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LINEAR ALGEBRA

by: Evert Christiansen

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14

LINEAR ALGEBRA MATH 304

Evert Christiansen
Texas A&M
GPA 3.92

Staff

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COURSE
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PAGES
14
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This 14 page Study Guide was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Study Guide belongs to MATH 304 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 42 views. For similar materials see /class/226045/math-304-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
Math 304 Solutions 3 1 Section 32 8 Let A be a particular vector in R2 Determine whether the following are subspaces in R2 a 81 B E RZXZlAB BA b 82 B E RZXZlAB a BA c 33 B E RZXZlBA 0 Answer a 81 is a subspace of R2 Proof First we show that 81 is closed under scalar multiplication Let B be an arbitrary matrix in 81 and let on be an arbitrary scalar Then since B E 31 we know that AB BA Consider the matrix dB We would like to show that dB is in 81 so we would like to show that AdB dBA This is clearly true since AB BA and or is a scalar Thus dB E 31 Thus 81 is closed under scalar multiplication Next we show that 81 is closed under addition Let B and C be arbitrary matrices in 31 Then since B E 81 and C E 31 we know that AB BA and AC CA Consider the matrix B C We would like to show that B C is in 31 so we would like to show that AB C B CA We can show this AB C AB AC distributive law BACA ABBAandACCA B CA distributive law Thus B C E 31 Thus 81 is closed under addtion Since 81 is closed under scalar multiplication and addition 81 is a subspace of R2 D 82 is not a subspace of R2 Proving that 82 is not a subspace is slightly more complicated than usual We can7t just create a counterexample because we don7t know what the matrix A is We would like to prove that for any matrix A the set 82 is not a subspace If A I then the set 82 is actually empty and there is no counterexample Instead the set is not a subspace because subspaces are required to be nonempty by de nition Here are 2 proofs Proof First if 82 is empty then 82 is not a subspace the de ni tion of subspace requires the set to be nonempty If 82 is nonempty then there exists some matrices in 82 Let B be a matrix in 82 Then AB a BA Then consider zero times B which is the zero matrix The zero matrix is not in 32 since AO O OA Thus 82 is not closed under scalar multiplication so 82 is not a subspace D Here is a different proof using the same idea Proof The zero matrix is not in 32 since AO O OA Thus 82 is not a vector space as it does not satisfy axiom 3 Since it is not a vector space it cannot be a subspace D 33 is a subspace of R2 Proof First we show that 33 is closed under scalar multiplication Let B be an arbitrary matrix in Sg and let on be an arbitrary scalar Since B E 33 we know that BA 0 Consider the matrix dB We would like to show that dB is in 33 so we would like to show that dBA O This is clearly true since BA O and or is just a scalar Thus dB E 83 Thus 33 is closed under scalarl multiplication Next7 we would like to show that 33 is closed under addition Let B and C be arbitrary matrices in 83 Since B E 33 and C E 337 we know that BA O and CA 0 Consider the matrix B C We would like to show that B C is in 337 so we would like to show that B CA 0 We can show this B CA BA CA distributive law 00 BAOandCAO O Thus7 B C E Sg Thus7 33 is closed under addition Since 33 is closed under scalar multiplication and addition7 it is a subspace of R2 D 12 Let x1L7 X27 7x19 be a spanning set for a vector space V a If we add an additional vector Xk1 to the set7 will we still have a spanning set Explain b If we delete one of the vectors7 say xk from the set7 will we still have a spanning set Explain Answer a Yes7 we will still have a spanning set Since X17 X27 7x19 spans V7 every vector in V can be written as a linear combination X17X27 7X19 If we add xkirL7 every vector in V will still be able to be written as a linear combination for example7 we could use the original linear combinations7 and add 0 X19 b There is not enough information to tell It could still be a spanning set7 or removing that vector could make it no longer a spanning set 13 In R2 let 10 E117lt00gt7 00 E217lt10gt7 ShOW E117E127E217E22 span RZXZ Answer Consider an arbitrary element of R2 ii We can write this element as a linear combination of E117E127E217E22 ab Cd Thus7 the matrices E117E127E217E22 span R2 gt aEll bE12 CE21 dE22 Let A be an n X 72 matrix Prove that the following statements are equivalent a MA 0 