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Vector Spaces

§4.2 Vector Spaces

Satya Mandal, KU

September 27

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Goals

◮ Give definition of Vector Spaces

◮ Give examples and non-examples of Vector Spaces

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Abstruct Definition of Vector SpacesExamples for §4.2A List of Important Vector SpacesProperties of Vector SpacesNon-Example

Operations on Sets

◮ On the set of integers Z, or on the set of real numbers Rwe worked with addition +, multiplication ×.

◮ On the n−space Rn, we have addition and scalarmultiplication.

◮ These are called operations on the respective sets. Suchan operation associates an oredered pair to an element inV , like,

(u, v) 7→ u+ v, or (c , v) 7→ cv

These are called binary operations, because they associatean ordered pair to an element in V .

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Abstruct Definition of Vector SpacesExamples for §4.2A List of Important Vector SpacesProperties of Vector SpacesNon-Example

Operations on Sets: Continued

◮ Likewise, in mathematics, we define the same on any setV . Given two sets V ,R , a binary operation ∗, o associatesan ordered pair to an element in V .

∗ : V × V −→ V (u, v) 7→ u ∗ v

o : R × V −→ V (c , v) 7→ cov

Such operations are mostly denoted by +,× and calledaddition, multiplication or scalar multiplication, dependingon the context.

◮ Examples of Binary operations include:(a) Matrixaddition, multiplication; (b) polynomial addition andmultiplication;(c) addition, multiplication andcomposition of functions.

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Abstruct Definition of Vector SpacesExamples for §4.2A List of Important Vector SpacesProperties of Vector SpacesNon-Example

Vector Spaces: Motivations

There are many mathematical sets V , with an addition + anda scalar multiplication, satisfy the properties of the vectors inn−spaces Rn, listed in Theorem 4.1 and 4.2. Examples ofsuch sets include

◮ the set all of matrices Mm,n of size m × n,

◮ set of polynomial,

◮ Set of all real valued continuous functions on a set.

In order to unify the study of all such sets, we define abstractvector spaces, by listing the properties in Theorem 4.1 and 4.2.

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Vector Spaces: Definition

Definition: Let V be a set with two operations (vectoraddition + and scalar multiplication). We say that V is aVector Space over the real numbers R if, for all u, v,w in Vand all scalars (reals) c , d , the following propertices arestistified:

◮ (1. Closure under addition): u+ v is in V .◮ (2. Commutativity): u+ v = v + u.◮ (3. Associativity I): (u+ v) +w = u+ (v +w)◮ (4. Additive Identity or zero):There is an element in V ,

denoted by 0 and to be called a (the) zero vector suchthat

u+ 0 = u, for every u ∈ V .

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Continued:

◮ (5. Additive inverse): For every u ∈ V there is and anelement in V , denoted −u such that u+ (−u) = 0

◮ (6. Closure under scalar multiplication): cu is in V

◮ (7. Distributivity I): c(u+ v) = cu+ cv.

◮ (8. Distributivity II):(c + d)u = cu+ du.

◮ (9. Associativity II): c(du) = (cd)u

◮ (10. Multiplicative Identity): 1u = u

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Four Entities of Vector Spaces

Note that a Vector Space as four entities:

◮ (1) A set of vectors V ,

◮ (2) a set of scalars,

◮ (3,4 ) two operations.

◮ In this class the set of scalars is R. Vector spaces overcomplex scalars C are also studied at higher level courses.

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Examples from the Textbook

Reading Assignment: §4.2 Example 1-5.Examples in the textbook lists most of the imporant andstandard examples of Vecotor spaces. We give a quick review.

◮ Example 1: The plane, R2 with standard addition andscalar multiplication is a Vector Space.

◮ Example 2: The n−space, Rn with standard additionand scalar multiplication is a Vector Space.

◮ Example 3: Let M2,3 be the set of all ×3. So,

M2,3 =

{[

a b c

x y z

]

: a, b, c , x , y , z ∈ R

}

Then, with standard addition and scalar multiplicationM2,3 is a Vector Space.

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Examples from the Textbook

◮ Example 4: Let P2 denote the set of all polynomials ofdegree less or equal to 2. So,

P2 = {a2x2 + a1x + a0 : a2, a1, a0 ∈ R}

Then, with standard addition and scalar multiplication P2is a Vector Space.

