1. Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications, Second Edition Otto Bretscher

2. Monday, Feb 17 Chapter 3.2 Page 118 Problems 16,28,34 Main Idea: The Column Space is the Forest. The columns are the trees. Try to ignore the trees and look at the forest. Key Words: Kernel, Null Space, Image, Linear Combination, Linearly Independence Goal: See how matrix actions can be described in terms of Linear Combinations.

3. Previous Assignment Page 106 Problem 24.Describe the images and kernels of the transformations. Orthogonal projection onto the plane x+2y+3z = 0 in R 3. It should be obvious that the range is the plane and the kernel is the normal to the plane through the origin.

4. Let us try to validate that with the numbers. | 1 | The projection is in the direction | 2 |. | 3 | | x | | x | | 1 | The point | y | ---> | y | + k | 2 | | z | | z | | 3 | with the k chosen so that the resulting image is on the plane.

5. Thus: (x+k) + 2 (y+2k) + 3(z+3k) = 0 (x+2y+3z)+k(14) = 0 k = - (1/14) (x+2y+3z) The mapping is described as: | x | | x | | 1| |13 x - 2 y - 3 z| | y | --> | y |-(x+2y+3z)/14 | 2 | =(1/14)| -2 x + 10 y - 6 z| | z | | z | | 3 | | -3 x - 6 y + 5 z|

6. The matrix for the projection is | 13 -2 -3 | 1/14 | -2 10 -6 | | -3 -6 5 | The range and kernel are given by solving AX = B for a general vector B.

7. | 13/14 -2/14 -3/14 a | | -2/14 10/14 -6/14 b | | -3/14 -6/14 5/14 c | | -2 10 -6 14b | | 13 -2 -3 14a | | -3 -6 5 14c | | 1 -5 3 -7b | | 13 -2 -3 14a | | -3 -6 5 14c | | 1 -5 3 -7 b | | 0 63 -42 14 a +91 b | | 0 -21 14 -21 b +14 c |

8. | 1 -5 3 -7 b | | 0 63 -42 14a +91 b | | 0 -21 14 -21 b + 14c | | 1 -5 3 - 7b | | 0 9 - 6 2a + 13 b | | 0 -3 2 -3 b + 2c | | 1 -5 3 -7 b | | 0 -3 2 -3 b + 2c | | 0 9 - 6 2a + 13 b | | 1 -5 3 -7 b | | 0 -3 2 -3 b + 2c | | 0 0 0 2 a + 4 b + 6c | So a+2b+3c = 0

9. | 1 -5 3 -7 b | | 0 1 -2/3 + b – 2/3 c| | 0 0 0 2 a + 4 b + 6 c | | 1 0 -1/3 -2 b + 10/3 c | | 0 1 -2/3 + b -2/3 c | | 0 0 0 2a + 4 b + 6 c | The first two columns of A span the Range. | 13 | | -2 | = Range of A | -3 | | -6 |

10. | -2b-3c | One can more easily describe the range as | b | | c | | 0 | The kernel is what ever maps to | 0 | which is the | 0 | the null space of the matrix A. | 1/3 | | 1 | Kernel = which is the same as. | 1 | | 3 |

11. Page 106 Problem 26 What is the image of a function f from R to R given by f(t) = t 3 + a t 2 + b t + c where a,b,c are arbitrary scalars? The image is (-infinity,infinity). Since the function is of odd degree, it is negatively unbounded for negative x, and positively unbounded for positive x. By continuity it passes through every value.

12. Page 106 Problem 44 Consider a matrix A, and let B = rref(A). (a) Is ker(A) necessarily equal to ker(B) Explain. Yes, it has to be. The solutions to AX = 0 and BX = 0 have to be the same since row operations preserve the solutions to simultaneous linear equations.

13. (b) Is im(A) necessarily equal to im(B). Explain. No. Not at all, row operations do not preserve the row space. For example: RCF | 0 0 | = | 1 0 | | 1 0 | | 0 0 | Image = | 0 | Image = | c | | c | | 0 | Y-axis X-axis

14. Discussion: We are working with stuff that has a lot of applications and possibilities. Every application has its own special terminology. So when you talk to somebody, expect them to be using the same basic properties that we are now learning. But they will call them by different names. So now I will give the "traditional“ words.

15. A vector is a single column. That is, a nx1 matrix. A matrix has more than one column. We sometimes think of the matrix as consisting of columns. If V 1, V 2,..., V n are vectors and c 1, c 2,..., c n are numbers, then the vector V = c 1 V 1 + c 2 V 2 +... + c n V n is called a linear combination of V 1, V 2,..., V n with coefficients c 1, c 2,..., c n.

