Mathematics 206 Solutions for HWK 13b Section 5.2
|
|
- Georgina Hodges
- 5 years ago
- Views:
Transcription
1 Mathematics 206 Solutions for HWK 13b Section 5.2 Section Problem 7ac. Which of the following are linear combinations of u = (0, 2,2) and v = (1, 3, 1)? (a) (2, 2,2) (c) (0,4, 5) Solution. Solution by inspection doesn t look feasible to me, so solve two systems of equations. We may as well solve them both at the same time. We want to know whether there exist scalars a 1, a 2 for which a 1 (0, 2, 2) + a 2 (1,3, 1) = (2, 2,2) and whether there exist scalars c 1, c 2 for which c 1 (0, 2, 2) + c 2 (1,3, 1) = (0, 4, 5). When these equations are rewritten as systems of equations, the two coefficient matrices are identical. Combining the two augmented matrices into a single matrix, we have the doubly augmented matrix The reduced row-echelon form for this matrix is From the bottom row and the last column we see that it s impossible to find the desired c 1, c 2, so the vector in part (c) is not a LC of u and v. However, the vector in (a) is a LC of u and v. From the first three columns we see that values for a 1 and a 2 = 2 should be be 2. In other words (2,2,2) = 2u + 2v. This result is easy to check. Section Problem 8ac. Express the following as linear combinations of u = (2,1, 4), v = (1, 1,3), and w = (3, 2,5). (a) ( 9, 7, 15) (c) (0,0, 0) Solution. Part (c) can be solved by inspection: one solution is 0 = 0u + 0v + 0w. Page 1 of 7 A. Sontag April 6, 2002
2 For part (a) we need to find scalars k 1, k 2, k 3 for which k 1 (2, 1, 4) + k 2 (1, 1, 3) + k 3 (3,2, 5) = ( 9, 7, 15). Rewrite this as a system of equations with augmented matrix The reduced row-echelon form is This gives k 1 = 2, k 2 = 1, and k 3 = 2 and ( 9, 7, 15) = 2u + v 2w. Note that this result is easy to check. Section Problem 9ac. Express the following as linear combinations of p 1, p 2, and p 3,given that p 1 (x) = 2 + x + 4x 2, p 2 (x) = 1 x + 3x 2, and p 3 (x) = 3 + 2x + 5x 2. (a) q, with q(x) = 9 7x 15x 2 (c) 0 (in other words the constant function 0) Solution. Again, part (c) is easily solved by inspection: 0 = 0p 1 + 0p 2 + 0p 3. For part (a), we need to find scalars k 1, k 2, k 3 such that, for every real number x, k 1 (2 + x + 4x 2 ) + k 2 (1 x + 3x 2 ) + k 3 (3 + 2x + 5x 2 ) = 9 7x 15x 2 Combining terms and equating the coefficients of like powers of x we find k k 2 = k 3 15 Notice that the augmented matrix is the same as that for 8(c). This gives us the single solution k 1 = 2, k 2 = 1, and k 3 = 2. In other words, q = 2p 1 + p 2 2p 3. Again, it s easy to check this result. Page 2 of 7 A. Sontag April 6, 2002
3 Section Problem 10ab. Which of the following are linear combinations of A =, B =, C = (a) M = 1 8 (b) Solution. A, B, C. Part (b) is easy: 0 = 0A + 0B + 0C, so (a) is a linear combination of the specficied For part (a), the question asks whether there exist scalars k 1, k 2, k 3 for which k 1 A + k 2 B + k 3 C = M. By equating individual entries, we see that this last equation is equivalent to the system of equations whose augmented matrix is The reduced row-echelon form is Thus M = A+2B 3C, a linear combination of the given three matrices. Again, it s easy to check this result. Section Problem 12. Let f and g be defined by f(x) = cos 2 x and g(x) = sin 2 (x). Which of the following lie in the space spanned by f and g? (a) cos 2x (b) 3 + x 2 (c) 1 (d) sin x (e) 0 Solution. Part (e) is easiest: 0 = 0f + 0g. Part (c) uses the familiar identity sin 2 x + cos 2 x = 1 for every x. Thus 1 = f + g. Similarly part (a) can be solved by using the double-angle formula for the cosine function: cos 2x = cos 2 x sin 2 x for every x, so cos 2x = f g. Page 3 of 7 A. Sontag April 6, 2002
4 The functions in (b) and (d) look like they won t work. For (b), here s one argument we could use. The linear combination af + bg will give a function whose values all lie between a b and a + b, since a cos 2 x + bsin 2 x a + b according to the triangle inequality. The function 3 + x 2, on the other hand will have values that are not bounded above: 3 + x 2 as x ±. So it s impossible to have 3 + x 2 = acos 2 x + bsin 2 x for every x. ( Other arguments are possible. For instance, you could use specific values of x to reach a contradiction.) For part (d) one could argue as follows. Suppose we did have sin x = a cos 2 x + bsin 2 x for every x. Using the value x = 0 we find a = 0 and the assumption reduces to sin x = bsin 2 x for every x. But then the left-hand side will have some values positive and some negative, while the right hand side will either be always 0 or always 0, according to whether b 0 or b 0. That s impossible. (Other arguments are possible.) Final conclusion: the functions in (a), (c), (e) lie in the specified space. Those in (b), (d) do not. Section Problem 15. u = ( 1, 1,1) and v = (3, 4,4). Find an equation for the plane spanned by the vectors Solution. The specified plane is the vector space consisting of all points