b A is nonsingular c For each b E R 7 the system Ax b has a unique solution Answer If you want to show that three statements are equivalent7 you need to show that each statement implies each of the other statements One way to do this is to show that a implies b7 b implies c7 and c implies a Then7 for example7 we also have that b implies a since b implies c and c implies a Another method would be to show that a implies b and b implies a7 and also show that b implies c and c implies There are clearly7 a lot of di erent combinations that would work to show that the three statements are equivalent In this problem there are many di erent ways to show the implications among the di erent statements One way would be to use Theorem 142 and Corollary 143 Here is a di erent method Proof We will prove that a implies b b implies c and c im plies a First we will prove that a implies By the de nition of nullspace statement a is equivalent to the following statement Ax 0 has only the trivial solution Consider the row reduced echelon form of A If the row reduced echelon form has a row of zeros then Ax 0 has in nitely many solutions Thus it cannot have a row of zeros so the matrix A must row reduce to the identity matrix Thus A is nonsingular Next we will show that b implies Suppose that A is a nonsingular matrix Then A has an inverse A l Let b be an arbitrary vector in R and consider the matrix equation Ax b We can multiply both sides of this matrix equation by A l We get x Ailb Thus the only solution to the equation Ax b is x A lb Thus the matrix equation has a unique solution Next we will show that c implies a Suppose that for all b Ax b has a unique solution This means in particular that Ax 0 has a unique solution Since 0 is a solution to Ax 0 we get that the only solution to Ax 0 is 0 Thus the nullspace of A is Thus we see that a implies b b implies c and c implies a so the three statements are equivalent D Review for Exam 3 41 Linear Transformations Key ideas Linear transformations preserve the operations of addition and scalar multiplication Lu1 u2 Lau1 buz aLu1 bLu2 If L V a W there are two subspaces of great importance The kernel kerL 1 E VlL1 0 and the image LS wlw Lv for some V E V If S V we have the range LV 42 Matrix Representations of Linear Transformations Key ideas HA is an m gtlt 71 matrix then Lx E Ax is a linear operator Furthermore if L is a linear operator from R a R then it must be multiplication by a matrix The ith column of the matrix are just Lei where 61 is the ith standard basis vector Matrix Representation Theorem If L V a W and E 11171127quot711n is a basis for V and F 11111027 wn is a basis for W then L can be represented as a matrix A where WWW AME The jth column of A is given by the coordinates of L1j with respect to F Euclidean Space case IfL R a R and u1u2 un is a basis for R and 51ng7 bn is a basis for R then the matrix representation of L is given by row reducing U1 uz un Lb1 Lb2 Lbm 1 A 43 Similarity If L V a V is a linear mapping from a vector space V to itself the matrix representation of L will depend on the basis chosen for V If A is the matrix representation of L with respect to the standard basis E 61 62 en and B is the representation with respect to the basis U u1u2 un and if S is the change of basis transformation from U to E then A and B are related by S E lU U and B S lAS U lAU General Result If E 01112 1n and F 11111027 wn are bases for V and W respectively and if A and B are the matrix representations of L with respect to E and F respectively then B S lAS where the change of basis matrix is given by S E lF A and B are similar if there is a matrix S such that B S lAS If this is the case then A and B represent the same linear operator The optimal case is when E is diagonal 51 The Scalar Product in R If we take two vectors x y E R then rTy is a scalar called the scalar product of z and y The length of a vector is given by T M 96 96 The two vectors form an angle 0 satisfying 96 MHzH 0089 where 6 is measured in radians IfrTy 0 then 6 7r27 37r2 so z is orthogonal to y There are two important projections the scalar projection of z onto y given by and the vector projection of z onto y given by 10 9 My H2 0 Know how to nd the distance of a point P to a line 0 Know how to nd the distance of a point P to a plane 0 Know how to nd the equation of a plane with a given normal passing through a given point 52 Orthogonal Subspaces Two subspaces X and Y are orthogonal if every vector x E X and every vector y E Y satis es zTy 0 The orthogonal complement of X consists of all vectors orthogonal to X YL x E Rnery 0y E Y There are two important subspaces the null space