Finite Di Mension Al Vector Spaces

8/11/2019 Finite Di Mension Al Vector Spaces

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Contents

1 Fields 1.1 Notes about Exercise Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.2.1 Properties of Scalars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 Vector Spaces 2.1 Definition of Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Linear Dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.3.1 Some observations and Commentary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Linear Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.5.1 Examples Observation and Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 The impact of the set of Scalars on the Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.7 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.8 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.8.1 Observations and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.2 Terminology for working with Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.3 Calculus of Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.4 Compliment. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.9 Dimension of a subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.10 Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.11 Dual Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12 Reflexivity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.13 Basis ofV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.14 Annihilators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.15 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1


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2 CONTENT


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List of Theorems and Definitions

1 Definition (Vector Space) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Definition (Vector Sum Notation) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Definition (Sum of Zero Vectors) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Definition (Linear Dependence) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Theorem (Linear Dependence 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Theorem (Linear Dependence 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Definition (Basis Definition) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Theorem (Linear Combination Uniqueness using Basis) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9 Theorem (Extension of linearly independent set to Basis) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Theorem (Cardinality of Basis) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Definition (Dimension of a Vector Space) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Theorem (Linear dependence of n+1 Vectors inn-dimensionalVector Space) . . . . . . . . . . . . . . . . . . . . 13 Definition (Isomorphic Vector Spaces) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Definition (Isomorphism) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Theorem (Isomorphism between equidimensional Vector spaces) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Theorem (Equinumerity Vector Space and its Field) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Definition (Subspace) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Theorem (Subspace Intersection) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Definition (Span of subset of vectors). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Theorem (Span of Vectors and Subspaces) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Theorem (Sum of Supspaces) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22 Theorem (Subspace Compliment) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Theorem (Dimension of a Subspace) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Theorem (Subspace Basis) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Lemma (Basis of Intersection and Union of Subspaces) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Theorem (Dimension of the Space of Annihilators) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3


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4 LIST OF THEOREMS AND DEFINITION


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Notes on Finite Dimentional Vector Spaces

Ramesh Kadambi

October 12, 2014


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2 LIST OF THEOREMS AND DEFINITION


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Chapter 1

Fields

1.1 Notes about Exercise Problems

Exercises that are neither imperative (Prove That) nor interrogative (is it true that ...?) but merely declarative, thenis intended as a challenge. For such exercises it is asked to discover if the assertion is true or false, prove it if it is true aconstruct a counter example if it is false, most of all discuss such alterations of hypothesis and conclusion as will make the tr

ones false and the false ones true. Second, the exercises, whatever their grammatical form, are not always placed so as to maktheir very position a hint to their solution. Frequently exercises are stated as soon as the statement makes sense, quite a bbefore machinery for a quick solution has been developed. A reader who tries (even unsuccessfully) to solve such a misplaceexercise is likely to appreciate and to understand the subsequent developments much better for his attempt.

1.2 Fields

All numbers going forward are referred to as scalars. Ascalar s {C}, {R}, {Z} , i.e ascalarmay be a complex number, rnumberor an integer etc..

1.2.1 Properties of Scalars

Throughout this study, scalars can be interpreted as reals or complexwithout loss of anything essential. The properties of tscalars are assumed to be as follows,

(A) For every pair (, ), ofscalars there corresponds a scalar, +, called the sum of and , such that:

1. addition/sumis commutative, += +,

2. addition/sum is associative, + (+) = (+) +,

3. a unique scalar0 (called zero) such that + 0 = for every scalar , and4. to everyscalar there corresponds a unique scalar such that + () = 0.

(B) For every pair (, ), ofscalars there corresponds a scalar, , called the product of and , such that,

1. multiplication/product is commitative, = ,

2. multiplication/product is associative, () = (),

3. there exists a uniquescalar1 such that 1 = , for every scalar , and4. for every non-zeroscalar there corresponds a unique scalar 1( or 1

) such that 1 = 1

(C) Multiplication is distributive with respect to addition,(+) = +

A set Sof objects (scalars) is a field if,

1. The operations ofaddition and multiplicationare defined on S.

2. The conditions (A), (B), (C) are all satisfied.

Ex: 1. The set of all rational numbersQ, with the usual definitions of sum and product.

2. The set ofrealsR, and the set ofcomplex numbersC.

3


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4 CHAPTER 1. FIELD

1.3 Exercises

1. Almost all the laws of elementary arithmetic are consequences of the axioms defining a field. Prove, in particular, thatFis a field, and if,, and belong to F, then the following relations hold.

a. 0 += follows from property A1.

b. If += +then= .

Proof. Let= += +, = + = + , using A1, A2 =

c. +() = ( here= +()). The result follows fromA1, A2. +() = +(+ ) = ()+ 0 +=

d .0 = 0.= 0

Proof. using the fact that 0 =+ () we have , .0 = .(+ ()) it follows fromC, .(+ ()) = ( ) = 0? Not really. The correct way of reasoning is, + 0 =.(+ 0) = . +.0 =. we also ha+ 0 =. The result follows using b. The result 0.= 0 follows from B1.

e. (

1)=

Proof. We have from d,.0 = 0 and 1+ (1) = 00 =.(1+(1)) = + (1), However fromA4, + () =Using b we have, (1)=

g If= 0 then either = 0 or = 0 or both.

Proof. We will look at the statement of the problem as = 0 = 0 or = 0 or both. The direction = 0 = 0 or both= 0 is straight forward from d.In the other direction we have, = 0 += , (1 +) = (1 +) = 1 from C3. 1 += 1= Similarly one can show that the statement implies = 0. Since= 0 += and + = ,= 0both = 0, = 0.

2 a. Is the set of all positive integers a field? (In familiar systems, such as the integels, we shall almost always use tordinary operations of addition and multiplication. On the rare occasions when we depart from this convention, wshall give ample warning. As for positive, by that word we mean, here and elsewhere in this book, greater than equal to zero. If 0 is to be excluded, we shall say strictly positive.)

No, the set of positive integers fails the axiom A4.

b. What about the set of all integers?

No, (multiplicative inverses require rational numbers)

c. Can the answers to these questions be changed by re-defining addition or multiplication (or both)?

Interesting question, I need to thing about it a bit. This interesting, if there exists an isomorphismf :S FfromsetSand a field (F, +, .) we can define (S, +, .) onSsuch thatx + y= f1(f(x) + f(y)) andx . y= f1(f(x). f(y

3. Let m be an integer, m2, and letZm be the set of all positive integers less than m, Zm={0, 1,...,m 1}. If andfare inZmlet + be the least positive remainder obtained by dividing the (ordinary) sum of and by m, and, similarlet and be the least positive remainder obtained by dividing the (ordinary) product of and by m. (Example:m= 12, then 3 + 11= 2 and 311= 9.) (a) Prove that Zm is a field if and only if m is a prime. (b) What is -1 in Z5 ? (What is 13 in Z7?

(a) Zm is a field iffm is a prime.


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1.3. EXERCISES

Proof. Ifm is a prime, then all the axioms are satisfied, since all the axioms follow from that of natural addition anmultiplication of numbers. Ifm is a prime thenZm is a field.

If Zm is a field then an inverse for every possible scalar obtained from operations of addition and multipliction. However, ifm is not prime then, for every pair (, ) such that = 0, there are no inverses.

Ex: If m = 4, Z4 ={0, 1, 2, 3}, 2.{0, 1, 2, 3}={0, 2, 0, 2} clearly 2 does not have an inverse.

We make use of the following two facts in order to prove the above statement.

1 If mod m = 0 and m is a prime then either or is divisible by m.

Proof. The proof is straight forward, if mod m = 0 then n such that = mn m = n

but this iscontradiction unless or is divisible bym. This can be extended to any i such that

n1i mod m = 0.

2 IfZm is a field and m is not a prime, then there exist , Zm such that mod m = 0.Proof. Ifm is not prime theniZm that are prime factors ofm. From which it follows, that(, )Zsuch that mod m = 0.

Now using (1) and (2) ifm is not a prime then from (2)a pair (, ) such that mod m = 0. Now ifZmis a fiethen has an inverse 1 such that 1 = 1, Consider the product 1, the product is either or 1 dependion how the product association is done. This contracdits axiom C2, and

Zm is not a field.

A better proof would be to show and do not have inverses.

(b) What is -1 in Z5?4 : 1 + 4 mod 5 = 0

(c) What is 13 in Z7?13 really does not belong to Zm. However, 3

1 = 5 since 3 . 5 mod 7 = 1

4. The example ofZp (where p is a prime) shows that not quite all the laws of elementary arithmetic hold in fields; in Zfor instance, 1 + 1 = 0. Prove that ifFis a field, then either the result of repeatedly adding 1 to itself is always differenfrom 0, or e1se the first time that it is equal to 0 occurs when the number of summands is a prime. (The characteristic the field Fis defined to be 0 in the first case and the crucial prime in the second.)

Proof. The proof is pretty straight forward. IfZm is a field then as proved earlier m is a prime. Given+= +mod m we have 1 + 1 + 1.....n times = 0 n11 = km n = km clearly the first time the sum is zero is whk= 1n = m the first time the sum is 0.

5. LetQ(2) be the set of all real numbers of the form + 2, where, are rational (ie. , Q). (a) IsQ(2) a fiel(b) What if and are required to be integers?