of the form s( 1,1, 1)+ t(3,4, 4), where s and t are real scalars. This gives a parametric description of the plane, but we want an equation of the form ax + by + cz = 0. Reclaiming our knowledge of Math 205, we recast the problem: find the equation for the plane through the points (0,0, 0),( 1, 1,1), (3,4, 4). Consult the Math 205 text or consult Section 3.5 in our text for ways to find the required equation. One way is as follows. Compute the cross product ( 1, 1,1) (3, 4,4), which works out to be 7j 7k. The vector j k = (0,1, 1) can therefore be used as a normal vector n for the desired plane and (0, 0,0) is a convenient point known to lie on the plane. The equation of the plane is therefore n (x 0, y 0, z 0) = 0 or (0,1, 1) (x, y, z) = 0 or y = z. Section Problem 16. Find parametric equations for the line spanned by the vector u = (3, 2,5). Solution. The line in question is the vector space consisting of all points of the form t(3, 2,5) for t a real scalar. So a set of parametric equations would be x = 3t, y = 2t, z = 5t, < t <. Other correct answers are possible. Page 4 of 7 A. Sontag April 6, 2002
5 Section Problem 19. Use Theorem to show that v 1 = (1, 6,4), v 2 = (2,4, 1), v 3 = ( 1,2, 5) and w 1 = (1, 2, 5), w 2 = (0, 8,9) span the same supspace of R 3. Solution. According to the cited theorem the given two sets of vectors will span the same subspace iff each of the vectors v 1, v 2, v 3 is a linear combination of w 1 and w 2 and each of the vectors w 1, w 2 is a linear combination of v 1, v 2, v 3. By inspection (or if need be, by solving a system of equations), we can see that v 1 = w 1 + w 2 v 2 = 2w 1 + w 2 v 3 = w 1 Also by inspection (or by solving the relevant system of equations) we see that w 1 = v 3 w 2 = v 1 + v 3 According to Theorem 5.2.4, span({v 1, v 2, v 3 }) = span({w 1, w 2 }). In other words every vector that is a linear combination of v 1, v 2, v 3 is also a linear combination of w 1, w 2, and vice versa. Section Problem 21c. Show that the set of all everywhere differentiable functions that satisfy f + 2f = 0 forms a subspace of F(, ). Solution. Let W denote the specified set of functions. In other words, a function f in F(, ) belongs to W iff it is differentiable everywhere and satisfies the condition f + 2f = 0. Step 1. The constant function 0 is differentiable everywhere and its derivative is the constant function 0. So when twice this function is added to the derivative of this function, the net result is still the constant function 0. In other words the constant function does belong to the set W. Step 2. Suppose that f and g both belong to W. Then we know that f and g are each differentiable everywhere. A calculus theorem tells us that if two functions are each differentiable at a point, then their sum is differentiable at that point and the sum of the derivatives is the derivative of the sum. Therefore f + g must be differentiable everywhere and satisfy (f + g) = f + g. From the fact that f and g belong to W we also know that f + 2f = 0 and g + 2g = 0. Therefore Page 5 of 7 A. Sontag April 6, 2002
6 (f + g) + 2(f + g) = f + g + 2f + 2g = f + 2f + g + 2g = = 0. We shown that f + g belongs to W. Since f and g were given arbitrarily, we ve shown that W is closed under addition. Step 3. Suppose that f belongs to W and k is a scalar. Since f belongs to W, f is differentiable everywhere. A theorem from calculus tells us that kf is differentiable everywhere and satisfies (kf) = kf. We also know that f + 2f = 0. It follows that (kf) + 2(kf) = kf + 2kf = k(f + 2f) = k0 = 0. We ve shown that kf belongs to W. Since k and f were given arbitrarily, we ve shown that W is closed under scalar multiplication. Steps 1,2, and 3 show that W is a subspace of F(, ). Section Problem 23ce. counterexample. True or False? Justify answer with an argument or a (c) If S is a finite set of vectors in a vector space V, then span(s) must be closed under addition and scalar multiplication. This is true, according to Theorem and the definition of span(s), which appears on page 217. The theorem tells us that span(s) is a subspace of V, and subspaces are always closed under addition and scalar multiplication. (e) If span(s 1 ) = span(s 2 ), then S 1 = S 2. This is definitely false. The same vector space can have lots of different spanning sets. A very simple example to show this is the following: V = R 2, S 1 = {(1,0), (0,1)}, S 2 = {(1, 0),(0, 2)}. In this example, span(s 1 ) = span(s 2 ) = V yet S 1 S 2. Lots of other examples, many of them more interesting than this one, are possible. Section Problem 24. (a) Under what conditions will two vectors in R 3 span a plane? A line? Two vectors in R 3 will span a plane iff neither is a linear combination of the other. (For then neither can be discarded without diminishing the span.) They span a line iff at least one of the two is nonzero and at least one is a linear combination of the other. (For then one can be discarded without diminishing the span. If one vector is the zero vector, discard it. Otherwise, each will be a linear combination of the other and one can discard either one. The remaining vector will have the same span as the original two vectors had. The span of a single nonzero vector is a line.) Page 6 of 7 A. Sontag April 6, 2002