16. ****************************************************** AV really means a way of combining the columns of A ****************************************************

17. Notice that if A is the matrix with columns V 1, V 2,..., V n, | c 1 | | c 2 | A = [V 1 V 2... V n ] and if V is the vector |. |, |. | | c n | then AV is simply the linear combination c 1 V 1 + c 2 V 2 +... c n V n. Thus matrix multiplication by a matrix A is actually looking at linear combinations of the columns of A.

18. We can solve AX = B if and only if B is a linear combination of the columns of A. The set of all vectors which are linear combinations of the columns of A is CS(A) = "the column space of A". This brings us to discuss the concept of a “vector space”.

19. ************************************************** Keep your eye on the Subspace *************************************************

20. In linear algebra we have to describe things. Fortunately: There are good ways to describe the objects. Unfortunately: There are many good ways to describe objects. Since an object can have several distinct names, we are often faced with the question of deciding if two given names refer to the same object.

21. When ever we try to describe something in detail, we have to choose the way we present it. There is no good rule for making this choice. Everybody can have different looking presentations. The same correct answer can appear very different. What we have to do is try to focus on the item itself, not on how it was written. The item itself will be an object called a "vector space".

22. Definition of Vector Subspace of R n. A subset W of R n is called a subspace of R n if it has the following properties: (a) W contains the zero vector in R n (b) W is closed under addition. (c) W is closed under scalar multiplication That is: If W 1 and W 2 are in W and k is any scalar, then W 1 +W 2 and k W are also in W

23. Common subspaces are the kernel and the image. Theorem: If A rxs is any matrix, then (1) ker(A) = {X in R s | AX = 0} is a subspace. (2) im(A) = {AX | X is in R s } is a subspace.

24. Proof that the kernel of A is a subspace. Part (a): A0 = 0 so 0 is in ker(A). Let X, X 1, X 2 be in ker(A), and k be any scalar: Part (b): A(X 1 +X 2 ) = AX 1 +AX 2 = 0+0 = 0. Part (c): A(kX) = k(AX) = k0 = 0.

25. Since ker(A) has 0, is closed under addition, and closed under scalar multiplication, ker(A) is a subspace.

26. Proof of (2). Im(A) is a subspace. Part (a): A0 = 0 so 0 is in im(A). Part (b): If AX 1 and AX 2 are elements in im(A), then AX 1 +AX 2 = A(X 1 +X 2 ) is also in the image of A. Part (c) If AX is in im(A) and k is any scalar, then k(AX) = A(kX) is also in the image of A.

27. The span of a set of vectors is everything that they generate.

28. Definition of Span: Consider the vectors V 1,V 2,..., V n in R m. The set of all linear combinations of the vectors V 1, V 2,..., V n is called their span; = {c 1 V 1 +c 2 V 2 +...+c n V n | c i are scalars}

29. | 1 | | 0 | What is the span of | 1 | and | 0 | ? | 0 | | 1 |

30. | 1 | | 0 | | 0 | What is the span of | 0 | | 1 | | 0 | ? | 0 |, | 0 |, | 1 |

31. | 1 | | 1 | | 0 | What is the span of | 0 |, | 1 |, | 1 | ? | 1 | | 2 | | 1 |

32. | 1 1 0 | Equivalently, what is the image of | 0 1 1 | ? | 1 2 1 | | a | The span simply asks, For which | b | do there exist | c | x 1, x 2, x 3 such that | 1 | | 1 | | 0 | | a | x 1 | 0 | + x 2 | 1 | + x 3 | 1 | = | b | | 1 | | 2 | | 1 | | c |

33. Solve this system by reducing it to Row Canonical Form. | 1 1 0 a | | 1 1 0 a | | 1 0 -1 a-b | | 0 1 1 b | | 0 1 1 b | | 0 1 1 b | | 1 2 1 c | | 0 1 1 -a+c | | 0 0 0 -a-b+c| The system is solvable when there is no stair step one in the last column. This means it is solvable when a+b=c.

34. So the span of these three vectors | a | is all vectors of the form | b |. Check that the |a+b| original vectors were of this form.

