Math 61CM - Solutions to homework 2
|
|
- Rosanna Vanessa McGee
- 5 years ago
- Views:
Transcription
1 Math 61CM - Solutions to homework 2 Cédric De Groote October 5 th, 2018 Problem 1: Let V be the vector space of polynomials of degree at most 5, with coefficients in a field F Let U be the subspace of V consisting of polynomials of the form az 5 + bz + c with a, b, c F Find a subspace W such that every element v V can be written in one and only one way as the sum of an element in U and another element in W Solution: We have V = { a 5 z 5 + a 4 z 4 + a 3 z 3 + a 2 z 2 + a 1 z + a 0 a 0,, a 5 F } U = { az 5 + bz + c a, b, c F } Since every polynomial can be written in a unique way as a sum of monomials, a good guess for W is We have several things to check W = { dz 4 + ez 3 + fz 2 d, e, f F } W is a subspace: it s clear that 0 W (taking d = e = f = 0), and for λ, µ F we have λ (dz 4 + ez 3 + fz 2 ) +µ (d z 4 + e z 3 + f z 2 ) = (d + d )z 4 + (e + e )z 3 + (f + f )z 2 W }{{}}{{} W W for any d, d, e, e, f, f F Any element v V can be written in some way as the sum of an element in U and another element in W : for any a 5, a 4, a 3, a 2, a 1, a 0 F, we have a 5 z 5 + a 4 z 4 + a 3 z 3 + a 2 z 2 + a 1 z + a 0 = (a 5 z 5 + a 1 z + a 0 ) + (a }{{} 4 z 4 + a 3 z 3 + a 2 z 2 ) }{{} U W This decomposition is unique: if (a 5 z 5 + a 1 z + a 0 ) + (a 4 z 4 + a 3 z 3 + a 2 z 2 ) = (a 5z 5 + a 1z + a 0) + (a 4z 4 + a 3z 3 + a 2z 2 ), then a i = a i for every 0 i 5, by the definition of equality of polynomials Hence this decomposition is unique 1
2 Problem 2: Prove that there exists a quadratic polynomial ax 2 + bx + c whose graph passes through the points (0, 1), (1, 0), (2, 3) Is such a polynomial unique? Solution: Let p(x) = ax 2 + bx + c We need to find a, b, c such that p(0) = 1, p(1) = 0 and p(2) = 3 Plugging in the definition of p(x), this yields the system c = 1 a + b + c = 0 4a + 2b + c = 3 Solving it, we find the unique solution a = 2 b = 3 c = 1 Therefore, such a polynomial is unique If you are interested: it is in general true that there exists a unique polynomial of degree n whose graph passes through n + 1 points so that no two of them have the same x-coordinate This is because the associated system of n + 1 equations has a non-zero determinant (see wiki/vandermonde_matrix) As we will see later in the class, this implies that the system has a unique solution Problem 3: Find a single homogeneous linear equation with unknowns x 1, x 2, x 3 such that the solution set is the span of the two vectors (1, 1, 1), (1, 2, 0) Solution: To answer this question, it is sufficient to guess a solution, and then explain why the solution set is the span of these two vectors However, I will show how to solve this problem without guessing We want to find a, b, c R such that (1, 1, 1) and (1, 2, 0) are solutions of ax 1 + bx 2 + cx 3 = 0 In other words, we want { a + b + c = 0 a 2b = 0 After a few manipulations, we see that this is equivalent to { c = 3b a = 2b Taking b = 1 for example, this gives the equation 2x 1 + x 2 3x 3 = 0 Now, we need to check that this is indeed the right solution Call S the solution set of 2x 1 + x 2 3x 3 = 0, and W the span of (1, 1, 1) and (1, 2, 0) W S: the above shows that the two vectors (1, 1, 1) and (1, 2, 0) are in S Since we know that S is a subspace of R 3 (Q6(b), HW1), it follows that W S In other words, we know that any linear combination of solutions is a solution, and W consists exactly in all the possible linear combinations of (1, 1, 1) and (1, 2, 0) 2
3 S W : any solution (x 1, x 2, x 3 ) of 2x 1 + x 2 3x 3 = 0 must satisfy x 2 = 3x 3 2x 1 So, we can write the general solution as (x 1, 3x 3 2x 1, x 3 ) = x 3 (1, 1, 1) + (x 1 x 3 )(1, 2, 0), which is a linear combination of (1, 1, 1) and (1, 2, 0) Hence, S W Problem 4: Let V be a vector space and suppose S = {v 1,, v k } is a finite set of linearly dependent vectors in V Prove that there is a proper subset of S whose span is equal to the span of S By a proper subset, we mean a subset T S such that T S Solution: By the Lemma 210 in the notes, there exists j {1,, k} and c 1,, c j 1, c j+1,, c k F such that v j = c 1 v c j 1 v j 1 + c j+1 v j c k v k Let T = S \ {v j } Clearly, T is a proper subset of S Now we show that span(t ) = span(s) span (T ) span (S): this is immediate, since T S span (S) span (T ): let v = a 1 v a k v k span (S) Then, replacing v j by the formula above, and regrouping the terms, we can write v = (a 1 + a j c 1 )v (a j 1 + a j c j 1 )v j 1 + (a j+1 + a j c j+1 )v j (a k + a j c k )v k, which is in span (T ) Hence span (S) span (T ) Problem 5: Use Gaussian elimination in R to show that the solution set of the homogeneous system x 1 + 2x 2 + 3x 3 + 4x 4 = 0 x 1 + 4x 2 + 3x 3 + 2x 4 = 0 2x 1 + 5x 2 + 6x 3 + 7x 4 = 0 x 1 + 3x 3 + 6x 4 = 0 is a plane through 0 (that is, it is the span of two linearly independent vectors), and find two linearly independent vectors whose span is the solution space Solution: We apply Gaussian elimination Recall that it consists in modifying the system of equations in order to find a set of variables, called the pivot variables, such that each one of them appears only once in the system of equations, with a coefficient 1, and two of them can not appear in the same equation Furthermore, this set should be maximal, ie