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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

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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 .

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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.

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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.

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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 .

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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

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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.

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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.

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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.

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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)).

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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).

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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.

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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.

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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.

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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).

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Homework

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

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