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Introduction to Linear Algebra

by: Braeden Lind

Introduction to Linear Algebra MA 405

Marketplace > North Carolina State University > Mathematics (M) > MA 405 > Introduction to Linear Algebra
Braeden Lind
GPA 3.93

L. Levy

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L. Levy
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This 6 page Study Guide was uploaded by Braeden Lind on Thursday October 15, 2015. The Study Guide belongs to MA 405 at North Carolina State University taught by L. Levy in Fall. Since its upload, it has received 34 views. For similar materials see /class/223696/ma-405-north-carolina-state-university in Mathematics (M) at North Carolina State University.

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Date Created: 10/15/15
Test 4 Review The test covers sections 41 Kernel and Range only 43 61 63 and the topics covered in class from 51 52 55 Know all theorems and de nitions presented in class Know how to work ill homework problems especially those not collected for grading De nitions Kernel range similar matrices eigenvalue eigenvector characteristic polynomial detI 7 A diagonalizable matrix exponential length distance angle dot product standard inner product orthogonal projection orthogonal complement orthogonal set orthonormal set Theorems 431 611 631 632 511 true for R 551 and 552 where V R Sample Problems H Range and Kernel see test 3 review Let T R2 a R2 be the linear transformation de ned by Tx1z2T 7z22x1T Consider the two bases A 111 11T12 01T and B wl 10Tw2 21T for R2 Find a The matrix A representing T with respect to A E0 b The matrix B representing T with respect to B c The matrix P such that B P lAP Consider the matrix 9 4 i5 1 A 1 0 71 0 1 71 a Find the eigenvalues and corresponding eigenspaces for A b Find a matrix P such that P lAP is diagonal c Find 6A 7 For what values of 04 will the following matrix fail to be diagonalizable 4 6 i2 71 71 1 0 0 04 9 page 341 12 Show that a nonzero nilpotent matrix is defective Hint Find the eigen values of a nilpotent matrix 6 If A2 04A for some scalor Oz 31 07 show that the eigenvalues of A must all be 0 or a 7 Reminder Similar matrices have the same eigenvalues because they have the same characteristic polynomial Do they necessarily have the same eigenvectors Prove or nd a counterexample Hint Consider a 2 gtlt 2 diagonalizable matrix to get an idea 8 a Show that a 2 gtlt 2 symmetric matrix must have real eigenvalues Hint For a general 2 gtlt 2 symmetric matrix nd the characteristic polynomial and compute the discriminant b Show that a 2 gtlt 2 skew symmetric matrix must have pure imaginary eigenvalues That is all eigenvalues are a multiple of 239 9 Consider u 17 000T and 1 1111T7 both vectors in R4 a Find and b Find lt um gt the standard inner product or dot product c Find the angle between u and 1 d Are u and 1 orthogonal e If S spanuv7 nd a basis for Si 10 Find the point on the line y 3x closest to the point 17 2 11 Consider the set S 111 110Te2 1 710T7v3 001T a Verify that S is an orthogonal set b ls S an orthonormal set Justify If S is not orthonormal7 then convert it to be an orthonormal set T u1uzU3 c Find 010203 such that 57 714T clul 02u2 63113 by using dot products7 NOT by solving a linear system of equations SOLUTIONS 1 See Test 3 review 2 a A lTlA lTvilAl lT 2lA T11 717 2T 7111 3112 and T12 717 0T 7111 112 so b B le3 lTw1l5l lTw2ls Tw1 07 2T 747111 21112 and Tw2 714T 79101 41 so 74 79 B lt 2 4 c P B a A That is P w1Al w2A wl 01 7 112 and LU2 2111 7 112 therefore 1 2 P lt 71 71 A74 5 71 71 A 1 0 71 A1 detI 7 A det 7 So the eigenvalues are A1 17 A2 2x3 0 73 5 71 73 A1 1 eigenspace N 71 1 1 and 71 0 71 2 0 7 N1 7 A span321T 72 5 71 72 A2 2 eigenspace N 71 2 1 and 71 0 71 3 0 7 NZI 7 A span731T 74 5 71 74 A3 0 eigenspace N 71 0 1 and 71 0 71 1 0 7 N31 7 A span1 1 1T 7 7 b Therefore P lAP D For example if A1 0 0 1 0 0 D 0 A2 0 0 2 0 One possible P is P 0 0 A3 0 0 0 0 In Class we that saw a Taylor expansion gives both P lAP D 5A1 0 0 51 0 0 and 6D 0 6A2 0 0 62 0 0 0 5A3 0 0 e e 0 0 77 Therefore 6AP 0 62 0 P l 37 0 0 1 82351 5 71 1 1 7 71 2 5 71 2 1 7 71 3 5 71 0 1 7 71 