Checking if the field axioms are satisfied,

A1 +

2 ++

2 = (+) + (+)

2 Q(2) since (+)Q and (+) QA2 follows from Associativity of rationals and A1

A3 = 0, = 00 + 0 . 2 Q(2) since 0QA4

+

2

2

Qsince

,

Q and +

2 + (

2) = 0

B1 , , , Q we have (+2)(+2) = + (+)2 + 2since + 2, (+) QB2 Associativity follows from associativity of rationals andB1, ,,,,,we have a = +2, b= +2, c

+

2 Q2 we have (ab)c= a(bc).(ab)c= (+ 2+ (+)

2)(+

2)

= ( + 2+ 2+ 2 ) + (+ )

2 + ( + 2)

2

= ( + 2+ 2+ 2 ) + (+ + + 2)

2 (1.3

a(bc) = (+

2)(+ 2+ ( +)

2)

=+ 2+ 2+ 2) + (+ 2+ +)

2 (1.3


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6 CHAPTER 1. FIELD

from (1.3.2) and (1.3.1) we see that associativity holds.

B3 Since 1 Q we have 1 + 02 Q2 such that +2. 1 =+2B4 The inverse, for every +

2 we have ( 2) 1

222 Q

2 such that (+

2)(( 2) 1222 ) = 1

C1 Distributive property: given ,,,,, we have

Proof.

(+

2)(+

2 ++

2) = ((+

2)(+

2) + (+

2)(+

2))

= (+ 2+ (+)

2) + (+ 2+ (+)

2)

=++ 2(+) + (+++)

2

=(++ (+)

2) +

2(++ (+)

2)

= (+

2)(++ (+)

2)

(a) what if and are integers? The inverse requires the coefficients be rational.

6 a Does the set of all polynomials with integer coefficients form a field?

Yes. The proof of addition, multiplication, commutative and associative properties of these operations, A1, A2, BB2 follow from the properties of integer addition and multiplication.

A3 The zero polynomial is the one with all zero coefficients. 0n

1xi

A4n1aixi P such thatn1aixi + n1 aixi = 0, since ifai Q ai(Q).B1-B2 given

n1aix

i,m

1 bjxj,n

1ckxk P we haven

1aixi .

m1 bjx

j P andn1aix

i(m

1 bjxj .

n1 ckx

k) = (n

1aixi .

m1 bjx

j)n

1 ckxk

B31x0 P such thatn1aixi . 1x0 =n1aixiB4 1n

1aixi

such thatn

1aixi. 1n

1aixi

= 1 but 1n1aixi

/ P The set of polynomials with integer coefficients is notfield. The same holds for real coefficients as well.

7 Let Fthe setOpof all ordered pairs (, ) of real numbers.

a If addition and multiplication are defined as

(, ) + (, ) = (+, +)

(, )(, ) = (,)

doesFbecome a field?

A1,A2 Follow from the properties of addition of real numbers.

A3 The pair (0, 0)Op such that (, ) + (0, 0) = (+ 0, + 0) = (, )A4(, )Op such that (, ) + (, ) = (0, 0) since, R.

B1,B2 These follow from the commutative and associative properties of reals. B1 follows from commutativity of reaB2 can be shown as follows,

(, )(, ) = (,)Op since, R,(, )((, )(, )) = (, )(,)

= (,) = ((), ()) using associativity of Reals

= (,)(, )

= ((, )(, ))(, )

B3 There exists (1, 1) inOp since 1 R, such that (, )(1, 1) = (, )B4 There exists ( 1

, 1

) Op such that (, )( 1 , 1 ) = (1, 1). Since 1 , 1 R, This trips you up, consider the pa(0, a), (a, 0), do not have a multiplicative inverse.


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1.3. EXERCISES

C1 The distributive property again follows from the distributive property of the reals.

(, )((, ) + (, )) = (, )(+, +)

= ((+), (+))

= (+, +)

= (,) + (,)

= (, )(, ) + (, )(, )

Since (0, a), (a, 0) fail to form an inverse this is not a field.

b If addition and multiplication are defined by,

(, ) + (, ) = (+, +)

(, )(, ) = ( ,+), is Fa field.

Yes F is a field. These are complex numbers. The properties of addition follow from the previous problem. Tadditive inverse is (, ) and the zero element is (0, 0) The properties of multiplication are as follows,

B1,B2 These follow from the fact that ,+R and therefore ( ,+)Op.B2 Givena = (, ), b= (, ) andc = (, ) we have to prove, a(bc) = (ab)c

Proof.

a(bc) = (, )((, )(, ))

= (, )( ,+)= ([ ] [ +], [ +] +[ )])= ([ ] [+], [+] +[ ]) using associativity of reals wrt multiplication= ([ ], [+])(, )= ((, )(, ))(, )

B3 There exists a unique scalar (1, 0)Op such that (, )(1, 0) = (1 0, 0 +1) = (, )

B4 There exists a unique scalar ( 12+2

, 12+2

) such that (, )( 12+2

, 12+2

) = ( 12+2 (

2+2), 12+2 (

)) = (1, 0)OpC1 Distributive property follows similarly by using a ton of algebra.

c What happens (in both preceding cases) if we consider ordered pairs of complex numbers instead?

a In the first case it will still remain a field since properties of addition follow from field properties ofC. In tcase of multiplication, since 1

x+iy C all the properties are satisfied..b It remains a field in the second case as well since the field proofs involved only properties of reals that we

field properties as well as commutativity and associativity of the operators. Since complex numbers have similproperties the field properties are retained.1

1Essentially as stated at the end of the chapter.


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8 CHAPTER 1. FIELD


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Chapter 2

Vector Spaces

2.1 Definition of Vector Space

Here we assume that we are working with a given particular field F; the scalars to be used are elements ofF.

Definition 1 (Vector Space). A vector space is a setV of elements called vectors satisfying the following axioms.

(A) To every pair, x, y Vthere corresponds a vectorx+y V, called the sum ofx andy , in such a way that,(1) addition is commutative, ie. x+y = y+x,

(2) addition is associative, ie. x+ (y+z) = (x+y) +z,

(3) there exists inV a unique vector0 (called the origin) such thatx + 0 =x for every vectorx V, and(4) x V, x|x+ (x) = 0, the vectorx is unique.

(B) To every pair Fandv V, there corresponds a vectorv inV, called the product of andv , such that,(1) multiplication by scalars is associative i.e(v) = ()v,

(2) 1x= x for every vectorx V and1 F.(C) (1) Multiplication by scalars is distributive w.r.t, vector addition,(x+y) = x+y,

(2) Multiplication by vectors is distributive w.r.t scalar addition(+)x= x+x.1

The axioms above are not necessarily logically independent. The name given to the vector space depends on the elements the field, ie. ifF {R,C,Q} it is called a real vector space, complex vector space, or rational vector space.

2.2 Exercises

1. 1. Prove that ifx and y are vectors and if is a scalar, then the following relations hold. (a) 0 +x = x. (b)0 = (c) . 0 = 0. (d) 0 . x = 0. (Observe that the same symbol is used on both sides of this equation; on the leftdenotes a scalar, on the right it denotes a vector.) (e) Ifx= 0 ,then either a= 0 or x= 0 (or both). (f)x= (1)(g)y+ (x y) = x (Herex y= x+ (y))

Proof. (a) 0 + x= x follows fom commutativity of addition and A(3).(b)0= 0 , Using A4, 0 + (0) = 0 0= 02(c) 0 = 0, using A3 we have, (x+0) = x, using C1 we have (x+0) = x+0 = x, it follows from A3 th

0= 0.

(d) 0 . x= 0. We have + 0 = (+ 0) . x= . x, using C2, we have . x+ 0 . x= . x, using A3, we ha0 . x = 0.

(e) Ifx = 0 then either = 0 or x = 0, this is pretty straight forward using (c) and (d) and the fact that the zeelement is unique in both the scalar fieldas well as vector space.

1Are the distributive properties derivable from other axioms? It does not look like it, can we prove it?2All bold case are vectors.

9


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10 CHAPTER 2. VECTOR SPACE

(f) x= (1)x, This essentially follws from field axiomsas well as C2, x + (1)x= (1 + (1))x= 0 . x = 0 x(1)x

(g) y+ (x y) = x, this follows from associativity of addition,y + (x y) = (y y) + x= 0 + x

2. Ifp is a prime, then Znp is a vector space over Zp, (cf.1, Ex. 3); how many vectors are there in this vector space?

Soln: pn

3. Let V be the set of all (ordered) pairs of real numbers. Ifx = (1, 2) and y = (1, 2) are elements ofV, write x+y(1+ 1, 2 + 2), x = (1, 0), 0 = (0, 0), x = (1, 2), Is V a vector space with respect to these definitions of tlinear operations? Why?

Proof. No,since 1x=x, 1x= (1, 0)=x

4. Sometimes a subset of a vector space is itaeIf a vector space (with respect to the linear operations already given). Considfor example, the vector space C3 and the subsets V ofC consisting of those vectors (1, 2, 3) for which (a) 1 R (1 = 0 (c) either 1 = 0 or 2 = 0 (d) 1+2 = 0 (e) 1+2 = 1 In which of these cases is V a vector space?