◮ Example 5. Let I be an interval and C (I ) deonotes theset of all real-valued continuous functions on I . Then,with standard addition and scalar multiplication C (I ) is aVector Space. For example,

C (−∞,∞),C (0, 1),C [0, 1],C (0, 1],C [1, 0)

are vector spaces.

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Formal Proofs

To give a proof we need to check all the 10 properties in thedefinition. While each step may be easy, students at this levelare not used to writing a formal proof. Here is a proof thatC (0, 1) is a vector space.

◮ So, the vectors are continuous functions f(x) defined onthe interval (0, 1).

◮ For vectors f, g ∈ C (0, 1) addition is defined as follows:

(f + g)(x) = f (x) + g(x)

◮ For f ∈ C (0, 1), c ∈ R scalar multiplication is defined as

(cf)(x) = c(f(x)).

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Formal Proofs: Continued

For this addition and scalar multiplication, we have to checkall 10 properties in the defintion. Suppose f, g,h ∈ C (0, 1)and c , d ∈ R.

◮ (1. Closure under addition): C (0, 1) is closed underaddition. This is because sum f + g of two continuousfunctions f, g is continuous. So, f + g ∈ C (0, 1).

◮ (2. Commutativity): Clearly, f + g = g + f.

◮ (3. Associativity I): Clearly (f + g) + h = f + (g + h)

◮ (4. The Zero): Let f0 denote the constant-zero function.So, f0(x) = 0 So, (f + f0)(x) = f(x) + 0 = f(x).Therefore f0 satiesfy the condition (4).

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Formal Proofs: Continued

◮ (5. Additive inverse): Let −f denote the function(−f)(x) = −f(x). Then (f + (−f)) = 0 = f0.

◮ (6. Closure under scalar multiplication): Clealry, cf iscontinuous and so cf ∈ C (0, 1).

◮ (7. Distributivity I): Clearly, c(f + g) = cf + cg.

◮ (8. Distributivity II): Clearly, (c + d)f = cf + df.

◮ (9. Associativity II): Clearly, c(df) = (cd)f

◮ (10. Multiplicative Identity):Clearly 1f = f.

All 10 properties are varified. So, C (0, 1) is continuous.

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A List of Important Vector Spaces

Here is a list of important vector spaces:◮ R the set of all real numbers◮ R

2 the set of all ordered pairs of real numbers. Thiscorresponds to the plane.

◮ R3 the set of all ordered triples of real numbers.This

corresponds to the three dimensional space.◮ R

n the set of all ordered n−tuples of real numbers.◮ C (−∞,∞),C (a, b),C [a, b],C (a, b],C [a, b) the set of

real valued continuous functions.◮ P the set of all polynomials, with real coefficients◮ Pn the set of all polynomials, with real coefficients, of

degree less or equal to n.◮ Mm,n the set of all matrices of size m× n with real entries.

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Theorem A: Uniqueness of zero and v

Theorem A. Suppose V is a vector space. Then,◮ There is exactly one vector satisfying the property of zero

(Condition 4.) We say the additive identity 0 is unique.◮ Given a vector u there is exactly on vector −u satisfying

condition 5. We say v has a unique additive inverse.

Proof. Suppose ϕ also satisfy condition 4. So, for any vectoru ∈ V we have

u+ 0 = u and u+ ϕ = u

Apply these two equations to 0, ϕ. We have

ϕ = ϕ+ 0 = 0+ ϕ = ϕ.

So, the first statement is establisehd.Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Proof: Continued:

Write v = −u. (−u is not a good noatation). To prove thesecond statement, assume, other then v, there is anothervector θ that satisfy condition 5. So.

u+ v = 0 and u+ θ = 0

We will prove that v = θ. Add θ to the first equation:

(u+ v) + θ = 0+ θ Or (u+ θ) + (v) = θ

Or0+ (v) = θ Or v = θ

The second statement is established.Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Properties of Scalar Multiplication

Theorem 4.4

Let v be a vector in a vector space V and c be a scalar. Then,

◮ (1) 0v = 0

◮ (2) c0 = 0

◮ (3)cv = 0 =⇒ c = 0 or v = 0

◮ (4) (−1)v = −v.

