UCD Linear Algebra II Sample Final Exam with Answers PDF

!!! Contains Solutions !!! University College Dublin An Coláiste Ollscoile Baile Átha Cliath Spring Examinations, Sample Final Exam with Final Answers Linear Algebra II...

!!! Contains Solutions !!! University College Dublin An Coláiste Ollscoile Baile Átha Cliath Spring Examinations, Sample Final Exam with Final Answers Linear Algebra II MST 20050 Instructions to Candidates § The time allowed for completing this paper is two hours. § Full marks will be awarded for complete and correct answers to all questions. § All questions carry equal marks. § Write legibly, and show all your work! Notation § We will only consider the field R of real numbers. § Vectors in Rn will be written as column vectors. § The zero vector in Rn is denoted 0n. § Mm,n pRq denotes the vector space of m ˆ n real matrices. § Mn pRq is shorthand for the vector space Mn,n pRq. § Rn rxs denotes the vector space of real polynomials in the variable x of maximal degree n. Page 1 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 1. In each of the following questions, indicate if the statement is True or False. If the statement is true, provide a short justification. If the statement is false, give a counterexample. (a) Let A P M2 pRq. If detpAq ‰ 0, then the homogeneous system of linear equations AX “ 02 only has the trivial solution. True. (b) Let A P M2 pRq and let B P R2. If detpAq “ 0, then the non-homogeneous system of linear equations AX “ B always has infinitely many solutions. False. (c) If B1 “ pv1 , v2 , v3 , v4 q is a basis of R4 , then B2 “ pv1 ` v2 , v2 ` v3 , v3 ` v4 , v4 ` v1 q is again a basis of R4. False. (d) Let D denote the formal derivative. The set U “ tppxq P R3 rxs | Dpppxqq “ 1u is a subspace of R3 rxs. False. (a) True. detpAq ­“ 0 ùñ A is invertible ùñ X “ A´1 02 “ 02. ȷ „ „ ȷ 1 0 1 (b) False. The system could be inconsistent. For example, let A “ and B “. 1 0 2 „ ȷ x (Note that detpAq “ 0.) Let X “. Then y AX “ B ðñ x “ 1 “ 2, contradiction. (c) False. The tuple B2 is linearly dependent (and so can’t be a basis): pv1 ` v2 q ´ pv2 ` v3 q ` pv3 ` v4 q ´ pv4 ` v1 q “ v1 ` v2 ´ v2 ´ v3 ` v3 ` v4 ´ v4 ´ v1 “ 04. (d) False. The polynomials x and 5 ` x are in U. But their sum is not: Dpx ` 5 ` xq “ Dp5 ` 2xq “ 2 ­“ 1. Page 2 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 2. In R3 , consider the ordered set of vectors ¨» fi » fi » fi˛ 1 2 0 S“ ˝ – ´1 , 1 , 1 fl‚. fl – fl – 1 ´7 ´3 Answer the following questions and in each case give a short justification of your answer: (a) Is S linearly independent? No. (b) Is S a spanning set for R3 ? No. (c) Is S a basis of R3 ? No. (d) What is the dimension of SpanpSq? 2 (a) Compute the determinant: 1 2 0 1 1 ´1 1 ´1 1 1 “ ´2 “ ´3 ` 7 ´ 2p3 ´ 1q “ 4 ´ 4 “ 0. ´7 ´3 1 ´3 1 ´7 ´3 Thus S is NOT linearly independent. (E.g., by Proposition 2.90.) (b) No. Because of (a) and Proposition 2.90. (c) No. From (a). (d) The span of S is the column space of the matrix A whose columns are the vectors in S. The dimension of SpanpSq is thus the column rank of A, which equals the row rank of A. So it suffices to compute a REF of A, and count the number of pivots. » fi » fi » fi 1 2 0 1 2 0 1 2 0 R2 ÑR2 `R1 – R ÑR `3R2 – –´1 1 1fl ÝÝÝÝÝÝÝÑ 0 3 1fl ÝÝ3ÝÝÝ3ÝÝÝÑ 0 3 1fl. R3 ÑR3 ´R1 