Proof. a. x + yVx, yVsince1+1 R.a.A1-A2 Since (1, 2, 3) Vthe properties A1 A2 follow from the field properties of reals and complex numbers.

a.A3 0= (0, 0, 0) Vsince 0 R,C and x + 0= x.a.A4x= (1, 2, 3) x= (1, 2, 3) Vsuch thatx + (x) = 0, since1 R and 2, 3 C.

a.B xVsince (1, 2, 3) = (1, 2, 3) and 1 Rwhich follows from field properties ofR and 2, 3 C,a.B1,B2 These follow from the field properties ofR,C.

a.C1,C2 These again follow from the fact thatV C3. Vis a vector space

b. Proof. The set V ={(0, x2, x3); x2, x3 C}is a vector space sincea. x, yV, x + yVsince x + y= (0, x2+y2, x3+y3)C3 and x1 = 0,

A1,A2 Follow from the vector space properties ofC3 and field properties ofC,

A3 0= (0, 0, 0) Vsince x1 = 0 and (0, 0, 0)C3,A4 xV; x= (0, x2, x3) C3; (0, x1, x2) + (0, x1, x2) = 0 x V,

B1-C3 Follow from properties ofC3 and the fact that x V xC3.

c either x1 = 0 or x2 = 0 (interpreting this as x1 and x2 cannot both be zero. This is not a vector space. Since 0 / V.d V ={x C3; x1 + x2 = 0} is a vector space since, (0, 0, 0) V and the remaining properties follow from the fact th

xV x C3.e V ={xC3; x1+x2 = 1}is not a vector space as the vector 0 = (0, 0, 0) / V.

2.3 Linear Dependence

Definition 2 (Vector Sum Notation). Vector sums are denoted byn

1xi over the set{xi} of vectors, ie. eachxi is a vectorDefinition 3 (Sum of Zero Vectors). In the case that the set{xi}=the sum

i xi = 0

Definition 4 (Linear Dependence). A finite set{xi} of vectors is linearly dependent if there exists a corresponding set{i} scalars, not all0, such that,

i

ixi = 0

, If on the other hand, the sum

iixi = 0i = 0 for eachi, the set{xi} is linearly independent.


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2.4. LINEAR COMBINATIONS

2.3.1 Some observations and Commentary

1. The definition (4) is intended to cover the case of the empty set. If there are no indices i, it is not possible to pick osome of them and assign to the selected vectors a non-zero scalar so as to make the sum vanish. Rephrasing the definiti

iixi = 0i = 0 as

iixi = 0 i such thati= 0. The conclustion is thereforeempty set of vectorsis linearindependent.

2. Linear dependence and independence are properties ofsets of vectors. However, they are used as an adjective to vecto

themselves. Ex: a set of linearly independent vectors instead of linearly independent set of vectors.3. We say that an infinite set of vectors X is linearly independent if every finite subset ofX is linearly indipendent.

4. Ifx, yC1 the the set{x, y} is a linearly dependent set of vectors. Sincexy+ (x)y = 0 since x, yC, the producare defined. Therefore any set containing more than 2 elements in C1 is linearly dependent set.

5. The set of polynomialsPis interesting. The finite set of vectors{1 t, t(1 t), 1 t2} are linearly dependent. (1 t)t(1 t) 1 +t2 = 0, However, the infinite set{1, t , t2, t3, .....} is a linearly independent set of vectors.

Proof.

0+1t+2t2 +3t

3 +......= 0n0

iti = 0i= 0

Since none of the ti can be expressed as a linear combination of the others, since|ti|


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2.5.1 Examples Observation and Comments

1. The setBP={tn; n {0, 1, 2, }} forms the basis of the set of polynomialsP. As seenBPforms an infinite basis for tsetPfor if there was a finite basis of size n, tn+1 such that, tn+1 is not a linear combination ofti; i {1, 2, 3, , n}

2. An example of a basis in Cn is the set of vectors xi, i {1, 2, 3, , n}, defined by the condition that the j th coordinate xi is ij . This is true since

(a) The vectors xi = ij are linearly independent, since if

iixi = 0 1(1, 0, 0, , 0) + 2(0, 1, 0, , 0) + n(0, 0, 0 , 1) = (1, 2, , n) = 0i = 0(b) any vectorx ={x1, x2, x3, , xn} can be written as (x1, 0, 0, , 0), (0, x2, 0, , 0), , (0, 0, , xn).(c) In general for a vector space Fn it can be seen that the basis is again of the form xji = ij, where x

ji is the j

component ofx.

2.5.2 Uniqueness

Theorem 8 (Linear Combination Uniqueness using Basis). In a general finite dimensional vector spaceV, with basis BV{x1, x2, , xn}. Every vectorx Vcan be uniquely represented as a linear combination of the basis vectorsBV.Proof. If there are two representations of a vector x V;

n1ixi and

n1ixi. We have by subtracting each other,

n1ixin

1ixi = x x= 0n

1 (i i)xi = 0 since the xi are linearly independent, we haven

1 (i i) = 0i i = 0i = iTheorem 9 (Extension of linearly independent set to Basis). IfV is a finite dimensional vector space and if{y1, , yn}any set of linearly independent vectors inV, then, unless the ys already form a basis, we can find vectorsym+1,....,ym+p that the totality ofys. ie.{y1, , ym, ym+1, , ym+p} form the basis for the vector spaceV. In other words, every lineaindependent set of vectors of a vector space can be extended to a basis.

Proof. The proof starts of with the assumption that we already know an existing basis {x1, x2, , xn}, this follows from tfact that ever vector space has a basis3. If we have a set of linearly independent vectors, then we can augment it with a vectfrom the existing basis, if the new augmented set is linearly independent we add the next vector from the basis and continutill we exhaust the basis. If the augmented set is not linearly independent we discard the vector from the basis and continwith the remander of the vectors in the basis. Once we have exhausted all the vectors in the basis we are left with a new bas

The proof that the new set will span the vector space is fairly straight forward, since every vector from the basis that was na linear combination of the original set is included in the new basis and every vector from the basis that was not included islinear combination of the vectors in new basis. Essentially we can go from our new basis to the old basis and from there spthe entire vector space.

2.6 Exercises

1a. Prove that the four vectorsx = (1, 0, 0), y = (0, 1, 0), z = (0, 0, 1), u = (1, 1, 1) in C3 form a linearly dependent set, bany three of them are linearly independent. To prove that a set of vectors are linearly dependent solve the linear systeof equations resulting from the equation x +y+x= 0.

Proof. (a) (1, 1, 1) = (1, 0, 0) + (0, 1, 0) + (0, 0, 1)

(b) The equations resulting from (1, 0, 0) +(0, 1, 0) +(0, 0, 1) = 0(,,) = 0 = 0, = 0, = 0.3The proof follows from Zorns Lemma, which states that for any set M= if every set C Mhas an upper bound then Mhas a maximal eleme

(may not be unique). Our proof now starts with setM = {set all the subsets ofV that are linearly independent}. This setM = since this set hatleast one vector. The set Mcan be ordered using the ordering if A, B M; A B if A B (note there are sets such that A, B M and adisjoint). Every set A Mhas an upper bound since A M; A V and the upper bound in the worst case will be V. Using Zorns lemmaM ha maximal element. Our claim is that this maximal element BV is the basis of the vector space V. IfBV does not span Vthen there exists a vectx Vthat is linearly independent ofBV and x /BV. This contradicts Zorns lemma ofBV being the maximal element. This proof is a bit wickThe question is why would a basis ever span the entire space V? The basis is actually defined as that set that spans the entire space so the questiis really not about whether or not it spans but it is not a basis if it does not span the vector space. The proof that there is a basis for every vecspace is purely based on well ordering and zorns lemma. At first sight it seems as though the notion of linear combination is never used, however tcontradiction to Zorns lemma is actually obtained by using the fact that either a vector v Vis either a linear combination of the vectors in the sBV or the v BV.


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2.6. EXERCISES

(c) The equations resulting from(1, 0, 0) + (0, 1, 0) + (1, 1, 1) = 0( + , + , ) = 0= 0, = 0, = 0. Trest are proved similarly.

1b. If the vectorsx, y, z, u inPare defined by x(t) = 1, y(t) =t, z(t) =t2, u(t) = 1 +t+t2, prove that x, y, z, u are lineardependent, but any three of them are linearly dependent.

Proof. 1b.1 u= x + y+ z the vectors x, y, z, u are linearly dependent.1b.2 x +y+z= 0 t +t+t2 =0, ,= 0Similarly one can prove the independence of the other combinations of vectors.

2. Prove that ifR is considered as a rational vector space (see3, (8)), then a necessary and sufficient condition that tvectors 1 and in R be linearly independent is that the real number be irrational.

Proof. This is an interesting problem. It sheds some light on the impact on the basis the choice of the underlying fiefrom which the scalars come from. R with scalars coming from R has just any real number as its basis. Usually 1 woudo it, since the scalar spans the entired vector space. If the field we choose from changes then the basis vector changes. Ithe case of the vector space (R,Q) our basis is (1, ) where is irrational.4

Given , Q, + = 0 always has the rational solution = if Q has to be irrational. On the flside if is irrational += 0 does not have solutions for , Q other than = 0, = 0.