of A NA rlAz 0 and the range of A RA Arm E 72 The following relationship holds NA MATL and NAT PML IfS T 0 and S U T R then we say the R is the direct sum ofS and T written as R S EB T Because of this we can write R NA ea RAT NAT ea RA 53 Least Squares Problems Key Ideas If one is trying to solve Ax b where A is an m gtlt 71 matrix with m gt n overdetermined that is more equations than unknowns then often we cannot nd a solution x such that Ax 7 b 0 In other words b is not in the range of A We resolve this situation by looking for a vector i which is closest in the sense that HAx 7 MP is minimized The vector i satis es the normal equations ATAi ATb The matrix ATA is invertible if A has full rank that is rankAn If the matrix ATA is invertible then 92 ATA 1ATb To t a polynomial p co clz 02x2 chN to the data we want to solve the equations xiv zivil 1 1 UN yl 9 9171 2 1 0N4L yg AC z v z vil 3 1 yg b 01 N N71 zm zm zm 1 00 gm The solution 6 gives the coef cients of the best tting polynomial of degreen N to the data The T183 calculator will do regression up to degree 4 54 Inner Product Spaces An inner product is an operation on a vector space V which has the following properties 1 ltz7xgt 2 0 and ltz7xgt 0 only form 0 2 lt7ygt lty gt for all my in V 3 ltaz 611 we 2gt my 2gt An example of an inner product is ltx ygt mTy Other examples include n ltA7 Bgt Z aijbij 1 j1 ltmgt Abfgd An inner product always de nes a norm by M2 E ltWcgt We have the scalar projection of u onto 1 a M and vector projection of u onto 1 W 11gt p v lt M2 The following inequalities are always true CauchySchwarz inequality lltu7vgtl S HUHHUH Triangle Inequality HUMH S M M Other norms not derived from an inner product are V L o W 2 Z W lnmm i1 Mp A n 117 Z W pnorm i1 o E i in nitynorm 0r maxnorm Math 304 Exam 3 Study Sheet April 137 2004 Since the notation in the book is not entirely the same or as useful as what we have done in class7 l7ve summarized the important things below At least for change of basis notationithis does not cover everything we did Change of Basis Notation Suppose B V1Vn is a basis for R We know that any vector V E R can be expressed uniquely in terms of B V clvl 02V2cnvn7 where 017 70 E R are unique ie7 once we choose V7 the 07s are uniquely determined We will write 01 02 V V5 I lenl g to mean that 017 0 are the coordinates of V in terms of the basis 8 Using the standard basis 8 1627H39 7en for R then our usual coordinates are the stame as the coordinates with respect to 8 So 1 1 2 2 a lxll Now if we let U5 be the n gtlt n matrix formed by taking U8 V1 V2 an where the columns of U5 are obtained by taking the column vectors for V1 Vn under the standard basis and lining them up The matrix U5 is called the transition matrics or change of basis matrix Then for any vector V E R we nd formulas7 lVls U8 39 lVl87 and likewise lVlB U 1lvls7 which allow us to change representations of V in terms of the standard basis and in terms of the basis 8 and back again If A 11171127 7un is another basis for R then we can interchange between B basis representations and A representations for any V E R NA UXIUB 39 MB and likewise lVlB U lUA 39 lVlB Now suppose that L R a R is a linear transformation Suppose that B V1V2 Vn is a basis for R and C W17 W27 7Wm is a basis for R Then there is an m gtlt 71 matrix ng so that lLXlc nglxl87 Vx E R In fact7 mg mt Louie Mme ltgt To think about what this means ng takes as input vectors in coordinates with respect to 8 applies L to them7 and then leaves as output vectors in coordinates with respect to C Example 1 We know that any linear transformation L R a R is represented by an m gtlt 71 matrix A Lx AX7 Vx E R and A Le1 Le2 Len Everything here is done in terms of the standard basis So what we have is Ll quotA L621 L622 men ltgt Example 2 Let I R a R be the identity transformation Ix x7 Vx E R We can represent I in terms of a basis 8 V1V2 Vn and we nd where S is the standard basis on R Likewise7 1113 112 1 U l lf L R a R is a linear transformation the representation of L in terms of the basis 8 is 1ng 111339 Lli Ilia lIl zSaV1 Lli 111293 For any linear transforrnation7 L R a Rm7 we can nd frorn Example 1 above If B is any basis of R and C is any basis of Rm7 Ll Ilgm LliZ 1112quot Example 3 Let L R3 a R3 be given by 1 1 2 3 1 L x2 0 1 0 2 3 1 0 9 3 Let B 751 7 7 7 and let C 781 7 7 Let S be the standard basis on R3 Things you should Check 71 2 4 1 3 9 mg 1 0 1 7 mg 2 i2 i3 0 12 71 2 4 and 1 3 3 7 73 Mi 1 0 7 WE 1 1 3 71 0 9 1 2 3 and Ll 77 1ng 77 Additional Study Problems 36 17 27 7 11 0 41 57 77 17 o 42 1 10 43 17 27 37 47 12

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