WEEK 5 - Vector Space, Subspace

7/31/2019 WEEK 5 - Vector Space, Subspace

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VECTOR

SPACES


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A vector space is a mathematical structure formed by a

collection of vectors: objects that may be addedtogether and multiplied ("scaled") by numbers, calledscalarsin this context.

The operations of vector addition and scalar

multiplication have to satisfy certain requirements, called

axioms.

Vector addition and scalar

multiplication: a vector v

(blue) is added to anothervector w (red, upperillustration). Below, w isstretched by a factor of 2,

yielding the sum v + 2w.
http://en.wikipedia.org/wiki/Mathematical_structurehttp://en.wikipedia.org/wiki/Vector_additionhttp://en.wikipedia.org/wiki/Scalar_multiplicationhttp://en.wikipedia.org/wiki/Scalar_%28mathematics%29http://en.wikipedia.org/wiki/Axiomhttp://en.wikipedia.org/wiki/Axiomhttp://en.wikipedia.org/wiki/Scalar_%28mathematics%29http://en.wikipedia.org/wiki/Scalar_multiplicationhttp://en.wikipedia.org/wiki/Vector_additionhttp://en.wikipedia.org/wiki/Mathematical_structurehttp://en.wikipedia.org/wiki/Mathematical_structurehttp://en.wikipedia.org/wiki/Mathematical_structure


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The concept of vector space relies on the

idea ofvectors. In general, the term vector isused for objects on which two operations canbe exerted.

The concrete nature of these operations

depends on the type of vector underconsideration, and can often be described bydifferent means, e.g. geometric oralgebraic.

In view of the algebraic ideas behind these

concepts explained below, the two operationsare called vector addition and scalarmultiplication.
http://en.wikipedia.org/wiki/Geometryhttp://en.wikipedia.org/wiki/Algebrahttp://en.wikipedia.org/wiki/Vector_additionhttp://en.wikipedia.org/wiki/Scalar_multiplicationhttp://en.wikipedia.org/wiki/Scalar_multiplicationhttp://en.wikipedia.org/wiki/Scalar_multiplicationhttp://en.wikipedia.org/wiki/Scalar_multiplicationhttp://en.wikipedia.org/wiki/Scalar_multiplicationhttp://en.wikipedia.org/wiki/Vector_additionhttp://en.wikipedia.org/wiki/Vector_additionhttp://en.wikipedia.org/wiki/Vector_additionhttp://en.wikipedia.org/wiki/Algebrahttp://en.wikipedia.org/wiki/Geometry


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Vector addition means that two vectors v and w can be"added" to yield the sum v + w, another vector. Thesum of two arrow vectors is calculated by constructing

the parallelogram two of whose sides are the givenvectors v and w. The sum of the two is given by thediagonal arrow of the parallelogram, starting at the

common point of the two vectors

v+w


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Scalar multiplication combines a numberalso

called scalarr and a vector v. In the example, a

vector represented by an arrow is multiplied by ascalar by dilating or shrinking the arrow accordingly:

if r= 2 (r= 1/4), the resulting vector r w has thesame direction as w, but is stretched to the double

length (shrunk to a fourth of the length, respectively)ofw (middle image above). Equivalently 2 w is thesum w + w. In addition, for negative factors, thedirection of the arrow is swapped: (1) v = v hasthe opposite direction and the same length as v(blue vector in the image).
http://en.wikipedia.org/wiki/Scalarhttp://en.wikipedia.org/wiki/Scalar


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Definition

A vector space is a nonempty set Vof objects, called vectors, on whichare defined two operations, called additionand multiplication by scalars(numbers), subject to the ten axioms (or rules) listed below. The axiomsmust hold for all vectors x, y, and z in Vand for all scalars k1 and k2.

1. The sum ofx and y, denoted by x+y, is in V2. x+y= y+x3. (x+y)+z=x+(y+z)4. There is a unique vector 0in Vsuch that x+0=x5. For each x in V, there is a vectorx in Vsuch that x+(x)=06. The scalar multiple ofx in k, denoted by kx, is in V.7. k(x+y)= kx+ky8. (k1 +k2) x= k1 x+ k2x9. k1(k2x) =(k1 k2)x10. 1 x = x

Vector Space

DEFINITION


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Remark: Depending on the application,

scalars may be real numbers or

complex numbers. Vector spaces in

which the scalars are complex numbersare called complex vector spaces, andthose in which the scalars must be real

are called real vector spaces.