there is no larger set of variables satisfying the same condition (this is equivalent to asking that the number of such variables is equal to the number of non-zero equations) To do this, you are allowed to do the three operations described at the beginning of the section 41 in the notes; the precise algorithm to follow in order to achieve this is described in the notes The other variables are called the free variables The reason for that name, and reason for which this whole Gaussian elimination thing is useful, is that once we have a system of equations as in the paragraph above, we can isolate each of the pivot variables, and express them only in terms of the free variables The free variables are free because any choice of values for them gives a unique solution for the corresponding pivot variables 3
4 We can then write the general solution of the system, and express it as a linear combination of vectors with coefficients given by the free variables This exhibits an explicit set of vectors that generates the solution set of the equation The matrix of our system is Substracting the appropriate multiple of the first equation to the second, third and fourth ones in such a way that the coefficient of x 1 is 0, we get Substracting 2 times the third equation from the second one, and ( 2) times the third one from the fourth one, we get Finally, we want to get rid of that 2 in the first equation, so that the variable x 2 appears only in the third one To achieve this, we substract 2 times the third equation from the first one, to get The second and the fourth equations are 0 = 0 (which is always true), so we can drop them The associated system is therefore { x1 + 3x 3 + 6x 4 = 0 x 2 x 4 = 0, which we rewrite as { x1 = 3x 3 6x 4 x 2 = x 4 in such a way that the pivot variables x 1 and x 2 are isolated Now it is clear that we can choose x 3 and x 4 in any way we want, and that x 1 and x 2 will be entirely determined by that choice So, we can write the general solution as ( 3x 3 6x 4, x 4, x 3, x 4 ) = x 3 ( 3, 0, 1, 0) + x 4 ( 6, 1, 0, 1), hence is in the span of ( 3, 0, 1, 0) and ( 6, 1, 0, 1) Furthermore, since ( 3, 0, 1, 0) and ( 6, 1, 0, 1) are not collinear, they are linearly independent It follows that ( 3, 0, 1, 0) and ( 6, 1, 0, 1) are two linearly independent vectors whose span is the solution space 4
5 Problem 6: In parts (a) and (b) we will show that multiplicative inverses exist in Z/pZ for p prime So, you should not use this fact in your answers to (a) and (b) However, feel free to use other basic facts about arithmetic in Z/pZ (a) Suppose a Z/pZ, with a 0 For any x, y Z/pZ, show that if ax = ay (with multiplication in Z/pZ, ie modulo p) then x = y [Hint: you will need to unwrap the definition of mod p multiplication, and make use of the hypotheses that p is prime and a 0] (b) By considering the set {a0, a1,, a(p 1)} over Z/pZ, or otherwise, show that there exists b Z/pZ such that ab = 1 (again, with multiplication mod p) (c) In the set of integers Z, solve the system of equations { 2x + y=2 mod 5 3x 2y=0 mod 5, by using Gaussian elimination in Z/5Z Solution: (a) Suppose that ax = ay This means that ax ay = 0 in Z/pZ, hence p divides ax ay = a(x y), now seen as an element of Z Since p is prime, we have that p divides a or p divides x y But we assumed that a 0 in Z/pZ, hence p does not divide a It follows that p divides x y, hence x = y in Z/pZ (b) Assume that there does not exist such an element b Then the elements a0,, a(p 1) take at most p 1 different values It follows by the pigeonhole principle that two of them must be equal, ie ak = al for k l in Z/p/Z By the point (a), this implies that k = l, a contradiction Another way of phrasing this is that the map from Z/pZ to Z/pZ given by multiplication by a is one-to-one, by (a) Since any one-to-one map from a finite set to itself also has to be onto, there must exist b Z/pZ such that the image of b is 1, ie ab = 1 (c) The associated matrix to that system is (including a last column for the right-hand side): ( ) We multiply the first row by 3 (which is the inverse of 2 in Z/5Z), and the second row by 2 (which is the inverse of 3 in Z/5Z) Remember that 2 ( 2) = 4 = 1 ( ) We substract the first row from the second one Remember that 1 3 = 2 = 3 and that 0 1 = 1 = 4 ( )
6 We want to get rid of the 3 in the first row For this, we substract the second row from the first one Remember that 1 4 = 3 = 2 ( ) Finally, we want to transform this 3 in the second row into a 1 To do this, we multiply by 2, which is the inverse of 3 Keep in mind that 2 4 = 8 = 3 ( ) The corresponding system of equations is { x=2 mod 5 y=3 mod 5 This is our solution Problem 7: If a, b R, prove: (a) There is a rational r (a, b); (b) There is an irrational c (a, b) (Hint: use part (a) and Q2 of HW1); (c) (a, b) contains infinitely many rationals and infinitely many irrationals Solution: (a) The idea is as follow: using Archimedes principle, we will produce a rational number which is smaller than b a We will them use Archimedes principle again to produce a multiple of it that is between a and b Here is how it works: with x = 1 and h = b a, Archimedes principle produces k 1 Z such that (k 1 1)(b a) 1 < k 1 (b a); It is clear that we must have k 1 > 0, since b a > 0 In particular, we get 1 k 1 < b a Now, apply Archimedes principle again with x = a and h = 1 k 1 to get k 2 Z such that (k 2 1) 1 k 1 a < k 2 1 k 1 Then, r = k 1 k 2 is rational, and larger than a by the line above We check that r < b: r = k 2 k 1 = (k 2 1) 1 k k 1 < a + (b a) = b 6