7 (b) Under what conditions will it be true that span{u} = span{v}? Either both vectors must be the zero vector, in which case the span is just {0}, or one must be a nonzero scalar multiple of the other, which will automatically make each a nonzero scalar multiple of the other. With u = kv and k 0, each vector au could be rewritten as akv while each bv could be rewritten as b ku. A compressed version of what s needed is that each of the vectors must be a scalar multiple of the other, the answer that s given in the text. (c) If Ax = b is a consistent system of m equations in n unknowns, under what conditions will it be true that the solution set is a subspace of R n? Explain. Given that Ax = b is consistent, we know that the solution set is a nonempty subset of R n. If b = 0, the system is homogeneous and Theorem tells us that the solution set is a subspace of r n. Ifb 0 it is pretty easy to check that the solution set can t be closed under either addition or scalar multiplication. So the solution set is a subspace of R n iff b = 0. Note that in this case the given system is automatically consistent. Alternate reasoning, with the same result, can be found in the text. Section Problem 26. (a) Let M 22 be the vector space of 2 2 matrices. Find four matrices that span M 22. Since a b = a c d + b c + d 0 1 we can use the four matrices, 0 1,,. 0 1 Many other correct answers are possible. (b) In words, describe a set of matrices that spans M nn. The set consisting of all n n matrices, each of which has exactly one of its entries equal to 1 and all of the remaining entries equal to 0. Page 7 of 7 A. Sontag April 6, 2002
Vector Spaces ปร ภ ม เวกเตอร
Vector Spaces ปร ภ ม เวกเตอร 5.1 Real Vector Spaces ปร ภ ม เวกเตอร ของจ านวนจร ง Vector Space Axioms (1/2) Let V be an arbitrary nonempty set of objects on which two operations are defined, addition and
More informationEXERCISE SET 5.1. = (kx + kx + k, ky + ky + k ) = (kx + kx + 1, ky + ky + 1) = ((k + )x + 1, (k + )y + 1)
EXERCISE SET 5. 6. The pair (, 2) is in the set but the pair ( )(, 2) = (, 2) is not because the first component is negative; hence Axiom 6 fails. Axiom 5 also fails. 8. Axioms, 2, 3, 6, 9, and are easily
More informationMath 314H EXAM I. 1. (28 points) The row reduced echelon form of the augmented matrix for the system. is the matrix
Math 34H EXAM I Do all of the problems below. Point values for each of the problems are adjacent to the problem number. Calculators may be used to check your answer but not to arrive at your answer. That
More informationChapter 3. Directions: For questions 1-11 mark each statement True or False. Justify each answer.
Chapter 3 Directions: For questions 1-11 mark each statement True or False. Justify each answer. 1. (True False) Asking whether the linear system corresponding to an augmented matrix [ a 1 a 2 a 3 b ]
More informationOHSX XM511 Linear Algebra: Multiple Choice Exercises for Chapter 2
OHSX XM5 Linear Algebra: Multiple Choice Exercises for Chapter. In the following, a set is given together with operations of addition and scalar multiplication. Which is not a vector space under the given
More informationChapter 2: Linear Independence and Bases
MATH20300: Linear Algebra 2 (2016 Chapter 2: Linear Independence and Bases 1 Linear Combinations and Spans Example 11 Consider the vector v (1, 1 R 2 What is the smallest subspace of (the real vector space
More informationMath 102, Winter 2009, Homework 7
Math 2, Winter 29, Homework 7 () Find the standard matrix of the linear transformation T : R 3 R 3 obtained by reflection through the plane x + z = followed by a rotation about the positive x-axes by 6
More informationNAME MATH 304 Examination 2 Page 1
NAME MATH 4 Examination 2 Page. [8 points (a) Find the following determinant. However, use only properties of determinants, without calculating directly (that is without expanding along a column or row
More informationMATH10212 Linear Algebra B Homework Week 4
MATH22 Linear Algebra B Homework Week 4 Students are strongly advised to acquire a copy of the Textbook: D. C. Lay Linear Algebra and its Applications. Pearson, 26. ISBN -52-2873-4. Normally, homework
More informationVector Spaces ปร ภ ม เวกเตอร
Vector Spaces ปร ภ ม เวกเตอร 1 5.1 Real Vector Spaces ปร ภ ม เวกเตอร ของจ านวนจร ง Vector Space Axioms (1/2) Let V be an arbitrary nonempty set of objects on which two operations are defined, addition
More informationMath 369 Exam #2 Practice Problem Solutions
Math 369 Exam #2 Practice Problem Solutions 2 5. Is { 2, 3, 8 } a basis for R 3? Answer: No, it is not. To show that it is not a basis, it suffices to show that this is not a linearly independent set.
