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# Introduction to Linear Algebra MATH 301

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This 22 page Class Notes was uploaded by Breanne Schaden PhD on Saturday October 3, 2015. The Class Notes belongs to MATH 301 at Boise State University taught by Charles Kerr in Fall. Since its upload, it has received 20 views. For similar materials see /class/218003/math-301-boise-state-university in Mathematics (M) at Boise State University.

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Date Created: 10/03/15

MATH 301 001 7 Diary 116 2707 7 Spring 2007 1 Mon Feb 12 135330 MST 2007 m301spO IhandoutsSO1Diary301Diarystar211 1 Current Diary Click here for the current diary page 2 11607 Tuesday Day One 1 We used Elementary Operations aka EROs 7 bottom of page 8 and the aug mented matrix to solve 1 32 3 1 2l72m3 4 1 331 2mg 44 8 m12 33 4 6 We got one solution The method we used is known as the GaussJordan Elimination Method a major MATH 301 workhorse 2 We also spent time on the simpler system 111 2112 3113 4111 5112 6113 7111 8112 9113 ll cacao Which has a nal augmented matrix 10 1 3 0123 0000 1 2 3 215 MATH 301 001 7 Diary 116 2707 7 Spring 2007 2 A 03 V V This happened because we could not eliminate dependencies among the variables We could not isolate each variable in its own equation as we were able to do in the rst four equations in four unknowns example In the last two or three minutes of class we looked at the system 111 2112 3113 i 3 4111 5112 6113 3 7111 8112 9113 We EROed this until we got 1 0 1 3 0 1 2 3 0 0 0 5277 which indicates that the given system has no solutions whatsoever Looking ahead we will need to check our answer to the 4 X 4 system pages 51 54 or is it 4 X 5 And we have to justify use of the EROs 11 problem 41 describes this MATH 301 001 7 Diary 116 2707 7 Spring 2007 3 3 11707 Wednesday Day Two 1 We showed a way to check the solution we found in our very rst example on Tuesday 2 We had to justify use of the EROs We did part of Theorem 11 which says that the application of an ERO to a system produces a system with exactly the same solution set We went through the Ei lt Ei kEj part of the proof 34 j 3 We must worry that the EROs might alter the solution set7 as happens when we use the transformation square both sides77 on an equation like m2 m 4 The proof used the fact that every ERO is reversible given a particular ERO7 one can nd an ERO Lalways di erent from the rst ERO which undoes the action of the rst ERO 5 On the notation front7 equation Ei is a linear equation 77 E airmr 7391 A G V The clock ran out on matrix indices Cf problems 11 7 10 A 1 V Looking ahead echelon matrix and echelon form of a matrix And the real deal the reduced echelon form of a matrix This is known also as the row reduced echelon form or the reduced row echelon form This is why your calculator has an rref command and in scilab we get gtA 123345637893 A 1 2 3 3 4 5 6 3 7 8 9 3 gtrrefA ans 1 O 1 3 O 1 2 3 O O O O MATH 301 001 7 Diary 116 2707 7 Spring 2007 4 4 11907 Friday Day Three 1 Page 10 two matrices are Row Equivalent 2 Without supporting computations it was stated that an Echelon Form of the aug V V mented matrix of 2111 6112 3113 2114 1 1 3112 2113 4 3111 9112 4113 4114 2 turns out to be 1 3 2 0 4 0 0 1 2 7 0 0 0 0 0 Another ERO step brings us to the Reduced Echelon Form aka REEF7 1 30 410 0012 7 00000 which is a real workhorse for subsequent developments Check out the de nition and be able to compute it by hand And note that scilab has an rref command So does Maple The above RREF shows that 2 and 4 are so called free variables because they have messy columns 1 and 3 are sometimes called basic variables They are the variables corresponding to left most non zero entries in the RREF matrix They have nice columns The RREF greatly simpli es writing the formula for the many solutions that our example has RREF leaves us with the two equations 1 3112 4114 3 24 77 which we crowded into this form 103u4v u 7 239v U MATH 301 001 7 Diary 116 2707 7 Spring 2007 Which we massaged into a column matrix form 1 10 3 2 7 0 1 3 7 u 0 1 4 0 0 MATH 301 001 7 Diary 116 2707 7 Spring 