3. Is it true that ifx, y and z linearly independent vectors, then so also are, x + y, y+ z, andz + x ?

Proof. If x, y and z are LI then ,,= 0; x + y + z = 0 we have (x + y) + (y+ z) + (z + x) = 0(+)x + (+)y+ (+)z= 0x + y, y+ z, z + xare linearly independent.

4a Under what conditions on the scalar are vectors (1 +, 1 ) and (1 , 1 +) C2 linearly dependent.

(1+, 1)+(1, 1+) = ((+)+(), (+)+()) = 0(+)+()= 0, (+)+()= 0=and =.

1 + 1 1 1 +

=

00

The condition for linear dependence is that, thedet

1 + 1 1 1 +

= 0(1 +)2 (1 )2 = 04= 0= 0

4b Under what conditions on the scalar are the vectors (, 1, 0), (1, , 1), (0, 1, ) R3 linearly dependent?

= 0(0, 1, 0), (1, 0, 1), (0, 1, 0) are linearly dependent. However we are looking for a more generic non-zero solution the system,

1 01 1

0 1

=

00

0

(2 1) = 0(2 2) = 0= 0, =

2

4c What is the answer to (b) for Q3(in place ofR3)?

= 0

4This is not true, your answer is wrong. (1, ) is not a basis of Rational Vector Space R.


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5a The vectors (1, 2) and (1, 2) in C2 are linearly dependent iff12 = 21

Proof. The vectors are linearly dependent if,, = 0 such that

1+1 = 0

2+2 = 0

solving for (, ),

(21

1 2) = 0, = 0 21

1 2 = 021 = 12

using the determinant methodology,

1 12 2

=

00

1 1

2 2= 021 = 12

On the other side, If21 = 12 then we have the following vectors (1, 2) and (1, 2) choose = 22 , = 1 clear(1, 2) +(1, 2) = 0

5b Find a similar necessary and sufficient condition for the linear dependence of two vectors in C3. Do the same for thrvectors inC3.

Proof. If two vectors are linearly dependent in C3

then the following equations should hold,1 12 2

3 3

=

00

0

Solving pairs of equations we end up with

12 = 21, 23 = 32,

now if =22

, = 1, we have (1, 2, 3) = (1, 2, 232 ) = (1, 2, 3), Now the other side of the iff, is easshown by the choice of, given the condition holds. For 3 vectors in C3 we have the following,

1 1 12 2 23 3 3

=

0001(23 23) 1(23 32) +1(13 32) = 0

23 = 23, 23 = 32, 23 = 32 (2.6

5c Is there a set of 3 linearly independent vectors in C2? No. Consider (x1, x2), (y1, y2), (z1, z2) C2 if they are linearindependent then,

x1 y1x2 y2

= 0

x1 y1x2 y2

is invertible. Hence there exists solutions to the equation

x1 y1x2 y2

=

z1z2

x1+y1= z1

x2+y2= z2 = z1y2z2y1

x1y2x2y1,

x1z2x2z1x1y2x2y1

.

6a Under what conditions on the scalars , are the vectors (1, ) and (1, ) in C2 linearly dependent.

If (1, ), (1, ) are linearly dependent then

1 1

is not invertible= from the condition for non- invertibility.

6b Under what conditions on the scalars , , are vectors (1, , 2), (1, , 2), (1, , 2) inC3 are linearly dependent.From the earlier condition from (2.6.1) in C3 we have, 2 =2, 2 =2,2 =2= , = , =


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2.6. EXERCISES

6c Guess and prove a generalization of (a) and (b) toCn. InCn the condition for linear dependency is,

Proof.

1 1 1 11 2 3 n

n11 n12 n1 n1nn1

n2

n3 nn

=0n1j ni =n1i nj1 = 2 = 3 = n

7a Find two bases in C4 such that the only vectors in common to both are (0, 0, 1, 1) and (1, 1, 0, 0)Soln: (0, 0, 1, 1)(1, 1, 0, 0)(1, 0, 0, 0)(0, 0, 0, 1) and (0, 1, 0, 0)(1, 1, 0, 0), (0, 0, 1, 0)(0, 0, 1, 1).

7b Find two bases inC4 that have no vectors in common so that one of them contains the vectors (1, 0, 0, 0) and (1, 1, 0, and the other one contains the vectors (1, 1, 1, 0) and (1, 1, 1, I).Soln: (1,0,0,0) (1,1,0,0), (0,0,1,0), (0,0,0,1) and (1,1,1,0), (1,1,1,1),(1,0,0,0),(0,1,0,0)

8a Under what conditions on the do the vectors (1, 1, 1) and (1, , 2) form a basis ofC3?Soln Two vectors cannot form basis ofC3. The other approach is that ever subset of a basis has to be linearly independe

1 11

1 2

=

00

0

= 1

8b Under what conditions on the scalar do the vectors (0, 1, ), (, 0, 1), and (, 1, 1 + ) form a basis ofC3?Soln

0 1 0 1 1 1 +

=00

0

(1 + ) += 0

0 = 0

The vectors will not form a basis for any value ofsince the condition for linear dependency, holds for these set of vectofor any value of. Since

01

+

0

1

=

1

1 +

9 Consider the set of all those veetors in C3 each of whose coordinates is either 0 or 1; how many different bases does thset contain? Soln : The vectors are (0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1),

(a){(0, 0, 1), (0, 1, 0), (1, 0, 0)}, {(0, 0, 1), (0, 1, 0), (1, 0, 1)}, {(0, 0, 1), (0, 1, 0), (1, 1, 1)}, {(0, 0, 1), (0, 1, 0), (1, 1, 0)} You cgather the other sets similarly.

10 IfX

is the set consisting of the six vectors (1, 1,0,0), (1,0, 1,0), (I, 0, 0, I), (0, 1, 1, 0), (0, 1, 0, 1), (0, 0, 1, 1) in C4 find twdifferent maximal linearly independent subsets ofX(A maximal linearly independent subset ofX is a linearly independesubsetY ofXthat becomes linearly dependent every time that a vector ofXthat is not already inYis adjoined toY.Soln : Any set of 4 linearly independent vectors forms a basis in C4. Since the basis is a maximal element of the linearindependent subsets ofX. We have,

(a){(1, 1, 0, 0), (1, 0, 1, 0), (0, 1, 1, 0), (0, 1, 0, 1)}, since

a

1100

+b

1010

+c

0110

+d

0101

=0

a+ba+c+d

b+cd

=0d = 0, b=c, a=c, a=b


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is a contradiction since a =c and a = +c, this is possible only if c = 0 a = 0, b = 0, c = 0, d = 0 tvectors are linearly independent. It is a maximal set can be proved by adding the vector (0, 0, 1, 1) to the set, {(1, 1, 0, 0), (1, 0, 1, 0), (0, 1, 1, 0), (0, 1, 0, 1), (0, 0, 1, 1)},

a

1100

+b

1010

+c

0110

+d

0101

+e

0011

=0

a+ba+c+db+c+e

d+e

=0

e=d, a=b, b+c= d, b+c=dc = 0, b= d, e=d, a=b

ifb = 1, d= 1, c= 0, e=1, a=1 we have (1, 1, 0, 0) + (1, 0, 1, 0) + (0, 1, 0, 1) + (0, 0, 1, 1) = 0(b){(1, 1, 0, 0), (0, 1, 1, 0), (0, 0, 1, 1), (0, 1, 0, 1)}. The proof of linear independence and maximality are done as before.

11 Prove that every vector space has a basis. (The proof of this fact is out of reach for those not acquainted with somtransfinite trickery, such as well-ordering or Zorns lemma.)

Proof. See the footnote under Theorem 9, uniqueness section.6

2.6.1 The impact of the set of Scalars on the Vector Space

1. The conditions for linear dependence will vary as seen in some of the problems in the problems section. The choice ofQ R had a significant impact on the range of values the elements of the vectors could take in order to be linearly independen

2. The basis vector sets are different.

2.7 Dimension

Theorem 10 (Cardinality of Basis). The number of elements in any basis of a finite dimensional vector spaceV is the same any other basis.

Proof. By definition and the proof that there exists a basis for every vector space indicates that it is a maximal element of the sof linearly independent vectors and the fact that adding another vector makes the set linearly dependent implies this theoremLet us say there are two basis

B1V

,

B2V

such that the

|B1V

| =

|B2V

|5. Without loosing generality let

|B1V

|0, n >0 and mn. Let M ={1, , n} Cn; 1 = 2 = = m= 0.4. Given m >0, n >0 such thatmn, the spacePn, and any m real numbers t1, , tm. Let

M ={x Pn: x(t1), x(t2), x(tm) = 0}.

5. Let M ={x Pn: x(t) = x(t)} is a subspace.

2.8.2 Terminology for working with Subspaces

1. Given any collections{Mp} of subsets of a gieen set, we write

pMp for the intersection of all Mp, i.e., for the set points commong to all Mp.

2. IfM,N are subsets of a set, we write MN, ifM is a subset ofN, i.e. every element inM is an element ofN. We not exclude the possibility that M = N; thusV Vas well as 0v V7.

3. Two subspaces M,Nare disjoint ifM N =0v.

2.8.3 Calculus of Subspaces

Theorem 18 (Subspace Intersection). The intersection of any collection of subspaces is a subspace.

Proof. This is easy to see.