35. To write this space in terms of a basis, separate on the coefficients. | a | | 1 | | 0 | | b | = a| 0 | + b | 1 | |a+b| | 1 | | 1 | | 1 | | 0 | The basis is: { | 0 |, | 1 | }. | 1 | | 1 |

36. Spanning sets without unnecessary elements are called bases.

37. Definition of Linear Independence; Consider a sequence V 1, V 2,..., V m of vectors in a subspace V of R n. The vectors V 1, V 2,..., V m are called linearly independent if none of them is a linear combination of the others. Otherwise, they are called linearly dependent.

38. Definition of Basis. We say that the vectors V 1, V 2,..., V m form a basis of V if they span V and are linearly independent. Definition of Linear Relation. Let V 1,V 2,..., V m in R n and c 1, c 2 … c m be scalars. An equation of the form c 1 V 1 + c 2 V 2 +... + c m V m = 0 is called a linear dependence relation among the vectors V 1, V 2,..., V m.

39. There is always the trivial relation, with c 1 = c 2 =... = c m = 0. Non trivial relations may or may not exist among the V 1, V 2,..., V m.

40. Find a linear dependence relation among the vectors | 1 | | 1 | | 5 | | 1 | | 1 | | 2 | | 5 | | 4 | | 8 | | 1 | | 9 | | 1 | | 9 | | 1 | | 1 | | 1 |, | 5 |, | 1 |, | 5 |, | 1 |

41. Solution: We solve for the coefficients : x 1, x 2, x 3, x 4, x 5 such that | 1 | | 1 | | 5 | | 1 | | 1 | | 0 | x 1 | 2 | + x 2 | 5 | + x 3 | 4 | + x 4 | 8 | + x 5 | 1 | = | 0 | | 9 | | 1 | | 9 | | 1 | | 1 | | 0 | | 1 | | 5 | | 1 | | 5 | | 1 | | 0 |

42. | 1 1 5 1 1 0 | | 1 1 5 1 1 0 | | 1 1 5 1 1 0 | | 2 5 4 8 1 0 | | 0 3 -6 6 -1 0 | | 0 3 -6 6 -1 0 | | 9 1 9 1 1 0 | | 0 -8 -36 -8 -8 0 | | 0 -8 -36 -8 -8 0 | | 1 5 1 5 1 0 | | 0 4 -4 4 0 0 | | 0 1 -1 1 0 0 | | 1 1 5 1 1 0 | | 1 0 6 0 1 0 | | 1 0 6 0 1 0 | | 0 1 -1 1 0 0 | | 0 1 -1 1 0 0 | | 0 1 -1 1 0 0 | | 0 3 -6 6 -1 0 | | 0 0 -3 3 -1 0 | | 0 0 1 -1 1/3 0 | | 0 -8 -36 -8 -8 0 | | 0 0 -44 0 -8 0 | | 0 0 -44 0 -8 0 | | 1 0 0 6 -1 0| | 1 0 0 6 -1 0 | | 1 0 0 6 -3/33 0 | | 0 1 0 0 1/3 0 | | 0 1 0 0 1/3 0 | | 0 1 0 0 1/3 0 | | 0 0 1 -1 1/3 0 | | 0 0 1 -1 1/3 0 | | 0 0 1 -1 1/3 0 | | 0 0 0 -44 20/3 0 | | 0 0 0 11 -5/3 0 | | 0 0 0 1 -5/33 0 | | 1 0 0 6 -1 0 | | 1 0 0 0 -3/33 0 | | 1 0 0 0 -3/33 0 | | 0 1 0 0 1/3 0 | | 0 1 0 0 1/3 0 | | 0 1 0 0 11/33 0 | | 0 0 1 0 6/33 0 | | 0 0 1 0 6/33 0 | | 0 0 1 0 6/33 0 | | 0 0 0 1 -5/33 0 | | 0 0 0 1 -5/33 0 | | 0 0 0 1 -5/33 0 |

43. | x1 | | 3 | | x2 | |-11 | | x3 | = a/33 | -6 | | x4 | | 5 | | x5 | | 33 |

44. | 1 | | 1 | | 5 | | 1 | | 1 | | 0 | 3| 2 |-11| 5 | - 6| 4 |+5 | 8 |+33| 1 | = | 0 | | 9 | | 1 | | 9 | | 1 | | 1 | | 0 | | 1 | | 5 | | 1 | | 5 | | 1 | | 0 | | 3 -11 - 30 + 5 + 33 | | 0 | | 6 -55 - 24 +40 + 33 | = | 0 | | 27 -11 - 54 + 5 + 33 | | 0 | | 3 -55 - 6 +25 + 33 | | 0 |