7 (b) By the point (a), there exists a rational number c such that a 2 < c < b 2 Adding 2, we get a < c + 2 < b The number c + 2 is irrational, by Q2 of HW1 (c) By (a), there exists a rational number x 1 between a and b Apply (a) again to find another rational number x 2 between a and x 1 Aply (a) again to find another rational number x 3 between a and x 2 This way, we construct infinitely many distinct rational numbers between a and b, as asked For irrational numbers, do the same as above, but using (b) to construct irrational numbers Problem 8: Let a n, b n and c n be three sequences such that a n b n c n for all n N, and the limits of a n and c n exist and coincide: lim a n = lim c n n n Show that then the sequence b n also converges, and lim a n = lim b n = lim c n n n n Solution: Let us call s = lim n a n = lim n c n Let ε > 0 Since a n converges to s, there exists N 1 N such that for every n N 1, we have a n s < ε 3 (we took ε 3 in the definition of convergence) Similarly, since c n converges to s, there exists N 2 N such that for every n N 2, we have c n s < ε 3 (we took ε 3 in the definition of convergence) Let N 3 = max{n 1, N 2 } Since N 3 is greater or equal to both N 1 and N 2, we have that for all n N 3, both a n s < ε 3 and c n s < ε 3 By the triangular inequality, we have a n c n = a n s + s c n a n s + s c n For n N 3, this is smaller than ε 3 + ε 3 = 2ε 3 Since a n c n, we have a n c n = c n a n We therefore get c n a n < 2ε 3, which is c n < a n + 2ε 3 Since a n b n c n, we get a n b n c n < a n + 2ε 3 Therefore, a n b n < a n + 2ε 3, which is 0 b n a n < 2ε 3 Therefore, b n a n < 2ε 3, for n N 3 Now, for n N 3, we have by the triangular inequality b n s = b n a n + a n s b n a n + a n s < 2ε 3 + ε 3 = ε This proves that b n converges to s Other method: Substracting s, we get a n s b n s c n s It is then easy to see that we must have b n s max{ a n s, c n s } Now, as in the solution before, given ε > 0, there exists N 3 N such that a n s < ε and c n s < ε for n N 3 Then, we also have b n s < ε for n N 3, proving that b n converges to s 7
8 Problem 9: Show that lim n n1/n = 1 Hint: show that for any δ > 0 there exists N (that depends on δ) so that n (1 + δ) n for all n N Solution: Let us prove the hint first Let δ > 0 Note that, for n 2, we have n + (1 + δ) n = n + n j=0 ( ) n δ j n nδ + n2 n δ 2 j 2 This last expression is larger than 0 for n large; indeed, it is a polynomial of degree 2 whose coefficient of n 2 is positive, hence it goes to + as n + Therefore there exists N N such that n+(1+δ) n 0 for n N, ie n (1 + δ) n for all n N This is equivalent to: for any δ > 0, there exists N N such that for all n N we have n 1/n 1 + δ Since n 1/n 1 for all n 1, this means that for every δ > 0, there exists N N such that 1 n 1/n < δ for n N It follows by the definition of limit that lim n n1/n = 1 8
Solutions to Homework 5 - Math 3410
Solutions to Homework 5 - Math 34 (Page 57: # 489) Determine whether the following vectors in R 4 are linearly dependent or independent: (a) (, 2, 3, ), (3, 7,, 2), (, 3, 7, 4) Solution From x(, 2, 3,
More informationMath 24 Spring 2012 Questions (mostly) from the Textbook
Math 24 Spring 2012 Questions (mostly) from the Textbook 1. TRUE OR FALSE? (a) The zero vector space has no basis. (F) (b) Every vector space that is generated by a finite set has a basis. (c) Every vector
More informationMath 61CM - Solutions to homework 6
Math 61CM - Solutions to homework 6 Cédric De Groote November 5 th, 2018 Problem 1: (i) Give an example of a metric space X such that not all Cauchy sequences in X are convergent. (ii) Let X be a metric
More informationSection 8.3 Partial Fraction Decomposition
Section 8.6 Lecture Notes Page 1 of 10 Section 8.3 Partial Fraction Decomposition Partial fraction decomposition involves decomposing a rational function, or reversing the process of combining two or more
More informationQuestionnaire for CSET Mathematics subset 1
Questionnaire for CSET Mathematics subset 1 Below is a preliminary questionnaire aimed at finding out your current readiness for the CSET Math subset 1 exam. This will serve as a baseline indicator for
More informationMath 109 HW 9 Solutions
Math 109 HW 9 Solutions Problems IV 18. Solve the linear diophantine equation 6m + 10n + 15p = 1 Solution: Let y = 10n + 15p. Since (10, 15) is 5, we must have that y = 5x for some integer x, and (as we
More informationMath 54. Selected Solutions for Week 5
Math 54. Selected Solutions for Week 5 Section 4. (Page 94) 8. Consider the following two systems of equations: 5x + x 3x 3 = 5x + x 3x 3 = 9x + x + 5x 3 = 4x + x 6x 3 = 9 9x + x + 5x 3 = 5 4x + x 6x 3
More informationPartial Fraction Decomposition Honors Precalculus Mr. Velazquez Rm. 254
Partial Fraction Decomposition Honors Precalculus Mr. Velazquez Rm. 254 Adding and Subtracting Rational Expressions Recall that we can use multiplication and common denominators to write a sum or difference
More informationThe value of a problem is not so much coming up with the answer as in the ideas and attempted ideas it forces on the would be solver I.N.