1 ´7 ´3 0 ´9 ´3 0 0 0 There are 2 pivots, so dim SpanpSq “ 2. Page 3 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 3. Consider the matrix » fi 5 3 1 4 A “ –´2 1 ´7 ´6fl P M3,4 pRq. 3 4 ´6 ´2 (a) Determine a basis of the row space RpAq. First 2 rows of the REF. (b) Determine a basis of the column space CpAq. First 2 columns of A. (c) Determine a basis of the row space RpAq that consists of rows of A. First 2 rows of A. (d) Determine a basis of the null space N pAq. ˆ„ ´5 ȷ „ ´4 ȷ˙ 3 , 20 1 0 1 Note: other answers are possible, depending on how you perform the computations. (a) We use the Row Space Algorithm; reduce A to REF B: » fi » fi » fi 5 3 1 4 ´2 1 ´7 ´6 ´2 1 ´7 ´6 R ØR2 R3 ÑR3 ´R2 – A “ –´2 1 ´7 ´6fl ÝÝÝ1ÝÝÝÝ Ñ– 3 4 ´6 ´2fl ÝÝÝÝÝÝÝÑ 3 4 ´6 ´2fl R2 ÑR2 `R1 3 4 ´6 ´2 3 4 ´6 ´2 0 0 0 0 » fi ´2 R2 ` 32 R1 1 ´7 ´6 11 ÝÝÝÝÝÑ – 0 2 ´ 33 2 ´11fl “: B. 0 0 0 0 The first two rows of B are a basis of RpAq. Note: other answers are possible, depending on how the student performs the row reduction. (b) We use the Column Space Algorithm. From (a) we see that the first two columns of B contain the pivots. Hence, the first two columns of A form a basis is CpAq. (c) We apply the Column Space Algorithm to the transposed of A: » fi » fi » fi 5 ´2 3 1 ´7 ´6 1 ´7 ´6 — 3 1 4ffi R ØR — 3 1 4ffi R2 ÑR2 ´3R1 , R3 ÑR3 ´5R1 — 0 22 22ffi AT “ — ffi 1 2 — –1 ´7 ´6fl ÝÝÝÝÝÑ –5 ffi Ý ÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑ — ffi ´2 3 fl R4 ÑR4 ´4R1 – 0 33 33fl 4 ´6 ´2 4 ´6 ´2 0 22 22 Page 4 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 » fi 1 ´7 ´6 scale R1 ,R2 ,R3 —0 1 1ffi ÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑ — ffi “: C. R3 ÑR3 ´R2 , R4 ÑR4 ´R2 – 0 0 0fl 0 0 0 The first two columns of C contain the pivots. Hence, the first two columns of AT are a basis of CpAT q. Thus, the first two rows of A are a basis of RpAq. (d) The null „ x ȷspace of A is the solution space of the homogeneous linear system AX “ 03 , with X “ yz. In (a) we have found a REF, B, of the coefficient matrix of this system. The w solutions of the system are thus determined by # ´2x ` y ´ 7z ´ 6w “ 0. y ´ 3z ´ 2w “ 0 Since there are 4 unknowns and 2 equations, there are 2 parameters: let z “ s and w “ t, with s, t P R. Then x “ ´5s ´ 4t, y “ 3s ` 2t. Hence, " „ ´5 ȷ „ ´4 ȷ * ˆ„ ´5 ȷ „ ´4 ȷ˙ 3 2 N pAq “ s 1 ` t 0 : s, t P R ùñ 3 , 20 is a basis of N pAq. 1 0 1 0 1 Page 5 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 4. Consider the upper triangular matrix » fi 1 3 1 A “ –0 2 4fl P M3 pRq. 0 0 3 (a) Determine the eigenvalues of A. 1,2,3 (b) For each eigenvalue of A, determine the corresponding eigenspace. ! ” ı ) ! ” ı ) ! ” 13{2 ı ) 1 3 E1 “ t 0 : t P R , E2 “ t 1 : t P R , E3 “ t 4 : t P R. 0 0 1 (c) Determine real matrices B and D with B invertible and D diagonal, such that B ´1 AB » “ D. fi » fi 1 3 13{2 1 0 0 B :“ 0 1 – 4 fl and D “ 0– 2 0fl. 0 0 1 0 0 3 (d) Determine the trace of A3. 