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Show that the set V of all 2x2 matrices with real entries is a vectorspace if vector addition is defined to be matrix addition and vectorscalar multiplication is defined to be matrix scalar multiplication.

SolutionIn this example we will find it convenient to verify the axioms in the

following order: 1, 6, 2, 3, 7, 8, 9, 4, 5, and 10. Let

u v

To prove Axiom 1, we must show that u+v is an object in V; that is, we

must show that u+v is a2x2 matrix.

But this follows from the definition of matrix addition, since

u+v

11 12

21 22

u u

u u

11 12

21 22

v v

v v

11 12 11 12 11 11 12 12

21 22 21 22 21 21 22 22

u u v v u v u v

u u v v u v u v

Example 1


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Similarly, Axiom 6 holds because for any real numbercwe have

cu

so that cu is a 2x2 matrix and consequently is an object in V.

Axiom 2 follows from Properties of Matrix Addition and ScalarMultiplication since

u+v v+u

Similarly, Axioms 3, 7, 8 and, 9 (form Properties of MatrixAddition and Scalar Multiplication).

To prove Axiom 4, we must find an object 0 in Vsuch that u+0=u for all u in V. This can be done by defining 0 to be

0

11 12 11 12

21 22 21 22

u u cu cu

c u u cu cu

11 12 11 12 11 12 11 12

21 22 21 22 21 22 21 22

u u v v v v u u

u u v v v v u u

0 0

0 0


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With this definition

u+0

u

To prove Axiom 5 we must show that each object u in Vhas a

negative u such that u+(u)=0. This can be done by defining

the negative ofu to beu

With this definition

u+(u) 0

Finally, Axiom 10 is a simple computation:

1u u

11 12 11 12

21 22 21 22

0 0

0 0

u u u u

u u u u

11 12

21 22

u u

u u

11 12 11 12

21 22 21 22

0 0

0 0

u u u u

u u u u

11 12 11 12

21 22 21 22

1u u u u

u u u u


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Consider the set Vof positive real numbers. If x and y denotepositive numbers, then we write vectors in V as x=x and y=y.Now addition of vectors is defined by

and scalar multiplication is defined by

Determine whetherVis a vector space.

Solution

-We shall go through all ten axioms.

1. For . Thus, the sum

is in V; Vis closed under addition.

xyyx

kxk x

0,0and0 xyyx yxyx yx

Example 2


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2. Since multiplication of positive real numbers is commutative,we have for all x=xand y=yin V,

Thus, addition is commutative.

3. For all x=x,y=y, z=z in V,

Thus, addition is associative.

4. Since and , the zerovector0 is1 = 1

5. If we define , then

Therefore, the negative of a vector is its reciprocal.

xyyx yxxy

zyxzyx )()()()( zxyyzx

xx xx11 xx xx11

x

1

x

01xx01xx 11

)(and11

)( xxx

x


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6. If k isany scalar and is any vector, then. Hence

Vis closed under scalar multiplication.

7. If kis any scalar, then

8. For scalars k1

and k2

,

9. For scalars k1 and k2 ,

10. .

Since all the axioms are satisfied, we conclude that Vis a vector

space

0 xx 0 kxkx

yxyx kkyxxykkkk

)()(

xxx21

)(

212121)( kkxxxkk

kkkk

xx)()()(

2121

2112

kkxxkk

kkkk

xx xx1

1


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Let V=R2and define addition and scalar multiplication operationas follows:

Ifx=(x1, x2) and y= (y1, y2), then definex+y

and ifkis any real number, then definekx

For example, ifx=(2,4), y=(-3,5), and k=7, then

x+y

kx=7x

1 1 2 2( , )x y x y

1( ,0)kx

(2 ( 3),4 5) ( 1,9)

(7.2,0) (14,0)

Example 3


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The addition operation is the standard addition operation on R2,

but the scalar multiplication operation is not the standard scalarmultiplication. There are values ofu for which Axiom 10 fails to

hold. For example, ifx=(x1, x2) is such that , then

1x x

Thus, Vis not a vector space with the stated operations.