Math 410 Homework Problems In the following pages you will find all of the homework problems for the semester. Homework should be written out neatly and stapled and turned in at the beginning of class
More information1. Solve each linear system using Gaussian elimination or Gauss-Jordan reduction. The augmented matrix of this linear system is
Solutions to Homework Additional Problems. Solve each linear system using Gaussian elimination or Gauss-Jordan reduction. (a) x + y = 8 3x + 4y = 7 x + y = 3 The augmented matrix of this linear system
More informationChapter 3. Vector spaces
Chapter 3. Vector spaces Lecture notes for MA1111 P. Karageorgis pete@maths.tcd.ie 1/22 Linear combinations Suppose that v 1,v 2,...,v n and v are vectors in R m. Definition 3.1 Linear combination We say
More informationMATH 1120 (LINEAR ALGEBRA 1), FINAL EXAM FALL 2011 SOLUTIONS TO PRACTICE VERSION
MATH (LINEAR ALGEBRA ) FINAL EXAM FALL SOLUTIONS TO PRACTICE VERSION Problem (a) For each matrix below (i) find a basis for its column space (ii) find a basis for its row space (iii) determine whether
More informationMATH 2050 Assignment 6 Fall 2018 Due: Thursday, November 1. x + y + 2z = 2 x + y + z = c 4x + 2z = 2
MATH 5 Assignment 6 Fall 8 Due: Thursday, November [5]. For what value of c does have a solution? Is it unique? x + y + z = x + y + z = c 4x + z = Writing the system as an augmented matrix, we have c R
More informationMath 54 HW 4 solutions
Math 54 HW 4 solutions 2.2. Section 2.2 (a) False: Recall that performing a series of elementary row operations A is equivalent to multiplying A by a series of elementary matrices. Suppose that E,...,
More informationLIMITS AT INFINITY MR. VELAZQUEZ AP CALCULUS
LIMITS AT INFINITY MR. VELAZQUEZ AP CALCULUS RECALL: VERTICAL ASYMPTOTES Remember that for a rational function, vertical asymptotes occur at values of x = a which have infinite its (either positive or
More informationSolutions for Homework Assignment 2
Solutions for Homework Assignment 2 Problem 1. If a,b R, then a+b a + b. This fact is called the Triangle Inequality. By using the Triangle Inequality, prove that a b a b for all a,b R. Solution. To prove
More informationPARTIAL FRACTION DECOMPOSITION. Mr. Velazquez Honors Precalculus
PARTIAL FRACTION DECOMPOSITION Mr. Velazquez Honors Precalculus ADDING AND SUBTRACTING RATIONAL EXPRESSIONS Recall that we can use multiplication and common denominators to write a sum or difference of
More informationSolutions to Practice Final
s to Practice Final 1. (a) What is φ(0 100 ) where φ is Euler s φ-function? (b) Find an integer x such that 140x 1 (mod 01). Hint: gcd(140, 01) = 7. (a) φ(0 100 ) = φ(4 100 5 100 ) = φ( 00 5 100 ) = (
More informationGRE Subject test preparation Spring 2016 Topic: Abstract Algebra, Linear Algebra, Number Theory.
GRE Subject test preparation Spring 2016 Topic: Abstract Algebra, Linear Algebra, Number Theory. Linear Algebra Standard matrix manipulation to compute the kernel, intersection of subspaces, column spaces,
More informationMath 4377/6308 Advanced Linear Algebra
2. Linear Transformations Math 4377/638 Advanced Linear Algebra 2. Linear Transformations, Null Spaces and Ranges Jiwen He Department of Mathematics, University of Houston jiwenhe@math.uh.edu math.uh.edu/
More informationANALYTICAL MATHEMATICS FOR APPLICATIONS 2018 LECTURE NOTES 3
ANALYTICAL MATHEMATICS FOR APPLICATIONS 2018 LECTURE NOTES 3 ISSUED 24 FEBRUARY 2018 1 Gaussian elimination Let A be an (m n)-matrix Consider the following row operations on A (1) Swap the positions any
More informationChapter Five Notes N P U2C5
Chapter Five Notes N P UC5 Name Period Section 5.: Linear and Quadratic Functions with Modeling In every math class you have had since algebra you have worked with equations. Most of those equations have
More informationMULTIPLYING TRINOMIALS
Name: Date: 1 Math 2 Variable Manipulation Part 4 Polynomials B MULTIPLYING TRINOMIALS Multiplying trinomials is the same process as multiplying binomials except for there are more terms to multiply than
More informationSYMBOL EXPLANATION EXAMPLE
MATH 4310 PRELIM I REVIEW Notation These are the symbols we have used in class, leading up to Prelim I, and which I will use on the exam SYMBOL EXPLANATION EXAMPLE {a, b, c, } The is the way to write the
More informationMath 1302 Notes 2. How many solutions? What type of solution in the real number system? What kind of equation is it?