More informationMAT 242 CHAPTER 4: SUBSPACES OF R n
MAT 242 CHAPTER 4: SUBSPACES OF R n JOHN QUIGG 1. Subspaces Recall that R n is the set of n 1 matrices, also called vectors, and satisfies the following properties: x + y = y + x x + (y + z) = (x + y)
More informationMath 3191 Applied Linear Algebra
Math 9 Applied Linear Algebra Lecture : Null and Column Spaces Stephen Billups University of Colorado at Denver Math 9Applied Linear Algebra p./8 Announcements Study Guide posted HWK posted Math 9Applied
More informationMATH 1553, C.J. JANKOWSKI MIDTERM 1
MATH 155, C.J. JANKOWSKI MIDTERM 1 Name Section Please read all instructions carefully before beginning. You have 5 minutes to complete this exam. There are no aids of any kind (calculators, notes, text,
More informationLinear Algebra, Spring 2005
Linear Algebra, Spring 2005 Solutions May 4, 2005 Problem 4.89 To check for linear independence write the vectors as rows of a matrix. Reduce the matrix to echelon form and determine the number of non-zero
More informationLecture 16: 9.2 Geometry of Linear Operators
Lecture 16: 9.2 Geometry of Linear Operators Wei-Ta Chu 2008/11/19 Theorem 9.2.1 If T: R 2 R 2 is multiplication by an invertible matrix A, then the geometric effect of T is the same as an appropriate
More informationGENERAL VECTOR SPACES AND SUBSPACES [4.1]
GENERAL VECTOR SPACES AND SUBSPACES [4.1] General vector spaces So far we have seen special spaces of vectors of n dimensions denoted by R n. It is possible to define more general vector spaces A vector
More informationProblem set #4. Due February 19, x 1 x 2 + x 3 + x 4 x 5 = 0 x 1 + x 3 + 2x 4 = 1 x 1 x 2 x 4 x 5 = 1.
Problem set #4 Due February 19, 218 The letter V always denotes a vector space. Exercise 1. Find all solutions to 2x 1 x 2 + x 3 + x 4 x 5 = x 1 + x 3 + 2x 4 = 1 x 1 x 2 x 4 x 5 = 1. Solution. First we
More informationSolutions of Linear system, vector and matrix equation
Goals: Solutions of Linear system, vector and matrix equation Solutions of linear system. Vectors, vector equation. Matrix equation. Math 112, Week 2 Suggested Textbook Readings: Sections 1.3, 1.4, 1.5
More informationMidterm 1 Review. Written by Victoria Kala SH 6432u Office Hours: R 12:30 1:30 pm Last updated 10/10/2015
Midterm 1 Review Written by Victoria Kala vtkala@math.ucsb.edu SH 6432u Office Hours: R 12:30 1:30 pm Last updated 10/10/2015 Summary This Midterm Review contains notes on sections 1.1 1.5 and 1.7 in your
More informationSolutions to Math 51 First Exam April 21, 2011
Solutions to Math 5 First Exam April,. ( points) (a) Give the precise definition of a (linear) subspace V of R n. (4 points) A linear subspace V of R n is a subset V R n which satisfies V. If x, y V then
More information1 Last time: inverses
MATH Linear algebra (Fall 8) Lecture 8 Last time: inverses The following all mean the same thing for a function f : X Y : f is invertible f is one-to-one and onto 3 For each b Y there is exactly one a
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 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 informationMatrix equation Ax = b
Fall 2017 Matrix equation Ax = b Authors: Alexander Knop Institute: UC San Diego Previously On Math 18 DEFINITION If v 1,..., v l R n, then a set of all linear combinations of them is called Span {v 1,...,
More informationWeek #4: Midterm 1 Review
Week #4: Midterm Review April 5, NAMES: TARDIS : http://math.ucsb.edu/ kgracekennedy/spring 4A.html Week : Introduction to Systems of Linear Equations Problem.. What row operations are allowed and why?...