1 2 1 11( , ) (1. ,0) ( ,0)x x x x

2 0x


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The set V of all triples of real numbers

(x,y,z) with the operations

Determine whether V is a vector space.

zykxzyxk

zzyyxxzyxzyx

,,,,

',','',',',,

Exercise 1


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The set V of all pairs of real numbers

(x,y) with the operations


, ', ' ' 1, ' 1

, ,

x y x y x x y y

k x y kx ky

Exercise 2


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Exercise 3

The set V of all pairs of real numbers

(x,y) with the operations


kykxyxk

yyxxyxyx

2,2,

','',',

Exercise 3


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A subspace of a vector space Vis a subset HofVthat has three

properties:

1. The zero vector ofVis in H.

2. H is closed under vector addition. That is, for each u and v in H, the

sum u+v is in H.

3. His closed under multiplication by scalars. That is, for each u in H

and each scalarc, the vectorcu is in H.

Note: The term subspaceis used when at least two vector spaces are in

mind, with one inside the other, and the phrase subspace of V

identifies Vas a larger space.

Subspaces

DEFINITION


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The vector spaceR2

is nota subspace ofR3

becauseR2

is noteven a subset ofR3. (The vectors in R3 all have three entries,whereas the vectors inR2 have only two.) The set

is a subset ofR3 that looks and acts like R2 , although it islogically distinct fromR2. Show that His a subspace ofR3.

Solution

The zero vector is in H, and H is closed under vector additionand scalar multiplication because these operations on vectors inHalways produce vectors whose third entries are zero (and sobelong to H). Thus His a subspace ofR3.

:

0

s

H t s and t are real

Example 4


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A plane inR3 notpassing through the origin is not a subspace ofR3, because the plane does not contain the zero vector ofR3.

Similarly, a line in R2not passing through the origin, such as inFigure 1, is nota subspace ofR2.

Figure 1: A line that is not a vector space

x2

x1

H

Example 5


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Exercise

Determine which of the following are

subspace ofR3

a) all vectors of the form (a,0,0)

b) all vectors of the form (a,1,1)

c) all vectors of the form (a,b,c) where b=a+c

Exercise 4


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Definition

Let v1, v2, , vp be a vector space V. A sum of the

form k1v1+k2v2++kpvp where k1, k2,

, kp arescalars, is called a linear combinationofv1, v2, ,vp.

Linear Combinations of Vectors

DEFINITION


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The set of all linear combinations ofv1, v2, , vp is calledthe spanof v1, v2, , vp. The span ofv1, v2, , vp will bedenoted by Span {v1, v2, , vp}

Theorem 1

Ifv1, , vp are in a vector space V, then Span{v1, , vp} isa subspaces ofV.

Note:

We call Span {v1,, vp} the subspace spanned (or generated) by {v1,, vp}. Given any subspaces Hof V, a spanning (or generated) set forH is a set

{v1,, vp} in Hsuch that H=Span{v1,, vp}

Spanning

Definition - Span


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Every vector inR3 is

expressible as a linear combination of the

standard basis vector

since

cb,a,v

1,0,0,0,1,0,0,0,1 kji

kjiv cbacbacba 1,0,00,1,00,0,1,,

),,(Span kjiv

Example 6


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Given v1 and v2 in a vector space V, let H=Span{v1 ,v2}. Show that His a

subspace ofV.

Solution

The zero vector is in H, since 0=0v1+0v2. To show that His closed undervector addition, take two arbitrary vectors in H, say,

u= s1v1 +s2v2 and w=t1v1 +t2v2

By Axioms 2, 3, and 8 for the vector space V,

u+w=(s1v1 +s2v2)+(t1v1+t2v2)=(s1+t1)v1+(s2+t2)v2

So u+w is in H. Futhermore, ifcis any scalar, then by Axioms 7 and 9,

cu=c(s1v1+s2v2)=(cs1)v1+(cs2)v2

which shows that cu is in Hand His closed under scalar multiplication. ThusHis a subspace ofV.

Example 7


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Let Hbe the set of all vectors of the form , where aand bare arbitrary scalars. That is, let

Show that His a subspace ofR4.

SolutionWrite the vectors in Has column vectors. Then an arbitrary vector in Hhas the form

This calculation shows that H=Span{v1, v2}, where v1 and v2 are thevectors indicated above. Thus His a subspace ofR4 by Theorem 1.

baabba ,,,3

RbabaabbaH inand,,,3

1

0

1

3

0

1

1

13

ba

b

a

ab

ba

Example 8


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Consider the vectoru=(1,2,-1) andv=(6,4,2) in .Show that w=(9,2,7) is alinear combination ofu and v and that

w=(4,-1,8) is not a linear combination ofu and v

Exercise 5

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WEEK 5 - Vector Space, Subspace