Math 1302 Notes 2 We know that x 2 + 4 = 0 has How many solutions? What type of solution in the real number system? What kind of equation is it? What happens if we enlarge our current system? Remember
More informationLinear Algebra Quiz 4. Problem 1 (Linear Transformations): 4 POINTS Show all Work! Consider the tranformation T : R 3 R 3 given by:
Page 1 This is a 60 min Quiz. Please make sure you put your name on the top right hand corner of each sheet. Remember the Honors Code will be enforced! You may use your book. NO HELP FROM ANYONE. Problem
More informationDepartment of Aerospace Engineering AE602 Mathematics for Aerospace Engineers Assignment No. 4
Department of Aerospace Engineering AE6 Mathematics for Aerospace Engineers Assignment No.. Decide whether or not the following vectors are linearly independent, by solving c v + c v + c 3 v 3 + c v :
More informationMATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS
MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS 1. HW 1: Due September 4 1.1.21. Suppose v, w R n and c is a scalar. Prove that Span(v + cw, w) = Span(v, w). We must prove two things: that every element
More informationProjections and Least Square Solutions. Recall that given an inner product space V with subspace W and orthogonal basis for
Math 57 Spring 18 Projections and Least Square Solutions Recall that given an inner product space V with subspace W and orthogonal basis for W, B {v 1, v,..., v k }, the orthogonal projection of V onto
More informationMATH FINAL EXAM REVIEW HINTS
MATH 109 - FINAL EXAM REVIEW HINTS Answer: Answer: 1. Cardinality (1) Let a < b be two real numbers and define f : (0, 1) (a, b) by f(t) = (1 t)a + tb. (a) Prove that f is a bijection. (b) Prove that any
More informationMath 240, 4.3 Linear Independence; Bases A. DeCelles. 1. definitions of linear independence, linear dependence, dependence relation, basis
Math 24 4.3 Linear Independence; Bases A. DeCelles Overview Main ideas:. definitions of linear independence linear dependence dependence relation basis 2. characterization of linearly dependent set using
More informationCHMC: Finite Fields 9/23/17
CHMC: Finite Fields 9/23/17 1 Introduction This worksheet is an introduction to the fascinating subject of finite fields. Finite fields have many important applications in coding theory and cryptography,
More informationAbstract Vector Spaces and Concrete Examples
LECTURE 18 Abstract Vector Spaces and Concrete Examples Our discussion of linear algebra so far has been devoted to discussing the relations between systems of linear equations, matrices, and vectors.
More informationMATH 225 Summer 2005 Linear Algebra II Solutions to Assignment 1 Due: Wednesday July 13, 2005
MATH 225 Summer 25 Linear Algebra II Solutions to Assignment 1 Due: Wednesday July 13, 25 Department of Mathematical and Statistical Sciences University of Alberta Question 1. [p 224. #2] The set of all
More informationHomework #2 solutions Due: June 15, 2012
All of the following exercises are based on the material in the handout on integers found on the class website. 1. Find d = gcd(475, 385) and express it as a linear combination of 475 and 385. That is
More information1.3 Linear Dependence & span K
( ) Conversely, suppose that every vector v V can be expressed uniquely as v = u +w for u U and w W. Then, the existence of this expression for each v V is simply the statement that V = U +W. Moreover,
More informationBeginning Algebra. 1. Review of Pre-Algebra 1.1 Review of Integers 1.2 Review of Fractions
1. Review of Pre-Algebra 1.1 Review of Integers 1.2 Review of Fractions Beginning Algebra 1.3 Review of Decimal Numbers and Square Roots 1.4 Review of Percents 1.5 Real Number System 1.6 Translations:
More informationLinear Independence. Consider a plane P that includes the origin in R 3 {u, v, w} in P. and non-zero vectors
Linear Independence Consider a plane P that includes the origin in R 3 {u, v, w} in P. and non-zero vectors If no two of u, v and w are parallel, then P =span{u, v, w}. But any two vectors determines a
More informationAlgebra I Unit Report Summary
Algebra I Unit Report Summary No. Objective Code NCTM Standards Objective Title Real Numbers and Variables Unit - ( Ascend Default unit) 1. A01_01_01 H-A-B.1 Word Phrases As Algebraic Expressions 2. A01_01_02
More informationProofs. Chapter 2 P P Q Q
