Vectors Picturing vectors Properties of vectors Linear combination Span. Vectors. Math 4A Scharlemann. 9 January /31

Transcription

1 1/31 Vectors Math 4A Scharlemann 9 January 2015

2 2/31 CELL PHONES OFF

3 3/31 An m-vector [column vector, vector in R m ] is an m 1 matrix: a 1 a 2 a = a 3. a m

4 3/31 An m-vector [column vector, vector in R m ] is an m 1 matrix: a 1 a 2 a = a 3. a m Can add two m-vectors in the obvious way, or multiply a vector by a real number: a 1 b 1 a 1 + b 1 a 2 b 2 a 1 + b 2 a 3 + b 3 = a 1 + b 3 ;... a m b m a 1 + b m

5 An m-vector [column vector, vector in R m ] is an m 1 matrix: a 1 a 2 a = a 3. a m Can add two m-vectors in the obvious way, or multiply a vector by a real number: a 1 b 1 a 1 + b 1 a 1 ca 1 a 2 b 2 a 1 + b 2 a 2 ca 2 a 3 + b 3 = a 1 + b 3 ; c a 3 = ca a m b m a 1 + b m a m ca m Do not multiply two vectors together like this. 3/31

6 4/31 Examples: = 7 ; 3 6 9

7 4/31 Examples: = 7 ; 1 2 =

8 4/31 Examples: = 7 ; 1 2 = = 3 4 0

9 4/31 Examples: = 7 ; 1 2 = = ( 1) =

10 4/31 Examples: = 7 ; 1 2 = = ( 1) =

11 5/31 iclicker question: Frequency AA = 3 4 0

12 5/31 iclicker question: Frequency AA = A) 10 B) 9 C) 10 D) 19 E)

13 6/31 Recall the real number line 0 a b

14 6/31 Recall the real number line 0 a b with addition (or subtraction): 0 a b a+b

15 6/31 Recall the real number line 0 a b with addition (or subtraction): 0 a b a+b (but it matters which side of 0 the numbers start out) a 0 a+b b

16 6/31 Recall the real number line 0 a b with addition (or subtraction): 0 a b a+b (but it matters which side of 0 the numbers start out) a 0 a+b b and multiplication (or division): 0 a 5a

17 6/31 Recall the real number line 0 a b with addition (or subtraction): 0 a b a+b (but it matters which side of 0 the numbers start out) a 0 a+b b and multiplication (or division): 0 a 5a You can do this without knowing the actual values of a and b!

18 7/31 You can combine addition and multiplication: 0 a b

19 7/31 You can combine addition and multiplication: 0 a b -3a 0 a b-3a b

20 7/31 You can combine addition and multiplication: 0 a b -3a 0 a b-3a b Easier to describe if you think of numbers as arrows from 0: 0 a b 0 a b -3a 0 a b -3a 0 a b-3a b

21 7/31 You can combine addition and multiplication: 0 a b -3a 0 a b-3a b Easier to describe if you think of numbers as arrows from 0: 0 a b 0 a b -3a 0 a b -3a 0 a b-3a b Addition: Put the head of one arrow at tail of the other.

22 7/31 You can combine addition and multiplication: 0 a b -3a 0 a b-3a b Easier to describe if you think of numbers as arrows from 0: 0 a b 0 a b -3a 0 a b -3a 0 a b-3a b Addition: Put the head of one arrow at tail of the other. Multiplication: Scale the length &, if negative, reverse direction

23 8/31 Use the same idea for 2- and 3-vectors: think of vector as arrow from 0

24 8/31 Use the same idea for 2- and 3-vectors: think of vector as arrow from 0 multiply number with vector via scaling.

25 8/31 Use the same idea for 2- and 3-vectors: think of vector as arrow from 0 multiply number with vector via scaling. x 2 add vectors head-to-tail; v u x 1

26 9/31 Use the same idea for 2- and 3-vectors: think of vector as arrow from 0 multiply number with vector via scaling. x 2 add vectors head-to-tail; v u x 1

27 10/31 Use the same idea for 2- and 3-vectors: think of vector as arrow from 0 multiply number with vector via scaling. add vectors head-to-tail; x 2 u + v v u x 1

28 11/31 Use the same idea for 2- and 3-vectors: think of vector as arrow from 0 multiply number with vector via scaling. add vectors head-to-tail: parallelogram rule; x 2 u + v v u x 1

29 12/31 Application: Gives more flexible way to describe a line.

