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# Introduction to vectors

## 1. Introduction to Vectors

Karashbayeva Zh.O.## 2. What are Vectors?

• Vectors are pairs of a direction and amagnitude. We usually represent a vector

with an arrow:

• The direction of the arrow is the direction

of the vector, the length is the magnitude.

## 3. Vectors in Rn

Vectors in Rn=1

n

R1-space = set of all real numbers

(R1-space can be represented geometrically by the x-axis)

n=2

( x1 , x2 )

R2-space = set of all ordered pair of real numbers

n=3

(R2-space can be represented geometrically by the xyplane)

R3-space = set of all ordered triple of real numbers ( x1 , x2 , x3 )

n=4

(R3-space can be represented geometrically by the xyzspace)

4

R -space = set of all ordered quadruple of real numbers ( x1 , x2 , x3 , x4 )

## 4. Multiples of Vectors

Given a real number c, we can multiply avector by c by multiplying its magnitude by

c:

2v

v

-2v

Notice that multiplying a vector by a

negative real number reverses the direction.

## 5. Adding Vectors

Two vectors can be added using theParallelogram Law

u

u+v

v

## 6. Combinations

These operations can be combined.2u

2u - v

u

v

-v

## 7. Components

To do computations with vectors, we placethem in the plane and find their

components.

v

(2,2)

(5,6)

## 8. Components

The initial point is the tail, the head is theterminal point. The components are

obtained by subtracting coordinates of the

initial point from those of the terminal

(5,6)

point.

v

(2,2)

## 9. Components

The first component of v is 5 -2 = 3.The second is 6 -2 = 4.

We write v = <3,4>

v

(2,2)

(5,6)

## 10. Magnitude

The magnitude of the vector is the lengthof the segment, it is written ||v||.

v

(2,2)

(5,6)

## 11. Scalar Multiplication

Once we have a vector in componentform, the arithmetic operations are easy.

To multiply a vector by a real number,

simply multiply each component by that

number.

Example: If v = <3,4>, -2v = <-6,-8>

## 12. Addition

To add vectors, simply add theircomponents.

For example, if v = <3,4> and w = <-2,5>,

then v + w = <1,9>.

Other combinations are possible.

For example: 4v – 2w = <16,6>.

## 13. Unit Vectors

A unit vector is a vector with magnitude 1.Given a vector v, we can form a unit vector

by multiplying the vector by 1/||v||.

For example, find the unit vector in the

direction <3,4>:

## 14. Special Unit Vectors

A vector such as <3,4> can be written as3<1,0> + 4<0,1>.

For this reason, these vectors are given

special names: i = <1,0> and j = <0,1>.

A vector in component form v = <a,b> can

be written ai + bj.

## 15.

## 16.

## 17. Spanning Sets and Linear Independence

Linear combination :A vector u in a vector space V is called a linear combination of

the vectors v1 , v 2 ,

, v k in V if u can be written in the form

u c1v1 c2 v 2

where c1 , c2 ,

ck v k ,

, ck are real-number scalars

## 18.

Ex : Finding a linear combinationv1 (1,2,3) v 2 (0,1,2) v 3 ( 1,0,1)

Prove (a) w (1,1,1) is a linear combination of v1 , v 2 , v 3

(b) w (1, 2,2) is not a linear combination of v1 , v 2 , v 3

Sol:

(a) w c1v1 c2 v 2 c3 v3

1,1,1 c1 1,2,3 c2 0,1,2 c3 1,0,1

(c1 c3 , 2c1 c2 , 3c1 2c2 c3 )

c1

c3

2c1 c2

3c1 2c2 c3

1

1

1

## 19.

1 0 1 1G.-J. E.

2 1 0 1

3 2 1 1

1 0 1 1

0 1 2 1

0 0 0 0

c1 1 t , c2 1 2t , c3 t

(this system has infinitely many solutions)

t 1

w 2 v1 3 v 2 v 3

t 2

w 3v1 5v 2 2 v 3

## 20.

(b)w c1 v1 c2 v 2 c3 v 3

1 0 1 1

G.-J. E.

2 1 0 2

3 2 1

2

1 0 1 1

0 1 2 4

0 0 0

7

This system has no solution since the third row means

0 c1 0 c2 0 c3 7

w can not be expressed as c1v1 c2 v2 c3 v3

## 21.

Thespan of a set: span(S)

If S={v1, v2,…, vk} is a set of vectors in a vector space V,

then the span of S is the set of all linear combinations of

the vectors in S,

span(S ) c1v1 c2 v 2

ck v k

ci R

(the set of all linear combinations of vectors in S )

Definition of a spanning set of a vector space:

If every vector in a given vector space V can be written as a

linear combination of vectors in a set S, then S is called a

spanning set of the vector space V

## 22.

Note: The above statement can be expressed as followsspan( S ) V

S spans (generates) V

V is spanned (generated) by S

S is a spanning set of V

Ex 4:

(a) The set S {(1, 0, 0), (0,1, 0), (0, 0,1)} spans R3 because any vector

u (u1 , u2 , u3 ) in R3 can be written as

u u1 (1, 0, 0) u2 (0,1, 0) u3 (0, 0,1)