Chapter Proofs In this chapter we develop three methods for proving a statement. To start let s suppose the statement is of the form P Q or if P, then Q. Direct: This method typically starts with P. Then,
More informationMath 601 Solutions to Homework 3
Math 601 Solutions to Homework 3 1 Use Cramer s Rule to solve the following system of linear equations (Solve for x 1, x 2, and x 3 in terms of a, b, and c 2x 1 x 2 + 7x 3 = a 5x 1 2x 2 x 3 = b 3x 1 x
More informationNumbers. 2.1 Integers. P(n) = n(n 4 5n 2 + 4) = n(n 2 1)(n 2 4) = (n 2)(n 1)n(n + 1)(n + 2); 120 =
2 Numbers 2.1 Integers You remember the definition of a prime number. On p. 7, we defined a prime number and formulated the Fundamental Theorem of Arithmetic. Numerous beautiful results can be presented
More informationACO Comprehensive Exam March 17 and 18, Computability, Complexity and Algorithms
1. Computability, Complexity and Algorithms (a) Let G(V, E) be an undirected unweighted graph. Let C V be a vertex cover of G. Argue that V \ C is an independent set of G. (b) Minimum cardinality vertex
More informationSolutions to Math 51 Midterm 1 July 6, 2016
Solutions to Math 5 Midterm July 6, 26. (a) (6 points) Find an equation (of the form ax + by + cz = d) for the plane P in R 3 passing through the points (, 2, ), (2,, ), and (,, ). We first compute two
More informationRow Space, Column Space, and Nullspace
Row Space, Column Space, and Nullspace MATH 322, Linear Algebra I J. Robert Buchanan Department of Mathematics Spring 2015 Introduction Every matrix has associated with it three vector spaces: row space
More informationMath 4377/6308 Advanced Linear Algebra I Dr. Vaughn Climenhaga, PGH 651A HOMEWORK 3
Math 4377/6308 Advanced Linear Algebra I Dr. Vaughn Climenhaga, PGH 651A Fall 2013 HOMEWORK 3 Due 4pm Wednesday, September 11. You will be graded not only on the correctness of your answers but also on
More informationSPRING 2006 PRELIMINARY EXAMINATION SOLUTIONS
SPRING 006 PRELIMINARY EXAMINATION SOLUTIONS 1A. Let G be the subgroup of the free abelian group Z 4 consisting of all integer vectors (x, y, z, w) such that x + 3y + 5z + 7w = 0. (a) Determine a linearly
More informationChapter 2. Polynomial and Rational Functions. 2.5 Zeros of Polynomial Functions
Chapter 2 Polynomial and Rational Functions 2.5 Zeros of Polynomial Functions 1 / 33 23 Chapter 2 Homework 2.5 p335 6, 8, 10, 12, 16, 20, 24, 28, 32, 34, 38, 42, 46, 50, 52 2 / 33 23 3 / 33 23 Objectives:
More informationLinear Algebra, Summer 2011, pt. 2
Linear Algebra, Summer 2, pt. 2 June 8, 2 Contents Inverses. 2 Vector Spaces. 3 2. Examples of vector spaces..................... 3 2.2 The column space......................... 6 2.3 The null space...........................
More informationSB CH 2 answers.notebook. November 05, Warm Up. Oct 8 10:36 AM. Oct 5 2:22 PM. Oct 8 9:22 AM. Oct 8 9:19 AM. Oct 8 9:26 AM.
Warm Up Oct 8 10:36 AM Oct 5 2:22 PM Linear Function Qualities Oct 8 9:22 AM Oct 8 9:19 AM Quadratic Function Qualities Oct 8 9:26 AM Oct 8 9:25 AM 1 Oct 8 9:28 AM Oct 8 9:25 AM Given vertex (-1,4) and
More informationAlgebra 1 Seamless Curriculum Guide
QUALITY STANDARD #1: REAL NUMBERS AND THEIR PROPERTIES 1.1 The student will understand the properties of real numbers. o Identify the subsets of real numbers o Addition- commutative, associative, identity,
More informationMATH 51H Section 4. October 16, Recall what it means for a function between metric spaces to be continuous:
MATH 51H Section 4 October 16, 2015 1 Continuity Recall what it means for a function between metric spaces to be continuous: Definition. Let (X, d X ), (Y, d Y ) be metric spaces. A function f : X Y is
More informationMATH CSE20 Homework 5 Due Monday November 4
MATH CSE20 Homework 5 Due Monday November 4 Assigned reading: NT Section 1 (1) Prove the statement if true, otherwise find a counterexample. (a) For all natural numbers x and y, x + y is odd if one of
More informationThis lecture: basis and dimension 4.4. Linear Independence: Suppose that V is a vector space and. r 1 x 1 + r 2 x r k x k = 0
Linear Independence: Suppose that V is a vector space and that x, x 2,, x k belong to V {x, x 2,, x k } are linearly independent if r x + r 2 x 2 + + r k x k = only for r = r 2 = = r k = The vectors x,
More informationMath 2142 Homework 5 Part 1 Solutions
Math 2142 Homework 5 Part 1 Solutions Problem 1. For the following homogeneous second order differential equations, give the general solution and the particular solution satisfying the given initial conditions.