36 NOTE: students may produce different solutions, depending on the order in which they consider the eigenvalues, the choices they make for the bases of the eigenspaces, etc. (a) Characteristic polynomial: pA pxq “ detpA ´ xI3 q “ p1 ´ xqp2 ´ xqp3 ´ xq. Easy since A is upper triangular. Thus, the eigenvalues are λ1 “ 1, λ2 “ 2, λ3 “ 3. ”xı (b) E1 “ N pA ´ I3 q ùñ we solve pA ´ I3 qX “ 03 with X “ y. Reduce coefficient matrix z to REF: » fi » fi » fi » fi 0 3 1 0 1 4 0 1 4 0 1 4 R1 ØR2 – R ÑR ´3R1 – scale R A ´ I3 “ 0 1 4 ÝÝÝÝÝÑ 0 – fl 3 1fl ÝÝ2ÝÝÝ2ÝÝÝÑ 0 0 ´11fl ÝÝÝÝÝÝÝ2ÝÑ –0 0 1fl R3 ÑR3 ´2R2 0 0 2 0 0 2 0 0 2 0 0 0 ! ”1ı ) ùñ y “ z “ 0, x “ t, t P R ùñ E1 “ t 0 : t P R. 0 E2 “ N pA ´ 2I3 q. This time we have » fi » fi ´1 3 1 ´1 3 1 scale R2 A ´ 2I3 “ – 0 0 4fl ÝÝÝÝÝÝÝ Ñ – 0 0 1fl ùñ x “ 3y, z “ 0, y “ t, t P R. R3 ÑR3 ´R2 0 0 1 0 0 0 ! ”3ı ) ùñ E2 “ t 1 : t P R. 0 Page 6 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 E3 “ N pA ´ 3I3 q. This time we have » fi ´2 3 1 13 A ´ 3I3 “ – 0 ´1 4fl ùñ x “ 2 t, y “ 4t, z “ t, t P R. 0 0 0 ! ” 13{2 ı ) ùñ E3 “ t 4 : t P R. 1 (c) The eigenvalues are all distinct, »so A is diagonalizable fi by»the Diagonalization fi Algorithm 1 3 13{2 1 0 0 in the Lecture Notes. Let B :“ –0 1 4 fl and D “ –0 2 0fl. Then B ´1 AB “ D 0 0 1 0 0 3 (d) trpA3 q “ sum of eigenvalues of A3. By Theorem 3.47, these are λ31 “ 1, λ32 “ 8, λ33 “ 27. Hence, trpA3 q “ 1 ` 8 ` 27 “ 36. Page 7 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 5. Consider the floor function f : R Ñ Z, defined by f pxq :“ greatest integer less than or equal to x. For example: f pπq “ 3. (a) Is f injective? Justify your answer. No. (b) Is f surjective? Justify your answer. Yes. (c) Determine the image of 4, i.e., determine f p4q. 4 (d) Determine the inverse image of 4, i.e., determine the set f Ð p4q. r4, 5q (a) No. f p3.1q “ f p3.2q “ 3. (b) Yes. For any n P Z, we have f pnq “ n. (c) f p4q “ 4. (d) f Ð p4q “ tx P R : f pxq “ 4u “ tx P R : 4 ď x ă 5u “ r4, 5q. Page 8 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 6. Consider the linear transformation „ ȷ „ ȷ 2 a b a f : M2 pRq ÝÑ R , ÞÝÑ. c d d For M2 pRq, consider the standard basis S “ pE11 , E12 , E21 , E22 q and also the basis ˆ „ ȷ „ ȷ „ ȷ „ ȷ˙ 1 0 0 2 0 0 0 0 B “ B11 “ , B12 “ , B21 “ , B22 “. 0 0 0 0 3 0 0 4 For R2 , consider the standard basis E “ pe1 , e2 q and also the basis ˆ „ ȷ „ ȷ˙ 1 0 C “ c1 “ , c2 “. 0 2 (a) Compute matS,E pf q, the matrix of f with respect to the standard bases S and E. „ ȷ 1 0 0 0 0 0 0 1 (b) Compute the change of basis matrix MCÑE. „ ȷ 1 0 0 1{2 (c) Compute the change of basis matrix MSÑB. » fi 1 0 0 0 —0 2 0 0ffi — ffi –0 0 3 0fl 0 0 0 4 (d) Using (a), (b) and (c), determine matB,C pf q, the matrix of f with respect to the bases B andȷC. „ 1 0 0 0 0 0 0 2 (a) f pE11 q “ r 10 s “ e1 ` 0e2 $ ’ ’ &f pE q “ r 0 s “ 0e ` 0e ’ „ ȷ 12 0 1 2 1 0 0 0 ùñ matS,E pf q “ ’f pE21 q “ r 00 s “ 0e1 ` 0e2 ’ 0 0 0 1 ’ 0 % f pE22 q “ r 1 s “ 0e1 ` e2 (b) Since MCÑE :“ matE,C pidR2 q we have # e1 “ r 10 s “ c1 ` 0c2 „ ȷ 1 0 ùñ MCÑE “ e2 “ r 01 s “ 0c1 ` 12 c2 0 1{2 Page 9 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 (c) Since MSÑB :“ matB,S pidM2 pRq q we have $ ’B11 “ E11 ` 0E12 ` 0E21 ` 0E22 » 1 0 0 0 fi ’ ’ &B “ 0E11 ` 2E12 ` 0E21 ` 0E22 —0 12 2 0 0ffi ùñ MSÑB “— –0 ffi ’ ’B21 “ 0E11 ` 0E12 ` 3E21 ` 0E22 0 3 0fl ’ % B22 “ 0E11 ` 0E12 ` 0E21 ` 4E22 0 0 0 4 (d) From Theorem 5.62 we know that » fi „ ȷ 1 0 0 0 „ ȷ 1 0 0 0 — 0 2 0 0ffi ffi “ 1 0 0 0. matB,C pf q “ MCÑE ¨matS,E pf q¨MSÑB “ ¨— 0 0 0 1{2 –0 0 3 0fl 0 0 0 2 0 0 0 4 Page 10 of 12 Spring, Sample Final Exam with Final Answers Linear Algebra II MST 20050 — o0o — Page 11 of 12