A x, ypMp, x+ypMp, since each of the Mp is a vector space and x +yMp p.A1-A2 This follows from the properties of vectors in a vector space and x, ypMpx, yMpp

A3p 0Mp0

pMp

A4

p, xMp

p

xMp,

x,

xpMp andx x= 0.

B To every pair (, x), where is a scalar and xpMp, we havexpMp, since xMpp.B1-B2 Follow from the properties of vectors inMp.

C1-C2 Follow from the properties of vectors inMp.

Definition 19 (Span of subset of vectors). If S is an arbitrary set of vectors from a vector spaceV, then the intersection every subspace ofVthat contains all the vectors ofS is called the span ofS, i.e. span(S) =pMp, S Mp.Theorem 20 (Span of Vectors and Subspaces). IfS is any set of vectors in a vector spaceVand ifM is the subspace spannbyS, thenM is the same as the set of all linear combinations of elements ofS.

Proof. From definition (19) the span of S is intersection of all subspaces that contain S. The span of S is the smallest scontaining all the elements ofS. Givenx, yS; x + yM, since M is a vector space. ConsiderMl={z; z= x + y, x, yS, (, ) F}. We would like to show that Ml= M. It is clear from the definition that MlM.IfzM andz / Ml, thenz= x+y, (x, y) V, (x, y) /S. But this is a contradiction since, MlMs ={Mp; (x, y)S, xyMp}, and Ml

pMp= M Ml= M.

Halmoss proof is a bit different. His claim is that since the set of all linear combinations of elements of S is a subspcontainingS and this contains M. On the flip side M containsS and is a subspace which implies it contains all linear combintions of vectors ofS. The issue I have is with the first conclusion. Why would the subspace containing all linear combinatioelements ofS contain M.

70v is the null vector space or the zero vector space


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Theorem 21 (Sum of Supspaces). IfH andKare two subspaces and ifM is the subspace spanned byH andK together, thM is the same as the set of all vectors of the formx + y, withx H andy K.Proof. This follows from Theorem 20and the fact that every vector in a vector space is a linear combination of vectors in thbasis.

2.8.4 Compliment

1 The notation H+ Kwill be used to represent the subspace M spanned by H and K.

2 We shall say a subspace Kof vector spaceVis a complement of a subspace KifH K=O8 andH+ K=V.Theorem 22 (Subspace Compliment). Every subspace has a complement.

Proof. Every subspace is a vector space. If the vector space is the entire spaceV then the complement isO. SinceV O=andV+O=V. For any other subspaceS, it has a basis, which is a subset of the basisBV. The subspace spanned by the relaticomplimentBV/BSis clearly a complement of the space spanned byBS. Since if it is not, it would mean the basis vectors wenot linearly independent.

2.9 Dimension of a subspace

Theorem 23 (Dimension of a Subspace). A subspaceM

in ann-dimensional vector spaceVis a vector space of dimensionProof. If the subspace is equal to the vector spaceVthen the dimension of the subspace is n. If not the subspaceS V, tbasisBS spansSand has to have a dimension less than or equal to the dimension of the basisBV. If not it would contradthe fact thatBV spansVand thatBVis the maximal set that spansV, ie. every setn + 1 vectors is linearly dependent.Halmos claims that the above proof does not prove that there exists a basis if at all. However, we have already proved that evervector space has a basis and this basis is a maximal set. His argument is as follows, IfM =Othen the only vector spanning tspace is 0, and M is 0-dimensional. IfM contains a non-zero vector x1 then, let M1(M) be the subspace spanned by x1. M1 = Mthen M is one-dimensional. IfM1= M, we find a subspace spanned by vectors x1, x2. The same logic now appliesM2 is the subspace spanned by the two chosen vectors, then ifM2 = MthenM is 2-dimensional, if not we continue the proceit will end at n steps since we cannot find n+ 1 linearly independent vectors. Halmos likes this second proof better as it donot make the assumption of an existing basis. The proof purely relies onTheorem 6, which does not require the assumptiof existence of a basis. However, looking into the proof carefully, theTheorem10on which the proof ofn+ 1 vectors formlinearly dependent set already assumes that there exists a basis. If you want to talk about basis of a vector space, one has to shothat it exists and it is a valid and a possible concept, there is no way of getting around the fact that the existence of a basis halready been proved. On second thought it may be because it is a constructive proof as opposed to an existence proof. The othissue is how would you verify that the subspace and the vector space are the same at each iteration?.

As a consequence ofTheorem 23we have,

Theorem 24 (Subspace Basis). Given anym-dimensional subspaceM in ann-dimensional vector spaceV, we can find a bas{x1, , xm, xm+1, , xn} inV, so that{x1, , xm} are inM and form, therfore, a basis ofM.Proof. This proof is pretty straight forward. From Theorem23we havedim(M)dim(V), and therefore the basis {x1, , xof the vector spaceV, spans M. Since M is a vector space, there exists a subset of the basis {x1, , xn} that spans M, if nit is a contradiction to our assumption that M V, and a subspace ofV.

2.9.1 Exercises

1. ifM,Nare finite-dimensional subspaces with the same dimension, and M Nthen M = N.

Proof. Given dim(M) =dim(N) and MN, we have the basis{y1, , yn} of the subspace Nspans M, and since tdimension ofM =n, consider{x1, , xn} M and is a basis ofM. Since{x1, , xn} N and are linearly independenwe can extend it to form basis ofN (Theorem9). Now the dimension of{x1, , xn} Nis n, so they already must a basis (Theorem12)ofN M = N9.

8Note that O is a zero vector space and not the zero vector 09Happy with this proof , better than the first iteration. One cannot just claim that{y1, , yn} M and hence its basis. Here we have show

that N M., by trying to extend the basis ofM to a basis ofN.


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2.9. DIMENSION OF A SUBSPACE

2. IfM,N are subspaces of a vectors spaceV, and if every vector inVbelongs to M or N or both, then either M =VN =V or both.

Proof. This is pretty straight forward. givenM,N Vconsider the basisBM={x1, , xn} M, BN ={y1, , yn}N. If it is true that every vector ofV Mor N or both. Then every vector ofV is a linear combination of the baBM, BN. SinceM/N are subspaces ofVwe can apply Theorem 9 to extend the basis from M/N toV. If every vectv Vcan be expressed as

n1aixi, clearlyxi form a basis ofV M =Vsimilarly if every vector v V can be express

asn1

biyi then, yi form a basis ofV

N =V

. If both are true then the third result follows.

3. Ifx, y, and z are vectors such that x + y+ z= 0, thenx and y span the same subspace as y and z.

Proof. LetSxy be the subspace spanned byx, yandSyz the subspace spanned by y, z. Every vectorax + by= a(z y)by= az + (b a)y Sxy=Syz =Szx .

4. Suppose x, y be vectors and Mbe a subspace in a vector spaceV, let H be the subspace spanned by Mand x, and letbe the subspace spanned by M and y. Prove that ifyH but y / Mthen x K.

Proof. Given,H =SMx={z= ax + bw, x V, wM}K=SMy ={z= ay+ bw, y V, wM}

We need to prove, if y / M but y H then x K. The proof is pretty straight forward. If y / M but y h, thy= ax + bwwherey, x VandwM. Also,x = 1

ay b

awx K. Essentially the point to be made is that ifx, y

andwMare linearly dependent, then x, yH,K.

5. Suppose that L,M,N, are subspaces of a vector space.

a Show that the equationL (M+N) = L M+ L Nis not necessarily true.Proof. Consider a vectorl L andl / M, or N or both, andl= am+bn, wherem M, n N clearly,l L(M+Nandl / L M+L N. The question is that of existence of such a vectorl. We can choose subspaces M,Nsuch thM N=O (easy to construct using a subset of a basis ofVand its complement). Choose a subspaceL V. Trest follows. Essentially, intersection is not distributive relative to the + operation.

b Prove thatL (M+ (L N)) = (L M) + (L N)Proof. We have

U= (M+ (L N)) ={x; x= y+ z; yM, z L N}L M+ L N ={x; x= y+ z, y L M, z L N}

L U={x; x L,M+ L N}We will first prove a lemma and use it to prove the above:

Lemma 25 (Basis of Intersection and Union of Subspaces). IfL,M are subspaces of the same vector spaceV thethe basis of the spaceL +M =BLBM and of the intersection subspaceL M =BLBM, whereBL, BM are subseof the basisBVof the vector spaceVspanning the appropriate subspaces.10

Proof. The proof is pretty straight forward. LetBV={v1, , vn} andBL ={vi, , vl} andBM ={vj , , vmGivenx L+M,

x= y+ z ; y L, zMx=

i

ivi+j

jvj ; vi BL, vj BM

x=k

ckvk where vk BL orBMvk BL BM

butvk are linearly independent and span the spacevk BL BM is the basis ofL +M.10The vector spaces need not be finite but countable.