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 informationMath 3013 Problem Set 4
(e) W = {x, 3x, 4x 3, 5x 4 x i R} in R 4 Math 33 Problem Set 4 Problems from.6 (pgs. 99- of text):,3,5,7,9,,7,9,,35,37,38. (Problems,3,4,7,9 in text). Determine whether the indicated subset is a subspace
More informationLecture 17: Section 4.2
Lecture 17: Section 4.2 Shuanglin Shao November 4, 2013 Subspaces We will discuss subspaces of vector spaces. Subspaces Definition. A subset W is a vector space V is called a subspace of V if W is itself
More information(b) The nonzero rows of R form a basis of the row space. Thus, a basis is [ ], [ ], [ ]
Exam will be on Monday, October 6, 27. The syllabus for Exam 2 consists of Sections Two.III., Two.III.2, Two.III.3, Three.I, and Three.II. You should know the main definitions, results and computational
More informationWorksheet for Lecture 15 (due October 23) Section 4.3 Linearly Independent Sets; Bases
Worksheet for Lecture 5 (due October 23) Name: Section 4.3 Linearly Independent Sets; Bases Definition An indexed set {v,..., v n } in a vector space V is linearly dependent if there is a linear relation
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 information10. Rank-nullity Definition Let A M m,n (F ). The row space of A is the span of the rows. The column space of A is the span of the columns.
10. Rank-nullity Definition 10.1. Let A M m,n (F ). The row space of A is the span of the rows. The column space of A is the span of the columns. The nullity ν(a) of A is the dimension of the kernel. The
More informationSpan & Linear Independence (Pop Quiz)
Span & Linear Independence (Pop Quiz). Consider the following vectors: v = 2, v 2 = 4 5, v 3 = 3 2, v 4 = Is the set of vectors S = {v, v 2, v 3, v 4 } linearly independent? Solution: Notice that the number
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 informationMATH 1553, SPRING 2018 SAMPLE MIDTERM 1: THROUGH SECTION 1.5
MATH 553, SPRING 28 SAMPLE MIDTERM : THROUGH SECTION 5 Name Section Please read all instructions carefully before beginning You have 5 minutes to complete this exam There are no aids of any kind (calculators,
More informationLecture 03. Math 22 Summer 2017 Section 2 June 26, 2017
Lecture 03 Math 22 Summer 2017 Section 2 June 26, 2017 Just for today (10 minutes) Review row reduction algorithm (40 minutes) 1.3 (15 minutes) Classwork Review row reduction algorithm Review row reduction
More informationDr. Abdulla Eid. Section 4.2 Subspaces. Dr. Abdulla Eid. MATHS 211: Linear Algebra. College of Science
Section 4.2 Subspaces College of Science MATHS 211: Linear Algebra (University of Bahrain) Subspaces 1 / 42 Goal: 1 Define subspaces. 2 Subspace test. 3 Linear Combination of elements. 4 Subspace generated
More informationMath 290, Midterm II-key
Math 290, Midterm II-key Name (Print): (first) Signature: (last) The following rules apply: There are a total of 20 points on this 50 minutes exam. This contains 7 pages (including this cover page) and
More informationLinear Algebra. Preliminary Lecture Notes
Linear Algebra Preliminary Lecture Notes Adolfo J. Rumbos c Draft date April 29, 23 2 Contents Motivation for the course 5 2 Euclidean n dimensional Space 7 2. Definition of n Dimensional Euclidean Space...........
More informationLinear Algebra. Preliminary Lecture Notes
Linear Algebra Preliminary Lecture Notes Adolfo J. Rumbos c Draft date May 9, 29 2 Contents 1 Motivation for the course 5 2 Euclidean n dimensional Space 7 2.1 Definition of n Dimensional Euclidean Space...........
More informationMath 4377/6308 Advanced Linear Algebra
1.4 Linear Combinations Math 4377/6308 Advanced Linear Algebra 1.4 Linear Combinations & Systems of Linear Equations Jiwen He Department of Mathematics, University of Houston jiwenhe@math.uh.edu math.uh.edu/
More informationAdvanced Linear Algebra Math 4377 / 6308 (Spring 2015) March 5, 2015
Midterm 1 Advanced Linear Algebra Math 4377 / 638 (Spring 215) March 5, 215 2 points 1. Mark each statement True or False. Justify each answer. (If true, cite appropriate facts or theorems. If false, explain
More informationMATH 2360 REVIEW PROBLEMS
MATH 2360 REVIEW PROBLEMS Problem 1: In (a) (d) below, either compute the matrix product or indicate why it does not exist: ( )( ) 1 2 2 1 (a) 0 1 1 2 ( ) 0 1 2 (b) 0 3 1 4 3 4 5 2 5 (c) 0 3 ) 1 4 ( 1
More informationMath 308 Spring Midterm Answers May 6, 2013
Math 38 Spring Midterm Answers May 6, 23 Instructions. Part A consists of questions that require a short answer. There is no partial credit and no need to show your work. In Part A you get 2 points per
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 informationChapter 6. Orthogonality and Least Squares
Chapter 6 Orthogonality and Least Squares Section 6.1 Inner Product, Length, and Orthogonality Orientation Recall: This course is about learning to: Solve the matrix equation Ax = b Solve the matrix equation
More informationwhich are not all zero. The proof in the case where some vector other than combination of the other vectors in S is similar.
