Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall

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 by Otto Bretscher

2. Wednesday, Feb 19 Chapter 3.3 Page 130 Problems 22, 30, 52 Main Idea: We label the vectors using an index system. Key Words: Dimension, basis. Goal: Learn the basic results about linearly independent spanning sets.

3. Previous Assignment Page 118 Problem 16 Use paper and pencil to decide whether the given vectors are linearly independent. | 1 | | 4 | | 7 | | 0 | a| 2 |+b| 5 |+c| 8 | = | 0 | | 3 | | 6 | | 9 | | 0 |

4. a b c RHS | 1 4 7 0 | | 2 5 8 0 | | 3 6 9 0 | | 1 4 7 0 | | 0 -3 -6 0 | | 0 -6 -12 0 | | 1 4 7 0 | | 0 1 2 0 | | 0 -6 -12 0 | | 1 0 -1 0 | | 0 1 2 0 | | 0 0 0 0 |

5. | a | | 1 | | b | = |-2 | | c | | 1 | They are not linearly independent. They satisfy this dependence relation: | 1 | | 4 | | 7 | | 0 | | 2 | -2 | 5 |+ | 8 | = | 0 | | 3 | | 6 | | 9 | | 0 |

6. Page 118 Problem 28. | 1 1 1 | Find a basis of the image of the matrix | 1 2 5 |. | 1 3 7 |

7. | 1 1 1 a | | 1 2 5 b | | 1 3 7 c | | 1 1 1 a | | 0 1 4 -a+b | | 0 2 6 -a+c | | 1 0 -3 2a - b | | 0 1 4 -a + b | | 0 0 -2 a -2b+c | | 1 0 -3 2a - b | | 0 1 4 -a + b | | 0 0 1 -a/2 +b-c/2 |

8. | 1 0 0 a/2 +2b -3c/2 | | 0 1 0 a - 3b +2c | | 0 0 1 -a/2+ b -c/2 | This says that you can get any vector you want. |1| |1| |1| |a| (a/2+2b-3c/2)|1|+(a-3b+2c)|2|+(-a/2+b-c/2)|5|=|b| |1| |3| |7| |c|

9. | 1 | | 1 | | 1 | One basis for the image is | 1 | , | 2 | , | 5 | | 1 | | 3 | | 7 | | 1 | | 0 | | 0 | Another basis for the image is | 0 |, | 1 |, | 0 |. | 0 | | 0 | | 1 |

10. Page 118 Problem 34 Consider the 5x4 matrix A = [V1 V2 V3 V4]. | 1 | Suppose that | 2 | is in the kernel of A. | 3 | | 4 | Write V4 as a linear combination of V1 V2 V3. We are given that V1 + 2 V2 + 3 V3 + 4 V4 = 0.

11. Thus V4 = -1/4 V1 -2/4 V2 -3/4 V3.

12. There are many many bases for a particular vector space. There is nothing unique about bases except for one thing. Two bases of the same space will always have the same number of vectors. We call this number the dimension of the space.

13. So our first task is to show that it is really true that two bases of the same vector space always have the same number of elements.

14. We actually show something stronger. An independent set is always smaller than a spanning set. Theorem. Consider vectors V1, V2, ..., Vi and W1, W2, ..., Ws in a subspace of Rn. If the V's are linearly independent and the W are a spanning set, then i <= s.

15. Proof: [V1 V2 ... Vi ] = [W1 W2 ... Ws] [X1 X2 ... Xi] nxi nxs sxi Independent Spanning set Coefficients If i > s the matrix [x1 x2 ... xi ] has this shape. .

16. Then there is a vector X =/= 0 such that [X1 X2 ... Xi] X = 0. But then : [V1 V2 ... Vi] X = 0 as well and this contradicts the fact that the Vi's are linearly independent. Thus we have to agree that i <= s. That means, the number of vectors in any spanning set must be at least as large as the number of vectors in any linearly independent set.

17. Theorem: All Bases of a subspace V of Rn have the same number of vectors. Proof: Given bases V1, V2, ..., Vp and W1, W2, ..., Wq, then p <= q and q <= p. Thus p = q.

18. Definition: The number of elements in any basis of a vector subspace V is called the dimension of V. Theorem: Consider a subspace V of Rn with dim(V) = m. (a) We can find at most m linearly independent vectors in V. (b) We need at least m vectors to span V. (c) If m vectors in V are linearly independent, then they form a basis of V. (d) If m vectors span V, then they form a basis of V.

19. Find a basis of the kernel of the matrix. A = | 1 2 0 3 0 | | 2 4 1 9 5 | a b c x y z w u | 1 2 0 3 0 | | 0 0 1 3 5 | | x | |-2 | | -3 | | 0 | | y | | 1 | | 0 | | 0 | | z | = a | 0 | + b | -3 | + c |-5 | | w | | 0 | | 1 | | 0 | | u | | 0 | | 0 | | 1 | <---These are a basis-- > of the Null Space

20. RCF(A) = | 1 2 0 3 0 | | 0 0 1 3 5 | x x Use these columns of A for a basis of the Range of A. The Range of A is also the Column Space of A. A basis of Range (A) is | 1 | | 0 | | 2 |, | 1 |

21. Notice that the elements of the null space provide the dependence relations for the columns which do not contain stair step ones.

22. When you do Row Canonical form, you are finding a basis of the row space. When you multiply on the left by an invertible matrix P, the row space is preserved. Proof: If P is any matrix, then the rows of PA are linear combinations of the rows of A. Thus RS(PA) c RS(A).

23. When P is invertible, then we get the reverse inclusion since RS(A) = RS(P-1 P A) c RS(PA) c RS(A). Thus RS(PA) = RS(A) when P is invertible.

24. In particular, RS(A) = RS(RCF(A)) and the non zero rows of RCF(A) are linearly independent due to the stair step ones. Thus the dimension of the RS(A) is the number of non zero rows in the RCF(A). The Dimension of the Row Space of A is called the row rank of A. The Dimension of the column space is called the column rank of A.

25. Theorem: The Dimension of the Row Space of a Matrix is always the same as the Dimension of the column space. Row Rank = Column Rank

26. Proof that Row Rank = Column Rank. Given a matrix A, choose columns C1 C2 ... Cr to be a basis of the column space. | x11 ... x1n | | | A = [ C1 C2 ... Cr ] | | | | | xr1 ... xr n | rxn

27. So A has a spanning set of the rows with r vectors. Thus Dimension of the Column Space is >= the Dimension of the Row Space. We can reverse the argument and prove the reverse inequality. Thus the dimension of the Row Space and Column Space are equal.