1. 1 Chapter 3 – Subspaces of R n and Their Dimension Outline 3.1 Image and Kernel of a Linear Transformation 3.2 Subspaces of R n ; Bases and Linear Independence 3.3 The Dimension of a Subspace of R n 3.4CoordinatesImage and Kernel of a Linear TransformationSubspaces of R n ; Bases and Linear IndependenceThe Dimension of a Subspace of R nCoordinates

2. 2 3.1 Image and Kernel of a Linear Transformation (Definition 3.1.1) The image of a function consists of all the values the function takes in its codomain. If f is a function from X to Y, then image(f)={f(x): x in X} ={b in Y: b=f(x), for some x in X}. Note that image(f) is a subset of the codomain Y of f.

3. 3 Image (Example 3) The image of the function f from R to R 2 given by is the unit circle. The function f is called a parametrization of the unit circle. More generally, a parametrization of a curve C in R 2 is a function g from R to R 2 whose image is C. (Example 4) If the function f from X to Y is invertible, then the image of f is Y, image(f)=Y. Indeed, for each b in Y, there is one (and only one) x in X such that b=f(x), namely, x=f -1 (b). b=f(f -1 (b)).

4. 4 Example 6 (Example 6) Describe the image of the linear transformation

5. 5 Example 7 (Example 7) Describe the image of the linear transformation

6. 6 Span (Definition 3.1.2) Consider the vectors in R n. The set of all linear combinations of the vector is called their span:

7. 7 (Fact 3.1.3) Image of a Linear Transformation (Fact 3.1.3) The image of a linear transformation is the span of the columns vectors of A. We denote the image of T by im(T) or im(A).

8. 8 (Fact 3.1.4) Some Properties of the Image (Fact 3.1.4) The image of a linear transformation T (from R m to R n ) has the following properties. –The zero vector in R n is in the image of T. –The image of T is closed under addition: If and are both in the image of T, then so is. –The image of T is closed under scalar multiplication: If is in the image of T and k is an arbitrary scalar, then is in the image of T as well.

9. 9 Example 8 (Example 8) Consider an matrix A. Show that im(A 2 ) is a subset of im(A), that is, each vector in im(A 2 ) is also in im(A).

10. 10 (Definition3.1.5) Kernel (Definition 3.1.5) The kernel of a linear transformation consists of all zero of transformation, that is, the solutions of the equation. In other words, the kernel of T is the solution set of the linear system. We denote the kernel of T by ker(T) or ker(A). For a linear transformation T from R m to R n, –im(T) is a subset of the codomain R n, and –ker(T) is a subset of the domain R m.

11. 11 Example 10 (Example 10) Find the kernel of the linear transformation

12. 12 Example 11 (Example 10) Find the kernel of the linear transformation from R 5 to R 4, where

13. 13 (Fact 3.1.6) Some Properties of the Kernel (Fact 3.1.6) Consider a linear transformation T from R m to R n. –The zero vector in R m is in the kernel of T. –The kernel is closed under addition. –The kernel is closed under scalar multiplication.

14. 14 Example 12 (Example 12) For an invertible n×n matrix A, find ker(A).

15. 15 Example 13 (Fact 13) For which n×m matrices A is ? Give your answer in terms of the rank of A.

16. 16 Fact 3.1.7 (Fact 3.1.7) When is ? –Consider an n×m matrix A. Then if (and only if) rank(A)=m. (This implies that m ≦ n, since m=rank(A) ≦ n.) –For a square matrix A, we have if (and only if) A is invertible.

17. 17 (Summary 3.1.8) Various Characterizations of Invertible Matrices (Summary 3.1.8) Let A be an matrix. The following statements are equivalent, that is, for a given A, they are either all true or all false. –A is invertible. –The linear system has a unique solution, for all in R n. –rref(A)=I n. –rank(A)=n. –im(A)=R n. –ker(A)={ }

18. 18 Equivalent

19. 19 3.2 Subspaces of R n ; Bases and Linear Independence (Definition 3.2.1) (Subspaces of R n ) A subset W of R n is called a (linear) subspace of R n if it has the following properties –W contains the zero vector in R n. –W is closed under addition: If and are both in W, then so is. –W is closed under scalar multiplication: If is in W and k is an arbitrary scalar, then is in W.

20. 20 (Fact 3.2.2) Image and Kernel Are Subspaces (Fact 3.2.2) If T is a linear transformation from R n to R m, then –ker(T) is a subspace of R n –im(T) is a subspace of R m

21. 21 Example 1 (Example 1) Is a subspace of R 2 ? Solution: Note that W consists of all vectors in the first quadrant, including the positive axes and the origin. The answer is no: W contains the zero vector, and it is closed under addition, but it is not closed under multiplication with negative scalars.