More informationReference Material /Formulas for Pre-Calculus CP/ H Summer Packet
Reference Material /Formulas for Pre-Calculus CP/ H Summer Packet Week # 1 Order of Operations Step 1 Evaluate expressions inside grouping symbols. Order of Step 2 Evaluate all powers. Operations Step
More informationA2 HW Imaginary Numbers
Name: A2 HW Imaginary Numbers Rewrite the following in terms of i and in simplest form: 1) 100 2) 289 3) 15 4) 4 81 5) 5 12 6) -8 72 Rewrite the following as a radical: 7) 12i 8) 20i Solve for x in simplest
More information(1) Consider the space S consisting of all continuous real-valued functions on the closed interval [0, 1]. For f, g S, define
Homework, Real Analysis I, Fall, 2010. (1) Consider the space S consisting of all continuous real-valued functions on the closed interval [0, 1]. For f, g S, define ρ(f, g) = 1 0 f(x) g(x) dx. Show that
More informationChapter 2 notes from powerpoints
Chapter 2 notes from powerpoints Synthetic division and basic definitions Sections 1 and 2 Definition of a Polynomial Function: Let n be a nonnegative integer and let a n, a n-1,, a 2, a 1, a 0 be real
More informationSeunghee Ye Ma 8: Week 2 Oct 6
Week 2 Summary This week, we will learn about sequences and real numbers. We first define what we mean by a sequence and discuss several properties of sequences. Then, we will talk about what it means
More informationProblems for M 10/12:
Math 30, Lesieutre Problem set #8 October, 05 Problems for M 0/: 4.3.3 Determine whether these vectors are a basis for R 3 by checking whether the vectors span R 3, and whether the vectors are linearly
More informationALGEBRA I FORM I. Textbook: Algebra, Second Edition;Prentice Hall,2002
ALGEBRA I FORM I Textbook: Algebra, Second Edition;Prentice Hall,00 Prerequisites: Students are expected to have a knowledge of Pre Algebra and proficiency of basic math skills including: positive and
More informationMath Analysis Notes Mrs. Atkinson 1
Name: Math Analysis Chapter 7 Notes Day 6: Section 7-1 Solving Systems of Equations with Two Variables; Sections 7-1: Solving Systems of Equations with Two Variables Solving Systems of equations with two
More informationEigenvalues and Eigenvectors
LECTURE 3 Eigenvalues and Eigenvectors Definition 3.. Let A be an n n matrix. The eigenvalue-eigenvector problem for A is the problem of finding numbers λ and vectors v R 3 such that Av = λv. If λ, v are
More informationMath Exam 2, October 14, 2008
Math 96 - Exam 2, October 4, 28 Name: Problem (5 points Find all solutions to the following system of linear equations, check your work: x + x 2 x 3 2x 2 2x 3 2 x x 2 + x 3 2 Solution Let s perform Gaussian
More informationMore Polynomial Equations Section 6.4
MATH 11009: More Polynomial Equations Section 6.4 Dividend: The number or expression you are dividing into. Divisor: The number or expression you are dividing by. Synthetic division: Synthetic division
More informationMATHEMATICAL INDUCTION
MATHEMATICAL INDUCTION MATH 3A SECTION HANDOUT BY GERARDO CON DIAZ Imagine a bunch of dominoes on a table. They are set up in a straight line, and you are about to push the first piece to set off the chain
More informationExtra Problems for Math 2050 Linear Algebra I
Extra Problems for Math 5 Linear Algebra I Find the vector AB and illustrate with a picture if A = (,) and B = (,4) Find B, given A = (,4) and [ AB = A = (,4) and [ AB = 8 If possible, express x = 7 as
More informationCHAPTER EIGHT: SOLVING QUADRATIC EQUATIONS Review April 9 Test April 17 The most important equations at this level of mathematics are quadratic
CHAPTER EIGHT: SOLVING QUADRATIC EQUATIONS Review April 9 Test April 17 The most important equations at this level of mathematics are quadratic equations. They can be solved using a graph, a perfect square,
More informationMath 3 Variable Manipulation Part 3 Polynomials A
Math 3 Variable Manipulation Part 3 Polynomials A 1 MATH 1 & 2 REVIEW: VOCABULARY Constant: A term that does not have a variable is called a constant. Example: the number 5 is a constant because it does
More informationFlorida Math Curriculum (433 topics)
Florida Math 0028 This course covers the topics shown below. Students navigate learning paths based on their level of readiness. Institutional users may customize the scope and sequence to meet curricular
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include
PUTNAM TRAINING POLYNOMIALS (Last updated: December 11, 2017) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationMath 61CM - Quick answer key to section problems Fall 2018
Math 6CM - Quick answer key to section problems Fall 08 Cédric De Groote These are NOT complete solutions! You are always expected to write down all the details and justify everything. This document is
More informationOrganization Team Team ID#
1. [4] A random number generator will always output 7. Sam uses this random number generator once. What is the expected value of the output? 2. [4] Let A, B, C, D, E, F be 6 points on a circle in that
More informationis equal to = 3 2 x, if x < 0 f (0) = lim h = 0. Therefore f exists and is continuous at 0.
Madhava Mathematics Competition January 6, 2013 Solutions and scheme of marking Part I N.B. Each question in Part I carries 2 marks. p(k + 1) 1. If p(x) is a non-constant polynomial, then lim k p(k) (a)
More informationCommon Core Algebra 2 Review Session 1
Common Core Algebra 2 Review Session 1 NAME Date 1. Which of the following is algebraically equivalent to the sum of 4x 2 8x + 7 and 3x 2 2x 5? (1) 7x 2 10x + 2 (2) 7x 2 6x 12 (3) 7x 4 10x 2 + 2 (4) 12x
More informationCheat Sheet for MATH461
Cheat Sheet for MATH46 Here is the stuff you really need to remember for the exams Linear systems Ax = b Problem: We consider a linear system of m equations for n unknowns x,,x n : For a given matrix A
More informationVector Spaces, Orthogonality, and Linear Least Squares
Week Vector Spaces, Orthogonality, and Linear Least Squares. Opening Remarks.. Visualizing Planes, Lines, and Solutions Consider the following system of linear equations from the opener for Week 9: χ χ
More information1 Systems of equations
Highlights from linear algebra David Milovich, Math 2 TA for sections -6 November, 28 Systems of equations A leading entry in a matrix is the first (leftmost) nonzero entry of a row. For example, the leading
More informationMATH 167: APPLIED LINEAR ALGEBRA Chapter 2
MATH 167: APPLIED LINEAR ALGEBRA Chapter 2 Jesús De Loera, UC Davis February 1, 2012 General Linear Systems of Equations (2.2). Given a system of m equations and n unknowns. Now m n is OK! Apply elementary
More informationMT804 Analysis Homework II
MT804 Analysis Homework II Eudoxus October 6, 2008 p. 135 4.5.1, 4.5.2 p. 136 4.5.3 part a only) p. 140 4.6.1 Exercise 4.5.1 Use the Intermediate Value Theorem to prove that every polynomial of with real
More informationDigital Workbook for GRA 6035 Mathematics
Eivind Eriksen Digital Workbook for GRA 6035 Mathematics November 10, 2014 BI Norwegian Business School Contents Part I Lectures in GRA6035 Mathematics 1 Linear Systems and Gaussian Elimination........................