30 12/31 Application: Gives more flexible way to describe a line. For a line through a point p, in direction d, use x = p + t d, t R Argument: x 2 p d x 1

31 13/31 Application: Gives more flexible way to describe a line. For a line through a point p, in direction d, use x = p + 1 d, t = 1 Argument: x 2 p p + d d x 1

32 14/31 Application: Gives more flexible way to describe a line. For a line through a point p, in direction d, use x = p + 1 d, t = 1 Argument: x 2 p p - d d x 1

33 15/31 Application: Gives more flexible way to describe a line. For a line through a point p, in direction d, use x = p + t d, t R Argument: x 2 p p + td d x 1

34 16/31 iclicker question: Frequency AA Pictured are points u, v R 2.

35 16/31 iclicker question: Frequency AA Pictured are points u, v R 2. Which point represents u 3 v? x 2 u v x 1

36 17/31 iclicker question: Frequency AA Pictured are points u, v R 2. Which point represents u 3 v? x 2 E D C B A u v x 1

37 18/31 iclicker question: Frequency AA Pictured are points u, v R 2. Which point represents u 3 v? x 2 u v x 1

38 19/31 iclicker question: Frequency AA Pictured are points u, v R 2. Which point represents u 3 v? x 2-3v -v u v x 1

39 20/31 For all w, x, y R n and s, t R, we have the following.

40 20/31 For all w, x, y R n and s, t R, we have the following. x + y = y + x (commutative)

41 20/31 For all w, x, y R n and s, t R, we have the following. x + y = y + x (commutative) ( x + y) + w = x + ( y + w) (associative)

42 20/31 For all w, x, y R n and s, t R, we have the following. x + y = y + x (commutative) ( x + y) + w = x + ( y + w) (associative) z + 0 = 0 + z = z

43 20/31 For all w, x, y R n and s, t R, we have the following. x + y = y + x (commutative) ( x + y) + w = x + ( y + w) (associative) z + 0 = 0 + z = z x + ( x) = x + x = 0

44 20/31 For all w, x, y R n and s, t R, we have the following. x + y = y + x (commutative) ( x + y) + w = x + ( y + w) (associative) z + 0 = 0 + z = z x + ( x) = x + x = 0 t( x + y) = t x + t y (distributive law)

45 20/31 For all w, x, y R n and s, t R, we have the following. x + y = y + x (commutative) ( x + y) + w = x + ( y + w) (associative) z + 0 = 0 + z = z x + ( x) = x + x = 0 t( x + y) = t x + t y (distributive law) (s + t) x = s x + t x (distributive law)

46 20/31 For all w, x, y R n and s, t R, we have the following. x + y = y + x (commutative) ( x + y) + w = x + ( y + w) (associative) z + 0 = 0 + z = z x + ( x) = x + x = 0 t( x + y) = t x + t y (distributive law) (s + t) x = s x + t x (distributive law) s(t x) = (st) x (associative)

47 20/31 For all w, x, y R n and s, t R, we have the following. x + y = y + x (commutative) ( x + y) + w = x + ( y + w) (associative) z + 0 = 0 + z = z x + ( x) = x + x = 0 t( x + y) = t x + t y (distributive law) (s + t) x = s x + t x (distributive law) s(t x) = (st) x (associative) 1 x = x

48 21/31 Definition Suppose {t 1, t 2... t k } are all real numbers. The vector y = t 1 v t k v k is called a linear combination of the vectors { v 1, v 2... v k }.

49 21/31 Definition Suppose {t 1, t 2... t k } are all real numbers. The vector y = t 1 v t k v k is called a linear combination of the vectors { v 1, v 2... v k }. Sample problem: Given vectors { a 1, a 2... a n, b} in R m, find real numbers {t 1, t 2... t n } so that t 1 a t n a n = b.

50 21/31 Definition Suppose {t 1, t 2... t k } are all real numbers. The vector y = t 1 v t k v k is called a linear combination of the vectors { v 1, v 2... v k }. Sample problem: Given vectors { a 1, a 2... a n, b} in R m, find real numbers {t 1, t 2... t n } so that t 1 a t n a n = b. Off hand, could have any number of {t 1, t 2... t n } solutions.