22. 22 Example 2 (Example 2) Show that the only subspaces of R 2 are R 2 itself, the, and any of the lines through the origin. Solution Suppose W is a subspace of R 2 that is neither the set nor a line through the origin. We have to show that W=R 2. Pick a nonzero vector in W. (We can find such a vector, since ) The subspace W contains the line L spanned by, but W does not equal L. Therefore, we can find a vector in W that is not on L. Using a parallelogram, we can express any vector in R 2 as a linear combination of and. Therefore, is contained in W. This shows that W=R 2, as claimed.

23. 23 Subspace A subspace of R n is usually presented either as the solution set of a homogeneous linear system (i.e., as a kernel) or as the span of some vectors (i.e., as an image).

24. 24 Example 3 (Example 3) Consider the plane V in R 3 given by the equation x 1 +2x 2 +3x 3 =0. –Find a matrix A such that V=ker(A). –Find a matrix B such that V=im(B).

25. 25 Example 4 (Example 4) Consider the matrix Find vectors in R 3 that span the image of A. What is the smallest number of vectors needed to span the image of A?

26. 26 (Definition 3.2.3) Redundant Vectors; Linear Independence; Basis (Definition 3.2.3) Consider vector in R n. –We say that a vector in the list is redundant if is a linear combination of the preceding vectors. –The vectors are called linearly independent if none of them is redundant. Otherwise, the vectors are called linearly dependent (meaning that at least one of them is redundant). –We say that the vectors form a basis of a subspace V of R n if they span V and are linearly independent. (Also, it is required that vectors be in V.)

27. 27 (Fact 3.2.4) Basis of the Image (Fact 3.2.4) To construct a basis of the image of a matrix A, list all the column vectors of A, and omit the redundant vectors from this list.

28. 28 Example 5 (Example 5) Are the following vectors in R 7 linearly independent?

29. 29 (Fact 3.2.5) Linear Independence and Zero Components (Fact 3.2.5) Consider vectors in R n. If is nonzero, and if each of the vectors (for i ≧ 2) has a nonzero entry in a component where all the preceding vectors have a 0, then the vectors are linearly independent.

30. 30 Example 6 (Example 6) Are the vectors linearly independent?

31. 31 (Definition 3.2.6) Linear Relations (Definition 3.2.6) Consider the vectors in R n. An equation of the form is called a (linear) relation among the vector. There is always the trivial relation, with c 1 =…=c m =0. Nontrivial relations (where at least one coefficient c i is nonzero) may or may not exist among the vectors.

32. 32 (Fact 3.2.7) Relations and Linearly Dependence (Fact 3.2.7) The vectors in R n are linearly dependent if (and only if) there are nontrivial relations among them.

33. 33 Example 7 (Example 7) Suppose the column vectors of an n×m matrix A are linearly independent. Find the kernel of matrix A.

34. 34 (Fact 3.2.8) Kernel and Relations The vectors in the kernel of an n×m matrix A correspond to the linear relations among the column vectors of A: The equation In particular, the column vectors of A are linearly independent if (and only if), or, equivalently, if rank(A)=m. This condition implies that m ≦ n. Thus, we can find at most n linearly independent vectors in R n.

35. 35 Example 8 (Example 8) Consider the matrix to illustrate the connection between redundant column vectors, relations among the column vectors, and the kernel.

36. 36 (Summary 3.2.9) Various Characterization of Linear Independence (Summary 3.2.9) For a list of vectors in R n, the following statements are equivalent: –Vectors are linearly independent. –None of the vectors is redundant, meaning that none of them is a linear combination of preceding vectors. –None of the vectors is a linear combination of the other vectors in the list. –There is only the trivial relation among the vectors, meaning that the equation has only the solution c 1 =…=c m =0.

37. 37 Example 9 (Example 9) If is a basis of a subspace V of R n, and if is a vector in V, how many solutions does the equation have?

38. 38 (Fact 3.2.10) Basis and Unique Representation (Fact 3.2.10) Consider the vectors in a subspace V of R n. The vectors form a basis of V if (and only if) every vector in V can be expressed uniquely as a linearly combination

39. 39 3.3 The Dimension of a Subspace of R n (Fact 3.3.1) Consider vectors and in a subspace V of R n. If the vectors are linearly independent, and the vectors span V, then.

40. 40 Number of Vectros in a Basis (Fact 3.3.2) All bases of a subspace V of R n consist of the same number of vectors.

41. 41 Dimension (Definition 3.3.3) Consider a subspace V of R n. The number of vectors in a basis of V is called the dimension of V, denoted by dim(V).

42. 42 Independent Vectors and Spanning Vectors in a Subspace of R n (Fact 3.3.4) Consider a subspace V of R n with dim(V)=m. –We can find at most m linearly independent vectors in V. –We need at least m vectors to span V. –If m vectors in V are linearly independent, then they form a basis of V. –If m vectors span V, then they form a basis of V.

43. 43 Example 1 (Example 1) Find bases of the image and kernel of the matrix

44. 44 Example 2 (Example 2) Find bases of the kernel and image of We are told that To keep track of the columns of A and B, it is sensible to denote the column vectors of A by, and those of B by.