It follows that S is linearly dependent since the equation is satisfied by which are not all zero. The proof in the case where some vector other than combination of the other vectors in S is similar. is
More informationSystem of Linear Equations
Math 20F Linear Algebra Lecture 2 1 System of Linear Equations Slide 1 Definition 1 Fix a set of numbers a ij, b i, where i = 1,, m and j = 1,, n A system of m linear equations in n variables x j, is given
More informationMATH 323 Linear Algebra Lecture 12: Basis of a vector space (continued). Rank and nullity of a matrix.
MATH 323 Linear Algebra Lecture 12: Basis of a vector space (continued). Rank and nullity of a matrix. Basis Definition. Let V be a vector space. A linearly independent spanning set for V is called a basis.
More informationLinear Algebra Exam 1 Spring 2007
Linear Algebra Exam 1 Spring 2007 March 15, 2007 Name: SOLUTION KEY (Total 55 points, plus 5 more for Pledged Assignment.) Honor Code Statement: Directions: Complete all problems. Justify all answers/solutions.
More informationLinear Algebra Practice Problems
Math 7, Professor Ramras Linear Algebra Practice Problems () Consider the following system of linear equations in the variables x, y, and z, in which the constants a and b are real numbers. x y + z = a
More informationMath 235: Linear Algebra
Math 235: Linear Algebra Midterm Exam 1 October 15, 2013 NAME (please print legibly): Your University ID Number: Please circle your professor s name: Friedmann Tucker The presence of calculators, cell
More informationMA 242 LINEAR ALGEBRA C1, Solutions to First Midterm Exam
MA 242 LINEAR ALGEBRA C Solutions to First Midterm Exam Prof Nikola Popovic October 2 9:am - :am Problem ( points) Determine h and k such that the solution set of x + = k 4x + h = 8 (a) is empty (b) contains
More informationLinear Algebra MATH20F Midterm 1
University of California San Diego NAME TA: Linear Algebra Wednesday, October st, 9 :am - :5am No aids are allowed Be sure to write all row operations used Remember that you can often check your answers
More informationSolution to Set 7, Math 2568
Solution to Set 7, Math 568 S 5.: No. 18: Let Q be the set of all nonsingular matrices with the usual definition of addition and scalar multiplication. Show that Q is not a vector space. In particular,
More informationHomework 1 Due: Wednesday, August 27. x + y + z = 1. x y = 3 x + y + z = c 2 2x + cz = 4
Homework 1 Due: Wednesday, August 27 1. Find all values of c for which the linear system: (a) has no solutions. (b) has exactly one solution. (c) has infinitely many solutions. (d) is consistent. x + y
More information4.3 - Linear Combinations and Independence of Vectors
- Linear Combinations and Independence of Vectors De nitions, Theorems, and Examples De nition 1 A vector v in a vector space V is called a linear combination of the vectors u 1, u,,u k in V if v can be
More informationMATH 2331 Linear Algebra. Section 1.1 Systems of Linear Equations. Finding the solution to a set of two equations in two variables: Example 1: Solve:
MATH 2331 Linear Algebra Section 1.1 Systems of Linear Equations Finding the solution to a set of two equations in two variables: Example 1: Solve: x x = 3 1 2 2x + 4x = 12 1 2 Geometric meaning: Do these
More informationLinear Algebra and Robot Modeling
Linear Algebra and Robot Modeling Nathan Ratliff Abstract Linear algebra is fundamental to robot modeling, control, and optimization. This document reviews some of the basic kinematic equations and uses
More informationQuizzes for Math 304
Quizzes for Math 304 QUIZ. A system of linear equations has augmented matrix 2 4 4 A = 2 0 2 4 3 5 2 a) Write down this system of equations; b) Find the reduced row-echelon form of A; c) What are the pivot
More informationDefinition 1. A set V is a vector space over the scalar field F {R, C} iff. there are two operations defined on V, called vector addition
6 Vector Spaces with Inned Product Basis and Dimension Section Objective(s): Vector Spaces and Subspaces Linear (In)dependence Basis and Dimension Inner Product 6 Vector Spaces and Subspaces Definition
More informationEXAM. Exam #2. Math 2360 Summer II, 2000 Morning Class. Nov. 15, 2000 ANSWERS
EXAM Exam # Math 6 Summer II Morning Class Nov 5 ANSWERS i Problem Consider the matrix 6 pts A = 6 4 9 5 7 6 5 5 5 4 The RREF of A is the matrix R = A Find a basis for the nullspace of A Solve the homogeneous
More informationAlgorithms to Compute Bases and the Rank of a Matrix
Algorithms to Compute Bases and the Rank of a Matrix Subspaces associated to a matrix Suppose that A is an m n matrix The row space of A is the subspace of R n spanned by the rows of A The column space
More informationFinal Exam Solutions June 10, 2004
Math 0400: Analysis in R n II Spring 004 Section 55 P. Achar Final Exam Solutions June 10, 004 Total points: 00 There are three blank pages for scratch work at the end of the exam. Time it: hours 1. True
More informationDEF 1 Let V be a vector space and W be a nonempty subset of V. If W is a vector space w.r.t. the operations, in V, then W is called a subspace of V.