2007 6 5 12207 Monday Day Four A H V We computed the crossing point of the lines 2 5y 3 4m 3y 8 2m 2y 3 by racking their equations up in an augmented matrix for which we found the RREF 10 72 01 2 00 0 which gave us the crossing point at 72 2 A N3 V Some folks might say that the previous system has in nitely many solutions7 owing to the row of zeros How would you talk them out of that View A 03 V We took three points 210 1 3 64 and found an equation of form m2y2AmByC0 for the circle through those points m22y 72 25 with center at 2 7 and radius 5 A 4 V We also did a partial fractions decomposition We guessed that 5 A B C 322 2 mm 1m1 Comparing the identically equal numerators on both sides7 Am 1 1 Bm 2 1 Cm 2 1 902ABCm3BC AB ZC yr 57 we arrive at a system of equations A B C 3BC AB 2C 7 010 for which A 2 B 1 and C 1 Thus MATH 301 001 7 Diary 116 2707 7 Spring 2007 7 5 Look ahead i On Tuesday we7ll begin by seeking the line of form Ax By C 0 Which passes through the three points 28610 434 15 and 146 5 This relates to example 8 in 13 What happens if we use the points 210 1 3 and 6 4 from the Circle problem above We7ll wax theoretical from the material in 137 pages 28 32 MATH 301 001 7 Diary 116 2707 7 Spring 2007 8 6 12307 Tuesday Day Five 1 We looked at the line of form Ax By C 0 through the points 28610 V V 434 15 and 146 5 Descartes says that these points give rise to the homogeneous system 0 0 286A 10B C 434A 15B C 146A 5BC0 We racked up an augmented matrix with the variables in CBA order and EROed into reduced echelon form 1 0 2 0 0 1 1445 0 0 0 0 0 From this we were able to see that the homogeneous system in question has non trivial solutions Our variable order choice makes A the independent free variable A t B 1145t C 2t OT A 1 5 B t 1445 u 144 C 2 10 We wrote down a line equation 5 1443 10 0 one of the in nitely many equations of this form for the line through these three points We know that the points 210 1 3 and 6 4 are non collinear Had we used them above7 the corresponding homogeneous system would have been consistent So7 what line equation would it give for these points That is7 how would this computation tip us o to non collinearity What would happen had we used our Monday method to nd an equation for the circle through 286 10 434 15 and 146 5 How would the calculation have tipped us o To what 4 We read through the section 13 remarks on page 29 5 Look ahead Example 77 page 33 and tra ic ow and Kircho MATH 301 001 7 Diary 116 2707 7 Spring 2007 7 12407 Wednesday Day Six 1 We did an example like example 77 page 33 What must we have for b1 b2 and by in order that the following system be consistent m2y3z b1 45y z b2 78y92 by We used variable order myzblbgbg to rack this up in an AUGmented matrix 1 2 3 1 0 0 4 5 6 0 1 0 7 8 9 0 0 1 We EROed this only until the left half was RREFed 1 0 1 53 23 0 0 1 2 43 13 0 0 0 0 1 2 1 The third row here shows us that consistency is guaranteed if b1 2b2b30 In that event7 a solution formula is given by 113 t We studied a tra ic ow example7 here using Maple to study the many solutions Click here MATH 301 001 7 Diary 116 2707 7 Spring 2007 8 12407 Friday Day Seven 1 We went through an establishment approved version of the proof in 11 40 2 Henceforth in MATH 301 001 auxiliary variables will be used in reporting the solutions V of a system with many solutions m1 2 3 1 m2 2 1 1 3 2 u 2 1 1 m4 0 1 0 m5 0 0 1 for 12 50 Then we began our look at the stuff in sections 15 and 16 with the matrix product A 17 where a A is anm X 71 matrix 7 b f is an n X 1 column matrix also known as an n vector and also f E R 7 c 27 e W d The ith entry of g is given by 71 yr 2 Gimme k1 We discussed the usual arm waving way of computing A The summation index formula needs to kept in mind for proofs and computer programs We used the above product to de ne AB where B is not a column matrix Let B have q columns b17 bq Then AB is the matrix AB 1413111413 IABq1IABq 71 so that entry i of column j of AB is given by Z aikbkj where ast denotes the k1 A entry in row 8 and column t 5 So AB makes sense only if A is m X n and B is n X q MATH 301 001 A Diary 116 2707 7 Spring 2007 11 6 We also looked at how one can take