45. 45 Constructing a Basis of the Image (Algorithm 3.3.5) To construct a basis of the image of A, pick those column vectors of A that correspond to the columns of rref(A) containing the leading 1’s.

46. 46 Dimension of the Image (Fact 3.3.6) For any matrix A, dim(im A)=rank(A).

47. 47 Rank-Nullity Theorem (Fact 3.3.7) For any n×m matrix A, the equation dim(ker A)+dim(im A)=m holds. The dimension of ker(A) is called the nullity of A, and in Fact 3.3.6 we ovserved that im(im A)=rank(A) Thus we can write the preceding euquation alternatively as (nullity of A)+(rank of A)=m. Some authors go so far as to call this the fundamental theorem of linear algebra.

48. 48 Example 3 (Example 3) Consider the orthogonal projection T onto a plane V in R 3. Here, the dimension of the domain is m=3, one dimension collapses (the kernel of T is the line V ⊥ orthogonal to V), and we are left with the two- dimensional im(T)=V. m-dim(ker T)=dim(image T): 3-1=2

49. 49 Bases of R n (Fact 3.3.8) The vectors in R n form a basis of R n if (and only if) the matrix is invertible.

50. 50 Example 4 (Example 4) For which values of the constant k do the following vectors form a basis of R 3 ?

51. 51 Various Characterizations of Invertible Matrices (Summary 3.3.9) Consider an n×n matrix A, the following statements are equivalent. –A is invertible. –The linear system has a unique solution, for all in R n. –rref(A)=I n. –rank(A)=n. –im(A)=R n. –ker(A)={ } –The column vectors of A form a basis of R n. –The column vectors of A span R n. –The column vectors of A are linearly independent.

52. 52 3.4 Coordinates (Example 1) Consider the vectors in R 3, and define the plane in R 3. Is the vector on the plane V? Visualize your answer.

53. 53 Coodinates in a Subspace of R n (Definition 3.4.1) Consider a basis of a subspace V of R n. By fact 3.2.10, any vector in V can be written uniquely as The scalars c 1, c 2,…,c m are called the B -coordinates of, and the vector is the B -coordinate vector of, denoted by. Thus Note that

54. 54 Linearity of Coordinates (Fact 3.4.2) If B is a basis of a subspace V of R n, then –

55. 55 Example 2 (Example 2) Consider the basis B of R 2 consisting of vector –

56. 56 Example 3 (Example 3)

57. 57 The Matrix of a Linear Transformation (Definition 3.4.3) Consider a linear transformation T from R n to R n and a basis B of R n. The n×n matrix B that transforms into is called the B -matrix of T: for all in R n. We can construct B column by column as follows: If, then

58. 58 Example 4 (Example 4) Consider two perpendicular unit vector and in R 3. Form the basis of R 3, where. (Take a look at Fact A.10 in the Appendix to review the basis properties of the cross product.) Note that is perpendicular to both and, and is a unit vector, since. –Draw a sketch to find –Find the B -matrix B of the linear transformation

59. 59 Example 5 (Example 5) As in Example 3, let T be the linear transformation from R 2 to R 2 that projects any vector orthogonally onto the line L spanned by the vector. In Example 3, we found that the matrix of B of T with respect to the basis is. What is the relationship between B and the standard matrix A of T (such that )? We introduced the standard matrix of a linear transformation back in Section 2.1; alternatively, we can think of A as the matrix of T with respect to the standard basis of R 2, in the sense of Definition 3.4.3. (Think about it!)

60. 60 Standard Matrix Versus B -Matrix (Fact 3.4.4) Consider a linear transformation T from R n to R n and a basis of R n. Let B be the B -matrix of T, and let A be standard matrix of T (such that for all in R n ). Then AS=SB, B=S -1 AS, and A=SBS -1, where

61. 61 Similar Matrices (Definition 3.4.5) Consider two n×n matrices A and B. We say that A is similar to B if there exists an invertible matrix S such that AS=SB, or B=S -1 AS.

62. 62 Example 6 (Example 6) Is matrix similar to ?

63. 63 Example 7 (Example 7) Show that if matrix A is similar to B, then its power A t is similar to B t, for all positive integers t. (That is, A 2 is similar to B 2, A 3 is similar to B 3, and so on.)

64. 64 Similarity is an Equivalence Relation (Fact 3.4.6) An n×n matrix A is similar to itself (Reflexivity). If A is similar to B, then B is similar to A (Symmetry). If A is similar to B and B is similar to C, then A is similar to C (Transitivity). Transitivity proof: There are invertible matrices P and Q such that B=P -1 AP and C=Q -1 BQ. Then, C=Q -1 BQ=Q -1 (P -1 AP)Q=(Q -1 P -1 )A(PQ)=(PQ) -1 A(PQ)=S -1 AS, where S=PQ. We have shown that A is similar to C.