6.2 SUBSPACES DEF 1 Let V be a vector space and W be a nonempty subset of V. If W is a vector space w.r.t. the operations, in V, then W is called a subspace of V. HMHsueh 1 EX 1 (Ex. 1) Every vector space
More informationVector Spaces 4.4 Spanning and Independence
Vector Spaces 4.4 and Independence Summer 2017 Goals Discuss two important basic concepts: Define linear combination of vectors. Define Span(S) of a set S of vectors. Define linear Independence of a set
More informationBASIC NOTIONS. x + y = 1 3, 3x 5y + z = A + 3B,C + 2D, DC are not defined. A + C =
CHAPTER I BASIC NOTIONS (a) 8666 and 8833 (b) a =6,a =4 will work in the first case, but there are no possible such weightings to produce the second case, since Student and Student 3 have to end up with
More informationMATH 304 Linear Algebra Lecture 20: Review for Test 1.
MATH 304 Linear Algebra Lecture 20: Review for Test 1. Topics for Test 1 Part I: Elementary linear algebra (Leon 1.1 1.4, 2.1 2.2) Systems of linear equations: elementary operations, Gaussian elimination,
More informationChapter 1. Vectors, Matrices, and Linear Spaces
1.6 Homogeneous Systems, Subspaces and Bases 1 Chapter 1. Vectors, Matrices, and Linear Spaces 1.6. Homogeneous Systems, Subspaces and Bases Note. In this section we explore the structure of the solution
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 2030: ASSIGNMENT 4 SOLUTIONS
MATH 23: ASSIGNMENT 4 SOLUTIONS More on the LU factorization Q.: pg 96, q 24. Find the P t LU factorization of the matrix 2 A = 3 2 2 A.. By interchanging row and row 4 we get a matrix that may be easily
More informationMath 3C Lecture 25. John Douglas Moore
Math 3C Lecture 25 John Douglas Moore June 1, 2009 Let V be a vector space. A basis for V is a collection of vectors {v 1,..., v k } such that 1. V = Span{v 1,..., v k }, and 2. {v 1,..., v k } are linearly
More informationVector Geometry. Chapter 5
Chapter 5 Vector Geometry In this chapter we will look more closely at certain geometric aspects of vectors in R n. We will first develop an intuitive understanding of some basic concepts by looking at
More informationSection 6.1. Inner Product, Length, and Orthogonality
Section 6. Inner Product, Length, and Orthogonality Orientation Almost solve the equation Ax = b Problem: In the real world, data is imperfect. x v u But due to measurement error, the measured x is not
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 informationMATH 304 Linear Algebra Lecture 10: Linear independence. Wronskian.
MATH 304 Linear Algebra Lecture 10: Linear independence. Wronskian. Spanning set Let S be a subset of a vector space V. Definition. The span of the set S is the smallest subspace W V that contains S. If
More informationb for the linear system x 1 + x 2 + a 2 x 3 = a x 1 + x 3 = 3 x 1 + x 2 + 9x 3 = 3 ] 1 1 a 2 a
Practice Exercises for Exam Exam will be on Monday, September 8, 7. The syllabus for Exam consists of Sections One.I, One.III, Two.I, and Two.II. You should know the main definitions, results and computational
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 informationSection 3.3. Matrix Equations
Section 3.3 Matrix Equations Matrix Vector the first number is the number of rows the second number is the number of columns Let A be an m n matrix A = v v v n with columns v, v,..., v n Definition The
More informationSpan and Linear Independence
Span and Linear Independence It is common to confuse span and linear independence, because although they are different concepts, they are related. To see their relationship, let s revisit the previous
More informationSolutions to Midterm 2 Practice Problems Written by Victoria Kala Last updated 11/10/2015
Solutions to Midterm 2 Practice Problems Written by Victoria Kala vtkala@math.ucsb.edu Last updated //25 Answers This page contains answers only. Detailed solutions are on the following pages. 2 7. (a)
More informationMultiple Choice Questions
Multiple Choice Questions There is no penalty for guessing. Three points per question, so a total of 48 points for this section.. What is the complete relationship between homogeneous linear systems of
More informationMath 2030 Assignment 5 Solutions
Math 030 Assignment 5 Solutions Question 1: Which of the following sets of vectors are linearly independent? If the set is linear dependent, find a linear dependence relation for the vectors (a) {(1, 0,
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 informationLinear Algebra (Math-324) Lecture Notes
Linear Algebra (Math-324) Lecture Notes Dr. Ali Koam and Dr. Azeem Haider September 24, 2017 c 2017,, Jazan All Rights Reserved 1 Contents 1 Real Vector Spaces 6 2 Subspaces 11 3 Linear Combination and
More informationLinear Combination. v = a 1 v 1 + a 2 v a k v k
Linear Combination Definition 1 Given a set of vectors {v 1, v 2,..., v k } in a vector space V, any vector of the form v = a 1 v 1 + a 2 v 2 +... + a k v k for some scalars a 1, a 2,..., a k, is called
More informationSolution: By inspection, the standard matrix of T is: A = Where, Ae 1 = 3. , and Ae 3 = 4. , Ae 2 =
This is a typical assignment, but you may not be familiar with the material. You should also be aware that many schools only give two exams, but also collect homework which is usually worth a small part
More information17. C M 2 (C), the set of all 2 2 matrices with complex entries. 19. Is C 3 a real vector space? Explain.