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similarly,

ifx L M x L and xMx=

i

aivi; vi BL ; x=j

bjvj; vj BMi

aivij

bjvj = 0

ifBL, BM are not disjoint then L M =O,i

aivij

bjvj =

BLBMk

(ak bk)vk+BLBM

i

aiviBMBL

j

bjvj = 0

since the vectors are linearly independent ak bk = 0, ai = 0, bj = 0. BLM=BL BM

Now the proof of the problem is pretty straight forward, the basis ofL (M+ (L N)) is given by:BU=BM (BL BN), andBLU =BL (BM (BL BN) = (BL BM) (BL BN),the basis of (LM) + (LN) is (BL) BM(BL BN), Similarly the basisB(LM)+(LN) =BLM BLN (BL BM) (BL BN), ie. the basis vectors are the same.

A simpler proof is that given x L (M+ (L N)x L,M+ (L N), since we have,x= y+ z; yM, z L but, x, z L, y= z xy L y L M. Conversely, ifx(L M) + (L N)x = y + z; y L M, zL N,y, z L, x L similarly, x = y+ z, yM, z L N xM+ (L N)x L (M+ (L N)).

6. A polynomial x is called even ifx(t) = x(t) identically in t (see10, 3, and it is called odd ifx(t) =x(t).

a Both class M of even polynomials and the class N of odd polynomials are subspaces of the spacePof all complpolynomials.

Proof. We need to prove that even and odd polynomials form a vector space.

i. Over the field of complex numbers clearlyx(t) + y(t)M x(t) + y(t) = x(t) + x(t).

A1-A2 The commutative and associative properties follow from the properties of general polynomials.A3 There exists a zero polynomial0 such thatx(t) + 0= x(t).

A4x(t) x(t)|x(t) + x(t) = 0, note thatx(t)Msincex(t) =1x(t)B To every pair Fand p(t) P, P P, sincep(t) = p(t).

B1-B2 The associative property follows from the properties of general polynomials.

C1-C2 These follow from the properties of general polynomials.

The above proof can be applied to show that odd polynomials are a subspace as well.

b Prove thatM and N are each others complements.

Proof. The first part is easy, i.e. M

N =0. The second part is to show that

p

P; p(t) = o(t) + e(t) this is ea

to show by considering the basis of even and odd polynomials.

6a Can it happen that a non-trivial subspace of a vector spaceV(i.e., a subspace different from bothOandV) has a uniqcomplement?Soln : Yes, it is possible I suppose. However the construction does not seem obvious. Clearly examples with non-uniqcomplements can be easily given. Consider in R2, the subspaces spanned by (0, 1), (1, 1) as well as (0, 1), (2, 1) is tentire space R2 and the subspaces are complementary to each other and non-unique. The question is an example fcomplementary subspaces where they are unique. One case where this is true is if the basisBV only one complementaset of subspaces, as in the case of polynomials of degreen.

6b IfM is an m-dimensional subspace in an n-dimensional vector space, then every complement ofM has dimension n m


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2.10. DUAL SPACES

Proof. The proof is pretty straight forward. Given a m-dimensional subspace M of a vector space its complement Nsuch that M N=O, M+N={x + y; xM, yN} span the vector spaceV. Clearly, given a basisBVof the vectspaceV, one can find a basisBM ={v1, , vm} BV such thatBM is the basis ofM, similarly one can find the basBN ={v1, , vn} BVand is a basis ofN. Since M

N =O, we haveBM

BN =. Using the fact that M +Nspathe vector space, we can see that dim(M+N) =dim(V) =dim(M) +dim(N). The result follows. The question is wis dim(M+N) = dim(V), Isnt that what we are trying to prove?

Approaching this a bit differently, since every vectorv = x +y: xM, y

N

v =

iaivi

+jbj

vj

dim(M+N)dim(V) but dim(M+N) = dim(M) +dim(N) since M

N =O. The result follows.

7a Show that if both M and N are three-dimensional subspaces of a five- dimensional vector space, then M and N are ndisjoint.

Proof. This follows from problem 6b. The dim(M) + dim(N) > dim(V) our argument is going to be similar to targument for problem 6b. The basis for M, BM ={vi BV; m M, m=

3

1vi}, similarly the basis for N, BN

{vi BV; n N, n =3

1vj}, now the given that the dimension dim(BM) = 3 and dim(BN) = 3 we have dim(BV)dim(BM) +dim(BN) BM

BN= MN =.7b IfM and N, are finite-dimensional subspaces of a vector space, thendim(M) +dim(N) = dim(M+N) +dim(M

N)

Proof. The proof follows from the lemma 25 we proved earlier. The basisBM+N =BMBN andBMN =BMBClearlyBM = (BM

BN)/BNBMN, Since (BMBN)/BNBMN =, we have card(BM) = card(BMBN)card(BN)+card(BMN)card(BM)+card(BN) = card(BM

BN)+card(BMN)dim(M)+dim(N) = dim(M+Ndim(M

N).

Corollary: Given an l-dimensional vector spaceVand m,n-dimensional vector spaces M and N such that m+n>l thM N O

Proof. The argument is similar to the argument given in problem (7a). Consider a basis of the vector spaceV now,BV{v1, , vn}. Since M,Nare subspaces ofVfrom theorem24there exist subsetsBM, BN BVsuch that the subsets fora basis of the respective subspaces. It follows that,card(BM) + card(BN)> card(BV) andBM, BN BV BM

BN=The conculsion follows.

2.10 Dual Spaces

Definition 26. A linear functional on a vector spaceV is a scalar-valued function y defined for every vector x, with tproperty that (identically in the vectorsx1 andx2 and the scalars1 and2) i.e: y:VF F.

y(

21

ixi) =

21

iy(xi)

Some examples and observations about linear functionals,

1. for x = (x1, , xn) in Cn, write y(x) = x1. More generally if{a1, , an} F the function y(x) =n

1 aixi islinear functional. Clearly y(1x1+ 2x2) = y(w) =

n1aiwi =

n1ai(1x1i+ 2x2i) = 1

n1 aix1i+ 2

n1 aix2i

1y(x1) +2y(x2)

2. We observe that y(0) = y(0.0) = 0.y(0) = 0. This is the reason why linear functionals are called homogeneous. particular inFn, ify is defined by y(x) = b+ n1 aixi is a linear functional iffb = 0.

3. For any polynomial x P, we write y(x) = x(0) is a linear functional. In general given scalars{a1, , an} areal numbers{t1, , tn} the function y(x) =

n1aix(ti) is a linear functional. This is pretty straight forward to se

y(x1 + x2) =n

1 ai(x1(ti) + x2(ti)) = n

1aix1(ti) + n

1 aix2(ti) = y(x1) + y(x2). A limiting case of the abois obtained as follows. Let (a, b) be any finite interval on the real t-axis, and be any complex-valued integrable functi

defined on (a, b); define y by y (x) =ba

(t)x(t)dt. The proof for the integral form is similar.

4. On an arbitrary vector spaceV, define y by writing, y (x) = 0x V. Clearly y is a linear functional.


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Definition 27. The setVF

, of all linear functionals on a vector spaceVF (which includes zero-functional defined in (4)) forma vector space.V is called the dual ofV.Proof. Given two linear functionals (y1, y2) on a vector space VF, any linear combinationy = y1+y2 is also a linear functionwhere1, 2 F. Consider [ax1+bx2, y] = [ax1+bx2, y1+y2] = [ax1+bx2, y1] + [ax1+bx2, y2] = [(ax1+bx2), y1][(ax1+bx2), y2] =a([x1, y1] +[x1, y2]) +b([x2, y1] +[x2, y2]) = a[x1, y1+y2] +b[x2, y1+y2] =a[x1, y] +b[x2, y]

A1-A2 Follows from the fact that the linear functional is scalar valued.

A3 The linear functional y0(x) = 0x V y(x) +y0(x) = y(x)y VFA4y((x)) VF, y(x) = y(1x) VF andy (x) +y(1x) = y(x) y(x) = 0.

B1-B2 These follow directly from the linearity property of linear functionals.

C1-C2 These properties follow from the linearity property of linear functionals.

We will use the following notation to indicate a linear functional, y(x) = [x, y]. With the new conventions we have the followin

1. [1x1+2x2, y] =1[x1, y] +2[x2, y]

2. [x, 1y1+2y2] =1[x, y1] +2[x, y2]

The relations1, 2are together expressed by saying that[x, y] is a bilinear functional of vectorsx V andy V.