More information2 so Q[ 2] is closed under both additive and multiplicative inverses. a 2 2b 2 + b
. FINITE-DIMENSIONAL VECTOR SPACES.. Fields By now you ll have acquired a fair knowledge of matrices. These are a concrete embodiment of something rather more abstract. Sometimes it is easier to use matrices,
More informationProof 1: Using only ch. 6 results. Since gcd(a, b) = 1, we have
Exercise 13. Consider positive integers a, b, and c. (a) Suppose gcd(a, b) = 1. (i) Show that if a divides the product bc, then a must divide c. I give two proofs here, to illustrate the different methods.
More informationToday. Polynomials. Secret Sharing.
Today. Polynomials. Secret Sharing. A secret! I have a secret! A number from 0 to 10. What is it? Any one of you knows nothing! Any two of you can figure it out! Example Applications: Nuclear launch: need
More informationLinear algebra I Homework #1 due Thursday, Oct Show that the diagonals of a square are orthogonal to one another.
Homework # due Thursday, Oct. 0. Show that the diagonals of a square are orthogonal to one another. Hint: Place the vertices of the square along the axes and then introduce coordinates. 2. Find the equation
More informationLinear Algebra. Min Yan
Linear Algebra Min Yan January 2, 2018 2 Contents 1 Vector Space 7 1.1 Definition................................. 7 1.1.1 Axioms of Vector Space..................... 7 1.1.2 Consequence of Axiom......................
More informationLinear Algebra Math 221
Linear Algebra Math 221 Open Book Exam 1 Open Notes 3 Sept, 24 Calculators Permitted Show all work (except #4) 1 2 3 4 2 1. (25 pts) Given A 1 2 1, b 2 and c 4. 1 a) (7 pts) Bring matrix A to echelon form.
More informationMATH 315 Linear Algebra Homework #1 Assigned: August 20, 2018
Homework #1 Assigned: August 20, 2018 Review the following subjects involving systems of equations and matrices from Calculus II. Linear systems of equations Converting systems to matrix form Pivot entry
More informationAlgebra Summer Review Packet
Name: Algebra Summer Review Packet About Algebra 1: Algebra 1 teaches students to think, reason, and communicate mathematically. Students use variables to determine solutions to real world problems. Skills
More informationMATH 167: APPLIED LINEAR ALGEBRA Chapter 3
MATH 167: APPLIED LINEAR ALGEBRA Chapter 3 Jesús De Loera, UC Davis February 18, 2012 Orthogonal Vectors and Subspaces (3.1). In real life vector spaces come with additional METRIC properties!! We have
More informationJUST THE MATHS UNIT NUMBER 1.9. ALGEBRA 9 (The theory of partial fractions) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1. ALGEBRA (The theory of partial fractions) by A.J.Hobson 1..1 Introduction 1..2 Standard types of partial fraction problem 1.. Exercises 1..4 Answers to exercises UNIT 1. -
More informationweb: HOMEWORK 1
MAT 207 LINEAR ALGEBRA I 2009207 Dokuz Eylül University, Faculty of Science, Department of Mathematics Instructor: Engin Mermut web: http://kisideuedutr/enginmermut/ HOMEWORK VECTORS IN THE n-dimensional
More informationRings. Chapter 1. Definition 1.2. A commutative ring R is a ring in which multiplication is commutative. That is, ab = ba for all a, b R.
Chapter 1 Rings We have spent the term studying groups. A group is a set with a binary operation that satisfies certain properties. But many algebraic structures such as R, Z, and Z n come with two binary
More informationRevision notes for Pure 1(9709/12)
Revision notes for Pure 1(9709/12) By WaqasSuleman A-Level Teacher Beaconhouse School System Contents 1. Sequence and Series 2. Functions & Quadratics 3. Binomial theorem 4. Coordinate Geometry 5. Trigonometry
More informationHomework 9 Solutions to Selected Problems
Homework 9 Solutions to Selected Problems June 11, 2012 1 Chapter 17, Problem 12 Since x 2 + x + 4 has degree 2 and Z 11 is a eld, we may use Theorem 17.1 and show that f(x) is irreducible because it has
More informationDay 6: 6.4 Solving Polynomial Equations Warm Up: Factor. 1. x 2-2x x 2-9x x 2 + 6x + 5
Day 6: 6.4 Solving Polynomial Equations Warm Up: Factor. 1. x 2-2x - 15 2. x 2-9x + 14 3. x 2 + 6x + 5 Solving Equations by Factoring Recall the factoring pattern: Difference of Squares:...... Note: There
More informationProofs. Chapter 2 P P Q Q
Chapter Proofs In this chapter we develop three methods for proving a statement. To start let s suppose the statement is of the form P Q or if P, then Q. Direct: This method typically starts with P. Then,
More informationMATH 167: APPLIED LINEAR ALGEBRA Least-Squares
MATH 167: APPLIED LINEAR ALGEBRA Least-Squares October 30, 2014 Least Squares We do a series of experiments, collecting data. We wish to see patterns!! We expect the output b to be a linear function of
More informationReview problems for MA 54, Fall 2004.
Review problems for MA 54, Fall 2004. Below are the review problems for the final. They are mostly homework problems, or very similar. If you are comfortable doing these problems, you should be fine on
More information