250 CHAPTER 4 Vector Spaces 14. On R 2, define the operation of addition by (x 1,y 1 ) + (x 2,y 2 ) = (x 1 x 2,y 1 y 2 ). Do axioms A5 and A6 in the definition of a vector space hold? Justify your answer.
More informationMODEL ANSWERS TO THE FIRST QUIZ. 1. (18pts) (i) Give the definition of a m n matrix. A m n matrix with entries in a field F is a function
MODEL ANSWERS TO THE FIRST QUIZ 1. (18pts) (i) Give the definition of a m n matrix. A m n matrix with entries in a field F is a function A: I J F, where I is the set of integers between 1 and m and J is
More informationChapter 2. Vector Spaces
Chapter 2 Vector Spaces Vector spaces and their ancillary structures provide the common language of linear algebra, and, as such, are an essential prerequisite for understanding contemporary applied mathematics.
More informationMidterm #2 Solutions
Naneh Apkarian Math F Winter Midterm # Solutions Here is a solution key for the second midterm. The solutions presented here are more complete and thorough than your responses needed to be - in order to
More informationElementary Linear Algebra
Elementary Linear Algebra Anton & Rorres, 10 th Edition Lecture Set 05 Chapter 4: General Vector Spaces 1006003 คณ ตศาสตร ว ศวกรรม 3 สาขาว ชาว ศวกรรมคอมพ วเตอร ป การศ กษา 1/2554 1006003 คณตศาสตรวศวกรรม
More informationExam in TMA4110 Calculus 3, June 2013 Solution
Norwegian University of Science and Technology Department of Mathematical Sciences Page of 8 Exam in TMA4 Calculus 3, June 3 Solution Problem Let T : R 3 R 3 be a linear transformation such that T = 4,
More informationMath Final December 2006 C. Robinson
Math 285-1 Final December 2006 C. Robinson 2 5 8 5 1 2 0-1 0 1. (21 Points) The matrix A = 1 2 2 3 1 8 3 2 6 has the reduced echelon form U = 0 0 1 2 0 0 0 0 0 1. 2 6 1 0 0 0 0 0 a. Find a basis for the
More informationInner Product Spaces
Inner Product Spaces Introduction Recall in the lecture on vector spaces that geometric vectors (i.e. vectors in two and three-dimensional Cartesian space have the properties of addition, subtraction,
More informationMath 250B Midterm II Information Spring 2019 SOLUTIONS TO PRACTICE PROBLEMS
Math 50B Midterm II Information Spring 019 SOLUTIONS TO PRACTICE PROBLEMS Problem 1. Determine whether each set S below forms a subspace of the given vector space V. Show carefully that your answer is
More informationWorksheet for Lecture 15 (due October 23) Section 4.3 Linearly Independent Sets; Bases
Worksheet for Lecture 5 (due October 23) Name: Section 4.3 Linearly Independent Sets; Bases Definition An indexed set {v,..., v n } in a vector space V is linearly dependent if there is a linear relation
More informationThe definition of a vector space (V, +, )
The definition of a vector space (V, +, ) 1. For any u and v in V, u + v is also in V. 2. For any u and v in V, u + v = v + u. 3. For any u, v, w in V, u + ( v + w) = ( u + v) + w. 4. There is an element
More informationAnd, even if it is square, we may not be able to use EROs to get to the identity matrix. Consider
.2. Echelon Form and Reduced Row Echelon Form In this section, we address what we are trying to achieve by doing EROs. We are trying to turn any linear system into a simpler one. But what does simpler
More information