51 22/31 How to think about solving for {t 1, t 2... t n } in the equation t 1 a t n a n = b :

52 22/31 How to think about solving for {t 1, t 2... t n } in the equation t 1 a t n a n = b : Let b 1 a 1j a 2j a 3j a mj b 2 b = b 3 ; a j =.. b m for 1 j n

53 22/31 How to think about solving for {t 1, t 2... t n } in the equation t 1 a t n a n = b : Let b 1 a 1j a 2j a 3j a mj b 2 b = b 3 ; a j =.. b m for 1 j n Then for any 1 i m, the i th row of the equation becomes: t 1 a i1 + t 2 a i2 + + t n a in = b i or a i1 t 1 + a i2 t a in t n = b i

54 In other words, solving t 1 a t n a n = b : is the same as solving this system of m linear equations: a 11 t a 1n t n = b 1 a 21 t a 2n t n = b 2. a m1 t a mn t n = b m We just learned how to do this! 23/31

55 24/31 Solving t 1 a t n a n = b : is the same as solving a system of m linear equations.

56 24/31 Solving t 1 a t n a n = b : is the same as solving a system of m linear equations. The system has augmented matrix a 11 a a 1n b 1 a 21 a a 2n b a m1 a m2... a mn b m

57 24/31 Solving t 1 a t n a n = b : is the same as solving a system of m linear equations. The system has augmented matrix a 11 a a 1n b 1 a 21 a a 2n b a m1 a m2... a mn b m Since each a j is a column of i numbers, can just write [ a 1 a 2... a n b ]

58 25/31 Example (motivated by last time): Suppose a 1 = 2 4 a 2 = 2 4 a 3 = 1 1 a 4 = and want to find c 1, c 2, c 3, c 4 so that 12 c 1 a 1 + c 2 a 2 + c 3 a 3 + c 4 a 4 =

59 26/31 This translates to the system of linear equations whose augmented matrix is

60 26/31 This translates to the system of linear equations whose augmented matrix is which we saw last time row reduces to:

61 26/31 This translates to the system of linear equations whose augmented matrix is which we saw last time row reduces to: and so has general solution c 2 = anything, c 1 = 3 2 c 2, c 3 = 2, c 4 = 1 2

62 27/31 Definition Given a collection { v 1, v 2,..., v k } of vectors in R m, the set of all linear combinations of these vectors, that is all vectors that can be written as c 1 v c k v k for some c 1,..., c k R is denoted Span { v 1,..., v k } and is called the span of { v 1,..., v k }.

63 27/31 Definition Given a collection { v 1, v 2,..., v k } of vectors in R m, the set of all linear combinations of these vectors, that is all vectors that can be written as c 1 v c k v k for some c 1,..., c k R is denoted Span { v 1,..., v k } and is called the span of { v 1,..., v k }. Easy example: If k = 1 so there is only one vector v, then Span{ v} is just all vectors that are multiples of v. That is, Span{ v} = {c v c R}

64 28/31 Picturing the span when m = 2, 3:

65 28/31 Picturing the span when m = 2, 3: When there is only one vector v then Span{ v} = {c v c R} is just the line that contains both 0 (take c = 0) and v (take c = 1).

66 28/31 Picturing the span when m = 2, 3: When there is only one vector v then Span{ v} = {c v c R} is just the line that contains both 0 (take c = 0) and v (take c = 1). With two vectors u and v, Span{ u, v} = {c 1 u + c 2 v} pictured via the parallelogram rule (Span = entire plane; c i 0 highlighted): u (3u+5v)/2 v

67 29/31 Repeat of example above: The matrix and reduced echelon form:

68 29/31 Repeat of example above: The matrix and reduced echelon form: has general solution: c 2 = anything, c 1 = 3 2 c 2, c 3 = 2, c 4 = 1 2

69 29/31 Repeat of example above: The matrix and reduced echelon form: has general solution: c 2 = anything, c 1 = 3 2 c 2, c 3 = 2, c 4 = 1 2 We can write this as c c 2 c 3 = 0 + t c

70 30/31 So we can think of the set of all solutions as Span

71 So we can think of the set of all solutions as Span So we can picture the solution as a line in the direction of the second vector, going through the point given by the first vector (but in R 4!) x 2 p p + td d x 1 30/31

72 31/31 Another from last time: Example with two free variables {x 2, x 4 } π e

73 31/31 Another from last time: Example with two free variables {x 2, x 4 } π e x 1 = 3 2 x 2 πx 4 x 3 = 2 ex 4 x 5 = 1 2

74 31/31 Another from last time: Example with two free variables {x 2, x 4 } π e x 1 = 3 2 x 2 πx 4 x 3 = 2 ex 4 x 5 = 1 2 which can be written: π x x 0 4 e

75 31/31 Another from last time: Example with two free variables {x 2, x 4 } π e x 1 = 3 2 x 2 πx 4 x 3 = 2 ex 4 x 5 = 1 2 which can be written: π 2 1 π x x 0 4 e 1 = Span 1 0 0, 0 e