, −1/ √ 6). (b) Find the QR-decomposition of A. (4 points) Solution. (b) The QR-decomposition of A is A =   1/ √ 2 −1/ √ 3 1/ √ 6 0 1/ √ 3 2/ √ 6 1/ √ 2 1/ √ 3 −1/ √ 6     √ 2 √ 2 √ 2 0 √ 3 − √ 3/3 0 0 2 √ 6/3   . 8 7. Let A =   −2 3 1 −2 1 −1   and b =   1 −1 2   . (a) Find the least squares solution of the linear system Ax = b. (16 points) Solution. The normal equation of the linear system Ax = b is � −2 1 1 3 −2 −1 �   −2 3 1 −2 1 −1   � x y � = � −2 1 1 3 −2 −1 �   1 −1 2   which simplifies to � 6 −9 −9 14 � � x y � = � −1 3 � . But the matrix on the left-hand side is invertible; hence � x y � = 1 3 � 14 9 9 6 � � −1 3 � and the least squares solution is � x y � = � 13/3 3 � . (b) Find the orthogonal projection of b on the column space of A. (4 points) Solution. The orthogonal projection of b on the column space of A is given by A � 13/3 3 � =   −2 3 1 −2 1 −1   � 13/3 3 � =   1/3 −5/3 4/3   . 9 8. Let A =   0 1 1 1 0 1 1 1 0   . (a) Find the eigenvalues of A. (5 points) Solution. The CE of A is det(λI − A) = det    λ −1 −1 −1 λ −1 −1 −1 λ    = (λ + 1)2(λ − 2) = 0. Hence the eigenvalues are λ1 = −1, with algebraic multiplicity 2, and λ2 = 2, with algebraic multiplicity 1. (b) Show that A is diagonalizable. (10 points) Solution. We find the eigenspaces of λ1 and λ2. For the first, consider λ1I − A =    −1 −1 −1 −1 −1 −1 −1 −1 −1    ∼    1 1 1 0 0 0 0 0 0    . Hence, nullspace(λ1I − A) = span         −1 1 0    ,    −1 0 1         . In a similar manner we find nullspace(λ2I − A) = span         1 1 1         . Since the set of vectors         −1 1 0    ,    −1 0 1    ,    1 1 1         is linearly independent, the matrix A is diagonalizable. (c) Find a matrix P that diagonalizes A and determine P −1AP. (5 points) Solution. a matrix P that diagonalizes A is P =    −1 −1 1 1 0 1 0 1 1    and P −1AP =   −1 0 0 0 −1 0 0 0 2   10 (d) Find A10 (5 points) Solution. We have A = P   −1 0 0 0 −1 0 0 0 2   P −1 = 11 9. Indicate whether each of the following statements is TRUE (T) or FALSE (F) without justifying your answer. (3 points each) —-(a) If A is an n×n matrix that satisfies AAT = I, then det(A) = 1. —-(b) If A2 = A and λ is an eigenvalue of A, then λ = 0 or λ = 1. —-(c) If A is an n × n matrix invertible matrix, then the orthogonal complement of its nullspace is Rn. —-(d) A square matrix is diagonalizable if and only if λ = 0 is an eigenvalue. —-(e) Any linear system Ax = b satisfies rank[A|b] = rank(A). —-(f) Any matrix A can be expressed as a product of elementary matrices. —-(g) If dim V < dim W < ∞, then there exists a one-to-one linear transformation T : V → W. —-(h) If a linear transformation T : R2 → R satisfies T(2, −1) ̸= 0, then it is onto. —-(i) The dimension of the vector space of 3 × 3 matrices is 10. —-(j) If A is an m × n matrix, then ATA is invertible if and only if the set of column vectors of A is linearly independent.