the rows 11 am of A and think of AB by rows as 01B 02B A B am1B amB Lookahead interpretation ofA as alinear combination ofthe columns 61 76m of A Also A 9 Identity matrix In A C7 NoncommutatiVity of matrix multiplication General index tWiddling AAA CDQIO VVVVVVVVV AssociatiVity Divisors of Zero Linear systems in form A b A H Transpose 90 MATH 301 001 7 Diary 116 2707 7 Spring 2007 12 9 10 12907 Monday Day Eight Click here for another POV on matrix multiplication An example showing matrix multiplication is noncommutative We reviewed the summation and index formulas for the entries of A and AB ATij 7 lndex Wise proof that ATBT BAT lndex Wise proof that matrix multiplication is associative De nition of linear combination A is a linear combination of the columns of A If am by E cmdy F has just one solution7 it is given by liladibcli leil Lookahead linear independence and associated matters see 17 MATH 301 001 7 Diary 116 2707 7 Spring 2007 13 10 13007 Tuesday Day Nine 1 Using the matrix A from 13 247 looked at expressing 3 7 1 2 and b2 7 2 1 3 as linear combinations of the columns of A Using the idea that 3 A93 2 mp4 11 where Ai is the ith column of A7 we saw that this problem was equivalent to the consistency of the systems Am b1 and Am b2 The second system came up inconsistent7 so B2 is not a linear combination of columns of A But the rst one is consistent 7 its RREF yielded a vector solution formula 1 12 2t 2 5 t 113 t from which we can get coe icients for the sought linear combination t 6 3 1 3 1 b1 2 0 1 1 2 6 0 1 3 7 1 We saw that we could write the zero vector 67 as a non trivial linear combination of the columns of A This led to the de nitions of Linear Dependence and Linear Independence And the observation that the columns of A above constitute a linearly dependent set We ended with a proof that a set of vectors in linearly dependent if and only if one of the vectors can be expressed as a linear combination of the other vectors in the set MATH 301 001 7 Diary 116 2707 7 Spring 2007 14 5 Lookahead In class7 as an all group project7 well look into the page 78 problems a 1 3 1107111 8 11171127113 b 276 d 11 f 14 MATH 301 001 7 Diary 116 2707 7 Spring 2007 11 13107 Wednesday Day Ten 1 We spent time on the page 78 problems from the Monday lookahead a Theorem 13 17 is linearly dependent ltgt there is a scalar k such that 11 k6 b From page 787 u1u2u3 is linearly dependent because 1 2 1 10 3 RREF 214 01 2 1 3 3 000 so that there must be not all Zero C7 1 2 3 such that 1 2 1 0 cl 2 02 1 03 4 0 6 3 3 0 c In looking at these problems we bumbled into the famous MoreVectors than Entries Theorem lf 1 p C R Withp gt 1717 then 1 p must be linearly dependent 2 The de nition of nonsingular matrix 3 We ended the hour in midst of looking at a characterization of nonsingular matrices Let A be n X n Then For all 3 e R A Bhasaunique ltgt solution A 5 has a unique solution MATH 301 001 7 Diary 116 2707 7 Spring 2007 16 12 2207 Friday Day Eleven 1 We nished the proof of the nonsingular matrix characterization theorem from Wednes day This proof called on a The distributiVity of matrix multiplication over matrix addition Which we didn7t prove in class V b c The too many Vectors theorem aka the more vectors than entries theorem Linear dependence V MATH 301 001 7 Diary 116 2707 7 Spring 2007 17 13 2507 Monday Day Twelve 1 We began with two examples a In problem 13 24a from Assignment 4 the deal was to determine conditions on b so that A b is consistent has a solution where 1 3 A 1 2 0 3 7 1 the text method was to work with the augmented matrix 1 3 1 b1 1 2 0 b2 3 7 1 b3 Another way is to work with the matrix augmented with the coe icients of the bi thus 1 1 0 0 0 0 1 0 1 0 0 1 and do EROs until the left side is in RREF 1 0 2 2 3 0 1 1 1 1 0 0 0 1 2 1 1 3 1 2 3 7 This showed the consistency criterion A b has a solution if and only if b1 2b2 by 0 That is A b has a solution only for some b not all b This means that A is a singular matrix Note that the above matrix puts us in position to write down the non trivial solutions of A 0 b When we treated I 1 1 A 1 1 1 1 2 4 from 12 50 Assignment 3 the same way the EROs brought us to 13 1 13 12 12 0 0 