2.10.1 Exercises

1. Consider the set C of complex numbers as a real vector space (88 in 3, (9)). Suppose that for each x = 1+i2(where and 1 and2 are real numbers and i =

1). Which of the y defined below are linear functionals,a y(x) = 1, is a linear functional since [ax1+bx2, y] = [(a1+b1) +i(a2+b2), y] =a1+b1 = a[x1, y] +b[x2, y]

b y(x) = 2, is a linear functional since [ax1+bx2, y] = [(a1+b1) +i(a2+b2), y] =a2+b2 = a[x1, y] +b[x2, y]

c y(x) = 21 , is not a linear functional,[ax1+ bx2, y] = [(a1+ b1) + i(a2+ b2), y] = (a1+ b1)2 =a[x1, y] + b[x2, y]

a21 + b21

d y(x) = 1i2is not a linear functional since, [ax1+bx2, y] = [(a1+b1)+i(a2+b2), y] = (a1+b1)i(a2+b2) /e y(x) =

21 +

22 , is not a linear functional since, [ax1+bx2, y] = [(a1+b1) +i(a2+b2), y] =

(a1+b1)2 + (a2+b2)2 =a[x1, y] +b[x2, y] =a

21+ 22+ b

(21+

22

2. Suppose that for each x = (x1, x2, x3) C3 the function y is defined as below, which ones are linear functionals.(a) y(x) = x1 + x2, is a linear functional since,[ax1 + bx2, y] = [((ax1,1 + bx2,1), (ax1,2 + bx2,2), (ax1,3 + bx2,3)), y]

ax1,1+bx2,1+ax1,2+bx2,2 = a[x1, y] +b[x2, y]

(b) y(x) = x1x2, is a linear functional since, [ax1+ bx2, y] = [((ax1,1 + bx2,1), (ax1,2 + bx2,2), (ax1,3 + bx2,3)), y] ax1,1+bx2,1 ax1,2 bx2,2 = a[x1, y] +b[x2, y]

(c) y(x) =x1+ 1, is not a linear functional since, [ax1+bx2, y] = [((ax1,1+bx2,1), (ax1,2+bx2,2), (ax1,3+bx2,3)), y] ax1,1+bx2,1+ 1=a[x1, y] +b[x2, y] =ax1,1+ 1 + bx2,1+ 1, also note that [0, y] = 1.11

(d) y(x) = x12x2 + 3x3, is a linear functional since, [ax1 + bx2, y] = [((ax1,1 + bx2,1), (ax1,2 + bx2,2), (ax1,3 + bx2,3)), y]ax1,1 + bx2,1 2(ax1,2 + bx2,2)+3(ax1,3 + bx2,3) = a[x1, y] + b[x2, y] =ax1,1 2ax1,2 + 3ax1,3 + bx2,1 2bx2,2 + 3bx23. Suppose for each x P the function y is defined as below, which of them are linear functionals?

(a) y(x) =21

x(t)dtis a linear functional since [ax1+bx2, y] =21[ax1(t) +bx2(t)] dt= a[x1, y] +b[x2, y]

(b) y(x) =20

x(t)2dt is not a linear functional since [ax1 + bx2, y] =20 [ax1(t) +bx2(t)]

2 dt= a[x1, y] +b[x2, y] 20(ax1(t))

2dt+20(bx2(t))

2dt

(c) y(x) =10

t2x(t)dt is a linear functional since [ax1+bx2, y] =10

t2 [ax1(t) +bx2(t)] dt= a[x1, y] +b[x2, y]

11A shifted linear functional does not remain linear functional anymore, it is like moving the line so that it does not pass throught the origin.


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2.11. DUAL BASES

(d) y(x) =10

x(t2)dtis a linear functional since [ax1+bx2, y] =10

ax1(t

2) +bx2(t2)

dt= a[x1, y] +b[x2, y]

(e) y(x) = dxdt

is a linear functional since [ax1+bx2, y] =d(ax1+bx2)

dt

= a dx1

dt +bdx2

dt =a[x1, y] +b[x2, y]

(f) y(x) = d2x

dt2|t=1is a linear functional since [ax1+bx2, y] =

d2(ax1+bx2)

dt2 |t=1

= a d

2x1

dt |t=1+bd2x2dt2|t=1 = a[x1, y]+b[x2,

4. If (a0, a1, a2, ) is an arbitrary sequence of complex numbers, and if x is an element ofP, x(t) =n

0iti, wr

y(x) =n0 aii. Prove thaty is an element of

P and that every element of

P can be obtained in this manner by

suitable choice ofai. 12

Proof. In order to show thatn

0aii P all I need to show is that it is a linear functional. Givenx1, x2 Pwe ha[ax1 + bx2, y] = [

n0 (ai + bi)t

i, y] =n

0ai(ai + bi) = an

0aii + bn

0 aii = a[x1, y] + b[x2, y]y P. The seconpart of the proof is pretty straight forward. For any y P choose your ai = y (ti) : i0, n, this works since ti forthe basis of the set of polynomials. The linear functional has to map each of the elements of the basis ti to some ai. Thcan be generalized to any vector space and linear functionals. 13

5. Ify is a non-zero functional on a vector spaceV, and if is an arbitrary scalar, does there necessarily exist a vector insuch that [v, y] = ?

Proof. The proof for this is very trivial. Consider a v V let [v, y] = now since , 1 F v1 = 1 v V a

[ 1

v, y] = 1

v1 =

v; [v1, y] =

6. Prove that ify andz are linear functionals (on the same vector space) such that [ v, y] = 0 whenever [v, z] = 0, then theexists a scalar such that y = z.

Proof. Consider the basisBV={v1, , vn}, nowv; [v, y] = [v, z] = 0 we haven

1 ii=n

1ii

2.11 Dual Bases

One more word before embarking on the proofs of the important theorems. The concept of dual space was defined without areference to coordinate systems; a glance at the following proofs will show a super- abundance of coordinate systems. We wito point out that this phenome- non is inevitable; we shall be establishing results concerning dimension, and dimension is tone concept (so far) whose very definition is given in terms of a basis .14

Theorem 28. IfVF is ann-dimensional vector space, if{v1, , vn} is a basis inVF, and if{1, , n} F is any set scalars, then there is one and only one functionaly onVF such that [vi, y] = i for i={1, , n}.Proof. The theorem claims that if two linear functionals map the basis vectors to the same set of scalars{1, , n}then thare the same.15 This is very clear since ify1, y2 are two linear functionals that map the basis vector to the same i then evegiven vector v =

n1 ivi and [v, y1] = [

n1 ivi, y1] =

n1 ii =

n1i[vi, y2] = [

n1 ivi, y2] = [v, y2]y1 =y2. Again if w

definedy as [x, y] =11+ +nn then, y is indeed a linear functional, and [xi, y] =i.Theorem 29. IfVF is an n-dimensional vectors space andBVF ={v1, , vn} is a basis inVF, then there is a uniquedetermined basisB

VFinVF,BVF ={y1, , yn}, with the property that [vi, yj ] = ij. Consequently the dual space of

n-dimensional space is n-dimensional.

Proof. The existence of such a functional is fairly obvious as well for any given vector x define [x, yi] = [n1ivi, yi] = i. W

have shown in one of the exercise problems that such a function is a linear-functional. It is clear as well that [vi, yi] = 1 a[vj , yi] = 0. Infact the existence is proven using Theorem28since we can always find a unique linear functional y for the s{1, , 0}. Clearly any given linear functionaly can be expressed as y =n1iyi where [vi, y] = i, To see that this workconsider a vector x, [x, y] = [

n1ivi, y] =

n1 i[vi, y] =

n1ii, but from our definition ofyi we have [x, yi] = i, the resu

follows. The setBVF

is a basis ofVF. We need to show that the set of yi are linearly independent. This is pretty straigforward to see, consider a vector x VF if

n1iyi = 0 we have [x,

n1iyi] =

n1 i[x, yi] = 0 for all x for x = vi we ha

0 =n

1 i[vi, yi] =n

1 iij = 0i = 012It is intersting that polynomials are already expressed in terms of their basis. The vectors ti are the elements of the basis.13I cannot believe I could not figure this out the second time.14This is a copy of the introductory statement from Halmoss book15The manner in which halmos has stated the theorem is a bit convoluted.


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Theorem 30. Ifu, w are two different vectors of then-dimensional space vector spaceV, then there exists a linear functiony onVsuch that [u, y]= [w, y]; or equivalently, to any non-zero vectorv V there corresponds a linear functionaly V suthat[v, y]= 0.Proof. The proof is pretty straight forward when you think of every vector as a linear combination of its basis. Given twvectors u, v V; such that u =ivi, w =n1ivi now let y be a linear functional onV, we have y =n1iyi, [u, y] [n

1 vi, y] =n

1i[vi, y] =n

1i[vi,n

1 jyj ] =i

1ii similarly [w, y] =n

1ii, if there is no y such that [u, y]= [w,and u

= w, we have [u, y] = [w, y]

y

n1 ii =n1 ii =n1i(i

i) = 0

i = i

u = w a contradiction.

simple choice ofy would be one of the basis linear functionals. The equivalence comes from the choice of the vector x = w The equivalence can be proven similarly, given [x, y] =

n1xii, if [x, y] = 0 for all y , then

n1xii = 0 for all y which impli

[x, yj ] = 0, j ={1, , n} x= 0. 16

2.12 Reflexivity

It is natural to think about linear functionals on the space ofDuals. We are esentially looking at the vector spaceV. Thse idis the reverse the application of the functional. We exploit the bilinearity of linear functionals do investigate linear functionaon the space of linear functionals on a vector spaceV. 17

Consider the linear functional [x0, y] where x0 is a fixed vector value and y takes values from the space of Duals ofV. Cleardue to the bilinearity of linear functionals [x0, y] is a linear functional.

Theorem 31. IfV is a finite dimensional vector space, then corresponding to every linear functionalz0 onV there is a vectx0 inV such thatz0(y) = [x0, y] =y(x0) for everyy V; the correspondencez0 x0 betweenV andV is an isomorphism.Proof. This is a really weird and a strange idea. The linear functionals onV are essentially all the linear functionals y V, valued at a chosen vector v V. So essentially the linear functionalsV are equinumerous with the vector spaceVconstruction. This is possible only because the linear functionals themselves preserve linearity i.e. [ x,y + z ] = [x, y] + [x,In order to prove isomorphism, we do the following, Ifx1, x2 are two vectors belonging toVand if [x1, z] = [x2, z]y V thx1 = x2. This is fairly clear since [x1, y] = [x2, y] for all y implies from theorem30, thatx1 = x2.