1 ll 00m WNO O 735 77529572 725 735M654 7623e72 E A373A22AAA72A71 10 7 01 0 0 10 7 01 0 0 10 7 01 0 0 3 71 2 31 111 P leAPeD 0 0 1 7e 5 Z766 3 45 57265 73 72 77 71 71 4 U detAI 7A A7 aA 71A 7 2 So the eigenvalues are A 172704 If 04 31 12 then we have three distinct eigenvalues7 hence A is diagonalizable lf remains to consider what happens if 04 is 1 or 2 Case 1 Oz 1 The eigenvalues are A 17 17 2 The eigenvalue of 2 will correspond to one independent eigenvector since its only a root of the characteristic polynornial once It remains to consider A 1 which will correspond to either one or two independent eigenvectors 73 76 2 11 7 A 1 2 71 0 0 0 1 2 0 77gt 0 0 1 0 0 0 which has nullity one7 and hence corresponds to one independent eigenvector This gives a total of 2 independent eigenvectors rather than 37 hence A is defective not diagonalizable Case 2 Oz 2 The eigenvalues are A 17 27 2 The eigenvalue of 1 will correspond to one independent eigenvector since its only a root of the characteristic polynornial once It remains to consider A 2 which will correspond to either one or two independent eigenvectors 72 76 2 217A 1 3 71 0 0 0 1 3 71 7 7gt 0 0 0 0 0 0 which has nullity two7 and hence corresponds to two independent eigenvectors This gives a total of 3 independent eigenvectors7 hence A diagonalizable Therefore A is defective7 only when 04 1 Suppose A 31 Q is nilpotent Claim A is defective Proof Suppose A is an eigenvalue with eigenvector X We have shown before that A will be an eigenvalue of A This is because AX AX i AZX AAX AAX AAX AAX AZX But AZX AZX i A3X AA2X AA2X A2AX A2AX A3X Continue inductively to get AmX AmX for any m E Zgt0 lfA is nilpotent then Ak Q for some k Since AkX AkX and Ak Q we get AkX Q By de nition of an eigenvector X 31 Q Therefore Ak 0 and so A 0 Therefore all eigenvalues are 0 Suppose A were diagonalizable This would imply that A1 0 0 0 1 0 A2 0 P AP t for some P nonsingular 0 0 0 An 5 3 But every A is 07 hence P lAP Q i A PQP l Therefore A must be defective Q7 which is a contradiction A2 04A and Oz 31 0 Claim All eigenvalues are 0 or a M Suppose A is an eigenvalue with eigenvector X AX AX i AZX AAX AAX AAX AAX AZX A2 04A AZX aAX aAX Together these give AZX aAX AZX aAX AZX 7 aAX Q i AA 7 aX Q and since X 31 Q7 this implies A 07 a No they do not always have the same eigenvectors 13 LetP724andletDi Also let the matrix A be A PDP l in other words P lAP D A and D both have eigenvalues 1 and 3 and A has eigenvectors 17 2T7 34T by the way we constructed it However since D is diagonal its eigenvectors are 61 and 62 a a b A 7 b d A2 7 trAA detA 0 Here trA a d and detA ad 7 b2 Eigenvalues must satisfy A2 7 a dA ad 7 b2 0 Using the quadratic formula to solve7 A being real or complex depends only on the discriminant the piece under the radical which is a d2 7 4ad 7 b2 Notice however that a d2 7 4ad 7 b2 12 d2 2ad 7 4ad 4b2 a2 7 2ad d2 4b2 a 7 d2 AbZ7 which is a sum of squares and therefore always non negative Therefore the eigenvalues must be real 10 03 Note There are more general proofs to show this for n gtlt n 0 b A 7b 0 So detAI 7 A A2 b2 which has roots A bz39 7bz39 a lluH 120202021andHUH1121212122 b ltuvgt10001 b c cos0 d lt um gt7 0 therefore u and 1 are not orthogonal also 6 31 1 77quot 7673 i u i HullHull e If x 6 SL then uTz vTx 0 That is if z x17x234T then for Alti27wemusthaveAxQori2 2 3 In other words SL NA span07110T707101T 10 Let X denote a vector along the line y 3x eg X 13T Let Y denote the vector from the origin to the point of interest7 ie Y 12T Therefore we want to nd pronY 39 7ltXYgt 716 i7 77 T7 7 21T pmJXY ltXXgtX 1T9X EX lt1v3l mm 11 a 1102 gt7JIT71217100 ltvlvg gt111T1130000 lt027113 gt12T1130000 Therefore the set is orthogonal b No7 the set is not orthonormal While HUSH 17 notice Hmll V12 12 02 xE and va 112 712 02 Make the set orthonormal by multiplying each vector by the reciprocal of its length7 hence making all vectors into unit vectors 7 1 U1 7 01 7 1 2 W U3 03 c For w clul 02u2 03 we get 01 lt w7ui gt since the ugs form an orthonormal set So T 1 1 T i 4 i C1lt5l4 7 7 70 1 1 T 6 CZ lt57 7174 E7 70 gt E 3 c3 lt 5714 001 gt 4


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