3A.5 Vector Spaces

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3A.5.1 Field (F)

Consider a set of scalars.

For any a and b from the set, if a þ b and ab are also elements in the set; if

1. a þ b ¼ b þ a and ab ¼ ba (commutativity)

2. ða þ bÞ þ g ¼ a þ ðb þ gÞ and ðabÞg ¼ aðbgÞ (associativity)

3. aðb þ gÞ ¼ ab þ ag (distributivity)

are satisfied; and if

1. Identity elements 0 and 1 exist in the set such that a þ 0 ¼ a and 1a ¼ a

2. Inverse elements exist in the set such that a þ ð2aÞ ¼ 0 and a·a21 ¼ 1

then the set is a field.

For example, the set of real numbers is a field.

3A.5.2 Vector Space (L)

Properties:

1. Vector addition ðx þ yÞ and scalar multiplication ðaxÞ are defined.

2. Commutativity: x þ y ¼ y þ x and associativity: ðx þ yÞ þ z ¼ x þ ðy þ zÞ are satisfied.

3. Unique null vector 0 and negation ð2xÞ exist such that x þ 0 ¼ x, x þ ð2xÞ ¼ 0:

4. Scalar multiplication satisfies

aðbxÞ ¼ ðabÞx ðassociativityÞ

aðx þ yÞ ¼ ax þ by

ða þ bÞx ¼ ax þ bx

)

ðdistributivityÞ

1x ¼ x; 0x ¼ 0

Special case: Vector space Ln has vectors with n elements from the field F:

Consider

x ¼

x1

x2

.. .

xn

2

66666664

3

77777775

; y ¼

y1

y2

.. .

yn

2

66666664

3

77777775

Then

x þ y ¼

x1 þ y1

.. .

xn þ yn

2

6664

3

7775

¼ y þ x

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and

ax ¼

ax1

.. .

axn

2

6664

3

7775

3A.5.3 Subspace S of L

1. If x and y are in S then x þ y is also in S:

2. If x is in S and a is in F then ax is also in S:

3A.5.4 Linear Dependence

Consider the set of vectors: x1; x2; …; xn:They are linearly independent if any one of these vectors cannot

be expressed as a linear combination of one or more remaining vectors.

Necessary and sufficient condition for linear independence:

a1x1 þ a2x2 þ · · · þ anxn ¼ 0 ð3A:21Þ

gives a ¼ 0 (trivial solution) as the only solution.

For example,

x1 ¼

1

2

3

2

664

3

775

; x2 ¼

2

21

1

2

664

3

775

; x3 ¼

5

0

5

2

664

3

775

These vectors are not linearly independent because x1 þ 2x2 ¼ x3:

3A.5.5 Bases and Dimension of a Vector Space

1. If a set of vectors can be combined to form any vector in L then that set of vectors is said to span

the vector space L (i.e., a generating system of vectors).

2. If the spanning vectors are all linearly independent, then this set of vectors is a basis for that vector

space.

3. The number of vectors in the basis ¼ the dimension of the vector space.

Note: The dimension of a vector space is not necessarily the order of the vectors.

For example, consider two intersecting third-order vectors. They will form a basis for the plane (twodimensional)

that contains the two vectors. Hence, the dimension of the vector space ¼ 2, but the order

of each vector in the basis ¼ 3.

Note: L n is spanned by n linearly independent vectors

) dimðLnÞ ¼ n

For example,

1

0

0

.. .

0

2

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3

77777777775

;

0

1

0

.. .

0

2

66666666664

3

77777777775

; · · ·;

0

0

.. .

0

1

2

66666666664

3

77777777775

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3A.5.6 Inner Product

ðx; yÞ ¼ yHx ð3A:22Þ

where H denotes the Hermitian transpose (i.e., complex conjugate and transpose). Hence, yH ¼ ðypÞT

where ( )p denotes complex conjugation.

Note:

1. ðx; xÞ $ 0 and ðx; xÞ ¼ 0 if and only if (iff) x ¼ 0

2. ðx; yÞ ¼ ðy; xÞp

3. ðlx; yÞ ¼ lðx; yÞðx; lyÞ ¼ lpðx; yÞ

4. ðx; y þ zÞ ¼ ðx; yÞ þ ðx; zÞ

3A.5.7 Norm

Properties

kxk $ 0 and kxk ¼ 0 iff x ¼ 0

klxk ¼ lllkxk for any scalar l

kx þ yk # kxk þ kyk

For example, the Euclidean norm:

kxk ¼ xHx ¼

Xn

i¼1

x2

i

􀁻 !1=2

ð3A:23Þ

Unit vector

kxk ¼ 1

Normalization

x

kxk ¼ x^

Angle between vectors

We have

cos u ¼ ðx; yÞ

kxkkyk ¼ ðx^; y^Þ ð3A:24Þ

where u is the angle between x and y:

Orthogonal vectors

iff ðx; yÞ ¼ 0 ð3A:25Þ

Note: n orthogonal vectors in L n are linearly independent and span Ln; and form a basis for Ln:

3A.5.8 Gram – Schmidt Orthogonalization

Given a set of vectors, x1; x2; …; xn; that are linearly independent in Ln; we construct a set of orthonormal

(orthogonal and normalized) vectors, y^1; y^2;…; y^n; which are linear combinations of x^i:

Start

y^1 ¼ x^1 ¼

x1

kx1k

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Then

yi ¼ xi 2

Xi21

j¼1 ðxi; y^jÞy^j for i ¼ 1; 2;…; n

Normalize yi to produce y^i:

3A.5.9 Modified Gram – Schmidt Procedure

In each step, compute new vectors that are orthogonal to the just-computed vector.

Initialization:

y^1 ¼

x1

kx1k

as before.

Then

xð1Þ i ¼ xi 2 ðy^1; xiÞy^1 for i ¼ 2; 3;…; n

y^i ¼

xð1Þ i

kxð1Þ i k

for i ¼ 2; 3; …n

and

xð2Þ i ¼ xð1Þ i 2 ðy^2; xð1Þ i Þy^2; i ¼ 3; 4;…; n

and so on.

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