18

What we showed is that the linear functionals of the form we picked is equinumerous with V. It remains to be shown ththe entire spaceV is indeed equinumerous withV. The above two sections show that the set of linear functionals z onVz :V Fis a subspace ofV. This subspace is isomorphic withV and therefore n dimensional. However, bothV, V an-dimensionalwhich implies thatV

isn-dimensionaland hence equinumerous withVand coincides with the subspace of linefunctionals.

2.13 Basis ofVGiven our definition ofV, every [y, z] = [x, y] for some, over the space ofV can be represented as [n1 ivi, y] where ea[zi, y] = [vi, y], wherezi form the basis ofV. Professor Halmos uses the isomorphicnature ofzi vi and just notate the basas vi i.eBV=BV.

2.14 Annihilators

Definition 32. The annihilator

S0 of any subset

Sof a vector space

V (

Sneed not be a subspace) is the set of all vectorsy

V such that[x, y] is identically zero for allx inS.S0 ={y V; v S V, [v, y] = 0}

Some observations and examples:

1.O0 =V andV0 =O( V)16In your original proof you forgot to use the statement y, which is extremely important for the existence proof to work. The proof works becau

your claim is that if there is no y such that .... is possible.17Well that is a mouthful...18Note that theorem30 is extremely important. If it were not true then V would have not only been a subspace ofV but not equinumerous w

V or V .


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2.15. EXERCISES

2. IfVis finite dimensional andS contains a non-zero vector, the Theorem30 shows thatS0 =V.Theorem 33 (Dimension of the Space of Annihilators). IfM is an m-dimensional subspace of an n-dimensional vector spaV, thenM0 is an (n-m)-dimensional subspace ofV.Proof. The proof is fairly trivial and straight forward. It is easy to see that M0 is a vector space since,

m1, m2M0, m1+m2M0 since [v, m2+m1] = [v, m1] + [v, m2] = 0 + 0 = 0, from the linearity properties of linefunctionals inV

.A1-A2 These follow from the vector space properties ofV.

A3 The zero vector 0 M0 since [m, 0] = 0v V, and m+ 0 =mm M0.A4 Sincem V there exists a linear functionalm. Since [v, m] =[v, m] = 0, m Mand m + (m) = 0.

mM0, mM0 since [v, m] =[v, m] = 0 follows from the linearity properties ofm.B1-B2 Follows from the properties of vectors inV.C1-C2 Follows from the properties of vectors inV.

From the above it is clear that, given M0 V, it is a subspace ofV.

The basis vectors of the spaceVare mapped to ij by the basis vectors of the dual spaceV. Consider a vector m =

m1 i

whereBM={v1, , vm}are the basis vectors of the subspace M. Let{y1, , yn} be the basis ofV.

[m, y] = [m, y] = [m,n1

iyi] =n1

i[v, yi]

=m1

i[m, yi] +n

m+1

j [m, yj ] =m1

i[m, yi] +n

m+1

j [m1

ivi, yj ]

clearlyn

m+1j [m

1 ivi, yj ] =n

m+1jm

1 i[vi, yj] = 0 since [vi, yj ] = 0i= j the subset of basis{ym+1, , yn} mathe vectorsvM to 0. It is now a matter to show that{ym+1, , yn}form a basis ofM0. This follows from the fact that M0a subspace ofV.19 Actually I have proved nothing here. The proof has to show that{ym+1, , yn}forms a basis ofM0

other wordsspan({ym+1, , yn}) = M0

. To do this, (a) let us consider a functional yM0

. Sincey V

, y=n

1iyi. Givthat [v, y] = 0 it is clear that y =n

m+1iyi y span({ym+1, , ym}). (b) On the other hand every linear combinatinm+1iyi maps every vector v M to 0, implies that every (y =

nm+1iyi) span({ym+1, , ym}) is also a member

M0. Together (a) and (b) imply that span({ym+1, , ym}) = M0.Theorem 34. IfMis a subspace in a finite dimensional vector spaceV, thenM00 = ((M0)0) = M .Proof. The theorem essentially claims that the dual space ofannihilators of the subspace M is the subspace itself using onotation resulting from the iso-morphism. The dual space of the space ofM0 are all the linear functionals of the form [z, y] [x0, y] where x0 M. Since [x0, y] = 0 for every y M0, M0 = M00. In order to show the iso-morphism, let M m-dimensional. This implies that M0 is (n-m)-dimesional. Now from Theorem 33 dim(M00) = n(nm) = m sindim(V) = n, proving that M and M00 are iso-morphic. 20

2.15 Exercises1. Define a non-zero linear functionaly onC3 such that ifx1 = (1, 1, 1) andx2 = (1, 1, 1), then [x1, y] = [x2, y] = 0.

Sol. Let x = (1, 2, 3) Consider the linear functionaly ; [x, y] =2 1 or1 2.2. The vectorsx1 = (1, 1, 1), x2 = (1, 1, 1) and x3 = (1, 1, 1) form a basis ofC3. If{y1, y2, y3} is the dual basis, and

x= (0, 1, 0), then find [x, y1], [x, y2] and [x, y3].

19This does not follow from the fact that M0 is a subspace ofV20Note again that in order to show that M00 = M we had to either use the subset argument or iso-morphism argument resulting from eq

dimensionality of vector spaces on the same field. First step to a proof is to understand what needs to be proved. It seems that I do not quite get whneeds to be proved.


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Sol. We have [x1, y1] = 1, [x2, y2] = 1 and [x3, y3] = 1. Givenx =3

1ixi we can solve for the coefficients i.1 1 11 1 1

1 1 1

12

3

=

01

0

1+2+3 = 0

1+2

3 = 1

1 2 3 = 02 = 1

2, 1 = 0, 3 =1

2

(a) [(0, 1, 0), y1] = [3

1ixi, y1] = [0x1+ 12x2 12x3, y1] = 0

(b) [(0, 1, 0), y2] = [3

1ixi, y1] = [0x1+ 12x2 12x3, y2] = 12

(c) [(0, 1, 0), y3] = [3

1ixi, y1] = [0x1+ 12

x2 12x3, y3] =123. Prove that if y is a linear functional on an n-dimensional vector spaceV, then the set of all those vectors x for whi

[x, y] = 0 is a subspace ofV; what is the dimension of that subspace?Sol. The proof that the set y0 ={x V; [x, y] = 0} is a subspace ofV, follows from the following facts,

(a) 0y0, since [0, y] = 0y V.(b) Linearity properties of vectorsx V.

To begin with every x V can be written in terms of the basis BV ={v1, , v2}. We therefore have for all xy0, [

n1aivi, y] = 0

n1 ai[vi, y] =

n1 aibi = 0ai = 0 or bi = 0 or both, where y =

n1 biyi, yi being the basis

V. Clearly if all thebi are non-zero, all ai have to be zero and that leaves us with dimension of the subspace being 0. Othe other hand if all the bi are zero the dimension of the subspace is n. If the number ofbi that are zero is m then, tdimension of the subspace is m.

Let us look at the space ofannihilatorson the space of linear functions on V. These are the set z V, such th[z, y] = [x, y] = 0y Y V ie. using the isomorphism betweenV andVit is the set ofx V. The set of annihilatoofY V is given byY0 ={z; [z, y] = 0, y Y0}. Applying Theorem33 we see that the dimension ofy0 =n 1 siny= dim(y) = 1dim(y

0

) = n 1.21

4. 4. If y(x) =3

1xi whenever x = (x1, x2, x3) is a vector inV, then y is a linear functional on C3; find a basis of tsubspace consisting of all those vectors x for which [x, y] = 0.

Sol. [x, y] = 0 the dimension of the vector spaceY0 =n 1. We construct a basis for this space since y takes all vectors the form x1+ x2+ x3 = 0x1 =(x2+ x3), x1 = (1, 12 , 12), x2 = (1, 1, 0) would form a basis ofY0.

5. Prove that ifm < n and if{y1, , ym} are linear functionals on an n-dimensional vector spaceV, then there existsnon-zero vector x V such that [x, yj ] = 0 for j = 1,...,m. What does this result say about the solutions of lineequations?

Sol. The existence follows from the solution to the problem3 and Theorem33. The dimension of the Y =span({y1, , ym}m implies the dimension ofY0 =n m x| [x, yj] = 0. Let us look at the representation of linear equations in usilinear functionals. A system of linear equations is represented as

x11 xn1x12 xn2 x1n xnn

a1a2 an

=

12 n

In essense we are looking for i that would form a linear combination of the vectors{x1, , xn}such that it is the vectv.

21I am unable to reconcile this argument with my earlier argument. I am not sure what I am missing. The earlier argument is wrong. On secothought, the first argument/proof is rubbish. The basis{y1, , yn} are not independent of the basis vectors. The second argument is what masense.


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2.15. EXERCISES

6. Suppose that m < n and that{y1, , ym} are linear functionaIs on an n-dimensional vector spaceV. Under whconditions on the scalars{1, , n} is it true that there exists a vector x Vsuch that [x, yj ] =j for j = 1, , mWhat does this result say about the solutions of linear equations?

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