(b) The set S {1, x, x 2 } spans P2 because any polynomial function

p( x) a bx cx 2 in P2 can be written as

p( x) a(1) b( x) c( x 2 )

## 23.

Ex 5: A spanning set for R3Show that the set S (1,2,3),(0,1,2),( 2,0,1) spans R3

Sol:

We must determine whether an arbitrary vector u (u1 , u2 ,u3 )

in R 3 can be expressed as a linear combination of v1 =(1,2,3),

v 2 =(0,1,2), and v 3 ( 2, 0,1)

If u c1v1 c2 v 2 c3 v3 c1

2c3 u1

2c1 c2

u2

3c1 2c2 c3 u3

The above problem thus reduces to determining whether this

system is consistent for all values of u1 , u2 , and u3

## 24.

1 0 2A 2 1

3 2

0 0

1

Ax u has exactly one solution for every u

span( S ) R3

## 25.

Definitionsof Linear Independence (L.I.) and Linear Dependence

(L.D.) :

S v1 , v2 ,

For c1v1 c2 v 2

, vk : a set of vectors in a vector space V

ck v k 0

(1) If the equation has only the trivial solution (c1 c2

then S (or v1 , v 2 ,

ck 0)

, v k ) is called linearly independent

(2) If the equation has a nontrivial solution (i.e., not all zeros),

then S (or v1 , v 2 ,

, v k ) is called linearly dependent (The name of

linear dependence is from the fact that in this case, there exist a v i

which can be represented by the linear combination of {v1 , v 2 ,

v i 1 ,

v k } in which the coefficients are not all zero.

, v i 1 ,

## 26.

Ex : Testing for linear independenceDetermine whether the following set of vectors in R3 is L.I. or L.D.

S v1 , v 2 , v 3 1, 2, 3 , 0, 1, 2 , 2, 0, 1

Sol:

c1

c1v1 c2 v 2 c3 v 3 0

2c3 0

2c1 c2

0

3c1 2c2 c3 0

1 0 2 0

1 0 0 0

0 1 0 0

G.-J. E.

2 1 0 0

3 2 1 0

0 0 1 0

c1 c2 c3 0 only the trivial solution

(or det(A) 1 0, so there is only the trivial solution)

S is (or v1, v 2 , v3 are) linearly independent

## 27.

• EX: Testing for linear independenceDetermine whether the following set of vectors in P2 is L.I. or L.D.

S v1 , v 2 , v 3 1 x 2 x 2 ,2 5 x x 2 ,x x 2

Sol:

c1v1+c2v2+c3v3 = 0

i.e., c1(1+x – 2x2) + c2(2+5x – x2) + c3(x+x2) = 0+0x+0x2

c1+2c2

=0

c1+5c2+c3 = 0

–2c1– c2+c3 = 0

1 2

1 2 0 0

G. E.

1 5 1 0

0 1

2 1 1 0

0 0

0

1

0

3

0 0

0

This system has infinitely many solutions

(i.e., this system has nontrivial solutions, e.g., c1=2, c2= – 1, c3=3)

S is (or v1, v2, v3 are) linearly dependent

## 28. Basis and Dimension

• Basis :Basis and Dimension

Spanning

Sets

V: a vector space

S ={v1, v2, …, vn} V

(1)

(2)

Bases

Linearly

Independent

Sets

S spans V (i.e., span(S) = V)

(For any u V ,

V

c v =Ax u has a solution (det(A) 0),

i

i

see Ex 5 on Slide 4.44)

S is linearly independent

(For ci vi = Ax 0, there is only the trivial solution (det(A) 0),

S is called a basis for V

Notes:

A basis S must have enough vectors to span V, but not so

many vectors that one of them could be written as a linear

combination of the other vectors in S

## 29.

Notes:(1) the standard basis for R3:

{i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)

n

(2) the standard basis for R :

{e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0),…, en=(0,0,…,1)

Ex: For R4, {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}

## 30.

(3) the standard basis matrix space:Ex: 2 2 matrix space:

1 0 0 1 0 0 0 0

,

,

,

0 0 0 0 1 0 0 1

(4) the standard basis for Pn(x):

{1, x, x2, …, xn}

Ex: P3(x)

{1, x, x2, x3}

## 31.

Ex 2: The nonstandard basis for R2Show that S ={v1 ,v 2 }={(1,1), (1, 1)} is a basis for R 2

c1 c2 =u1

(1) For any u=(u1 ,u 2 ) R , c1v1 +c2 v 2 =u

c1 c2 =u 2

2

Because the coefficient matrix of this system has a nonzero determinant, the

system has a unique solution for each u. Thus you can conclude that S spans R2

c1 c2 =0

(2) For c1v1 +c2 v 2 =0

c1 c2 =0

Because the coefficient matrix of this system has a nonzero determinant, you

know that the system has only the trivial solution. Thus you can conclude that S is

linearly independent

According to the above two arguments, we can conclude that S

is a (nonstandard) basis for R2