1 0 16 12 13 0 0 1 The shows that A b is consistent for all b and hence that A is non singular The right pocket matrix called B here does AB I MATH 301 001 7 Diary 116 2707 7 Spring 2007 18 2 V The de nition of Matrix Inverse We set about to show that B above7 and any ma trix arising the way B did7 is an inverse for A above The de nition of invertibility 3 The Product SingularDominance Lemma lt7s proof had instance of decoding matrix singularity 4 The NonsingularityInvertibility Equivalence Theorem We almost nished the proof A U V Lookahead nish the proof Some famous invertible matrices And more 19 stu MATH 301 001 7 Diary 116 2707 7 Spring 2007 19 14 2607 Tuesday Day Thirteen 1 V 2 V We nished reading through the texts proof of the NonsingularityInvertibility Equivalence Theorem This capped our work on the matrix A from 12 50 We had 13 1 13 1 2 6 2 B 12 12 0 7 3 3 0 6 16 12 13 1 3 2 for which we had AB I by Virtue of the way B was computed We now also have BA I so that B is an inverse for A We did problem 19 79 to show that if a matrix has any inverse7 it has exactly one inverse The inverse of A is denoted by A l We ended the hour looking at Theorem 18 on things equivalent to nonsingularity Lookahead famous invertible matrices7 the elementary matrices And the inter section of almost parallel lines MATH 301 001 7 Diary 116 2707 7 Spring 2007 20 15 Assignment 5 Comment on Grading 1 17 48 Its best to work with the basic de nition in the gray box on page 73 Like directly The hint says Exhibit so you could just write down a choice of a1 a2 and 03 then show why it works That is can you tell me values for al a2 and 03 that are guaranteed to make 01171 02172 03173 9 into a true equation Can you tell me values for al a2 and 03 which are not all of them zero but which still guarantee that the equation is true Watch out for the Circular Reasoning Police You are forbidden to begin your problem 48 solution with Let S 171172173 be a linearly dependent set of vectors That7s what you want to prove The Circ Police will see this as moving the nish line up to the start line You cannot issue such a let for set S because the author has already issued a let for S in the problem statement You cannot change the nature of S lt7s as if you were asked to write down all divisors of n 1728 And you solved the problem by saying Let n 1009 then the divisors are 1 and 1009 49 See if you can complete this nice proof by contradiction Suppose the assertion is not true then 171172173 must be linearly dependent Thus a non trivial linear combination of these vectors is the zero vector This means that we can express one of the 17 in terms of the rest Without loss of generality it could be 173 there are scalars 1 and 2 such that 3903 101 202 And so With this set of more than two vectors it is not enough to show that none of them is a scalar multiple of one of the others here7s a set with that property which is none the less linearly dependent MATH 301 001 7 Diary 116 2707 7 Spring 2007 21 c Here7s another nice start given the set 171172173 of the problem hypothesis7 let V be the matrix with these vectors as columns Then VTV is a diagonal matrix with strictly positive entries on the main diagonal pl 0 0 VTV 0 pg 0 0 0 P3 where pi 7 We can render the dependence equation7 013901 02 02 1303 9 as V6 0 where 01 6 12 03 Then T 6 V6 V6 3 17 50 a How7s about this By hypothesis we have 11 a2 and 13 not all zero7 such that 11171 12172 13173 Let a4 07 then we have 01171 02172 03173 14174 5 where 11 a2 a3 and 14 are not all zero Thus 171172173174 is linearly dependent b Another approach was to look at the matrix V which has the vi 1 2 3 as columns When we go for the RREF of V0 we get an augmented matrix which has columns indicating free variables The same variables will be free in the solution of V1740 and so MATH 301 001 7 Diary 116 2707 7 Spring 2007 22 16 2707 Wednesday Day Fourteen 1 The agenda group work around some linear independence proofs 2 Click here for the sheet of problems some typos xed 3 Friday Will spend time on the above sheets 1 Cdef and 2abc 17 Current Diary Click here for the current diary page

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