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Proof.

Proof.

◮ (1) By distributive property, We have0v + 0v = (0 + 0)v = 0v. By (property 5), there is anadditive inverse −(0v) of 0v. We add the same to bothsides of the above equation

(0v + 0v) + (−(0v)) = 0v + (−(0v)) OR

0v + (0v + (−(0v))) = 0 OR

0v + 0 = 0 Or 0v = 0

So, (1) is established.

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Proof: Continued

Proof.

◮ (2) First, by distributivity

c0 = c(0+ 0) = c0+ c0

Now add −(c0) to both sides:

c0+(−(c0)) = (c0+ c0)+(−(c0)) OR 0 = c0.

So, (2) is established.

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Proof: Continued

Proof.

◮ (3) Suppose cv = 0. Suppose c 6= 0. Then we canmultiply the equation by 1

c. So,

1

c(cv) =

1

c0 = 0 (by (2))

By axion (10), we have

v = 1v =

(

1

cc

)

v =1

c(cv) =

1

c0 = 0

So, either c = 0 or v = 0 and (3) is established.

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Proof: Continued

Proof.

◮ (4) We have, by distributivity and axiom (10)

v + (−1)v = 1v + (−1)v = (1− 1)v = 0v = 0 by (1).

So, (−1)v satisfies the axiom (5) of the definition. So(−1)v = −v. So, (4) is established.

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Examples of Sets with Operations

that are not Vector Spaces

Given a set W and two operations (like addition and scalarmyltiplication), it may fail to be a Vecor Space for failure ofany one of the axioms of the definition.Reading Assignment: §4.2 Example 6, 7, 8

◮ Example 6 from the Textbook The set of integers Z,with usual addition and scalar multiplication is not avector space. Reason: Z is not closed under scalarmultiplication .5(1) /∈ Z not a Vector Space.

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◮ Example 7 from the Textbook: Set S of polynomial ofdegree (exactly) 2 with usual addition and scalarmultiplication is not a vector space. Reason:Z is notclosed under Addition: f (x) = 2x2 +3x +1, g(x) = −2x2

are in S . But f + g = 3x + 1 is not in S .

◮ Example: Let R21be a the set of all ordered pairs of

(x , y) in the first quadrant. SoR

2 = {(x , y) ∈ R2 : x ≥ 0, y ≥ 1}. Under usual additionans scalar multiplication R2

1is not a vector space.

Reason: (1, 1) does not have a additive inverse in R21.

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Exercise 4

Exercise 4 Describle the zero vector (additive identity) ofM1,4.Solution.

0 =[

0 0 0 0]

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Exercise 15

Exercise 15 Let X be the set of all polynomials of degree 3.Is X a vector space? If not, why?Solution. X is not a vector space. (Here, by ”degree 3”means, exactly of degree 3.)

◮

Let f(x) = x3 + x + 3, g(x) = −x3 − x2 + 7x + 4

◮

Then (f + g)(x) = −x2 − 8x + 7 has degree 2

◮

So, f, g ∈ X , but f + g /∈ X .

So, X is not closed under addition.◮ Therefore X is not a vector space.Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Exercise 27 (edited)

Exercise 27 (edited)

Let S =

0 0 a0 b 0c 0 0

: a, b, c ∈ R

Is it a vector space?

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Solution.

Yes, it is a vector space, because all the 10 conditions, of thedefinitions is satisfied:(1) S is closed under addition, (2) addition commutes, (3)additions is associative, (4) S has the zero, (5) each matrixv ∈ S , its −v ∈ S (6) S is claosed under scalar multiplication,(7) Distributivity I works, (8) Distributivity II works (9)Associativity II works, (10) 1v = v.

Remark. A theorem will be proved in the next section, whichstates that we only need to check 2 contions: that S is closedunder addition and scalar multiplication (condition 1 and 6).

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

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Homework

Homework: §4.2 Exercise 5, 6, 15, 25, 26, 33

Satya Mandal, KU Vector Spaces §4.2 Vector Spaces

PreviewVector SpacesAbstruct Definition of Vector SpacesExamples for §4.2A List of Important Vector SpacesProperties of Vector SpacesNon-Example

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