aarjaneiro/planetmath huggingface dataset:
complementary subspace

complementary subspace

Direct sum decomposition.

Let

U

be a

vector space

, and

V

,

W

⊂

U

subspaces

. We say that

V

and

W

span

U

, and write

U

=

V

+

W

if
every

u

∈

U

can be expressed as a sum

u

=

v

+

w

for some

v

∈

V

and

w

∈

W

.

If in

addition

, such a

decomposition

is unique for all

u

∈

U

, or
equivalently if

V

∩

W

=

{

0

}

,

then we say that

V

and

W

form a

direct sum

decomposition

of

U

and write

U

=

V

⊕

W

.

In such circumstances, we also say that

V

and

W

are

complementary

subspaces, and also say that

W

is an

algebraic complement

of

V

.

Here is useful

characterization

of complementary subspaces if

U

is

finite-dimensional

.

Proposition 1

Let

U

,

V

,

W

be as above, and suppose that

U

is
finite-dimensional. The subspaces

V

and

W

are complementary if
and only if for every basis

v

1

,

…

,

v

m

of

V

and
every basis

w

1

,

…

,

w

n

of

W

, the combined list

v

1

,

…

,

v

m

,

w

1

,

…

,

w

n

is a basis of

U

.

Remarks

.

•

Since every linearly independent subset of a vector space can be extended to a basis, every subspace has a

complement

, and the complement is necessarily unique.

•

Also, direct sum decompositions of a vector space

U

are in a one-to correspondence fashion with

projections

on

U

.

Orthogonal decomposition.

Specializing somewhat, suppose that the ground field

𝕂

is either
the real or

complex numbers

, and that

U

is either an

inner product
space

or a

unitary space

, i.e.

U

comes equipped with a
positive-definite

inner product

⟨

,

⟩

:

U

×

U

→

𝕂

.

In such circumstances,
for every subspace

V

⊂

U

we define the

orthogonal complement

of

V

, denoted by

V

⟂

to be the subspace

V

⟂

=

{

u

∈

U

:

⟨

v

,

u

⟩

=

0

,

for all

⁢

v

∈

V

}

.

Proposition 2

Suppose that

U

is finite-dimensional and

V

⊂

U

a subspace.
Then,

V

and its orthogonal
complement

V

⟂

determine a direct sum decomposition of

U

.

Note: the

Proposition

is false if either the finite-dimensionality
or the positive-definiteness

assumptions

are violated.

Title

complementary subspace

Canonical name

ComplementarySubspace

Date of creation

2013-03-22 12:52:16

Last modified on

2013-03-22 12:52:16

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

11

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A03

Synonym

algebraic complement

Defines

complementary

Defines

direct sum

Defines

decomposition

Defines

orthogonal complement

Generated on Fri Feb 9 18:18:12 2018 by

LaTeXML
coordinate vector

coordinate vector

Let

V

be a

vector space

of

dimension

n

over a field

K

.  If

(

b

1

,

…

,

b

n

)

is a basis of

V

, then any element

v

of

V

can be uniquely expressed in the form

v

=

ξ

1

⁢

b

1

+

…

+

ξ

n

⁢

b

n

with

ξ

1

,

…

,

ξ

n

∈

K

.  The

n

-tuplet (

http://planetmath.org/OrderedTuplet

)

(

ξ

1

,

…

,

ξ

n

)

is called the

coordinate vector

of

v

with respect to the basis in question.  The scalars

ξ

i

are the

coordinates

(or the

components

of

v

).

It’s evident that the correspondence

v

↦

(

ξ

1

,

…

,

ξ

n

)

provides a

linear isomorphism

between the vector space

V

and the vector space formed by the

Cartesian product

K

n

.

Title

coordinate vector

Canonical

name

CoordinateVector

Date of creation

2013-03-22 19:02:16

Last modified on

2013-03-22 19:02:16

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Definition

Classification

msc 15A03

Related topic

ListVector

Defines

coordinates

Defines

components

Generated on Fri Feb 9 18:18:14 2018 by

LaTeXML
diagonalization of quadratic form

diagonalization of quadratic form

A

quadratic form

may be diagonalized by the following procedure:

1.

Find a variable

x

such that

x

2

appears in the quadratic form. If no such variable can be found, perform a linear change of variable so as to create such a variable.

2.

By completing the square, define a new variable

x

′

such that there are no cross-terms involving

x

′

.

3.

Repeat the procedure with the remaining variables.

Example

Suppose we have been asked to

diagonalize

the quadratic form

Q

=

x

2

+

x

⁢

y

-

3

⁢

x

⁢

z

-

y

2

/

4

+

y

⁢

z

-

9

⁢

z

2

/

4

in three variables. We could proceed as follows:

•

Since

x

2

appears, we do not need to perform a change of variables.

•

We have the cross terms

x

⁢

y

and

-

3

⁢

x

⁢

z

. If we define

x

′

=

x

+

y

/

2

-

3

⁢

z

/

2

, then

x

′

2

=

x

2

+

x

⁢

y

-

3

⁢

x

⁢

z

+

y

2

/

4

+

9

⁢

z

2

/

4

-

3

⁢

y

⁢

z

/

2

Hence, we may re-express

Q

as

Q

=

x

′

2

-

y

⁢

z

/

2

•

We must now repeat the procedure with the remaining variables,

y

and

z

. Since neither

y

2

nor

z

2

appears, we must make a change of variable. Let us define

z

′

=

z

+

2

⁢

y

.

Q

=

x

′

2

-

y

2

-

y

⁢

z

′

/

2

•

We have a cross term

-

y

⁢

z

′

/

2

. To eliminate this term, make a change of variable

y

′

=

y

+

z

′

/

4

. Then we have

y

′

2

=

y

2

+

y

⁢

z

′

/

2

+

z

′

2

/

16

and hence

Q

=

x

′

2

-

y

′

2

+

z

′

2

/

16

The quadratic form is now

diagonal

, so we are done. We see that the form has rank 3 and

signature

2.

Title

diagonalization of quadratic form

Canonical name

DiagonalizationOfQuadraticForm

Date of creation

2013-03-22 14:49:34

Last modified on

2013-03-22 14:49:34

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

7

Author

rspuzio (6075)

Entry type

Algorithm

Classification

msc 15A03

Related topic

DiagonalQuadraticForm

Generated on Fri Feb 9 18:18:18 2018 by

LaTeXML
dimension (vector space)

dimension (vector space)

Let

V

be a

vector space

over a field

K

. We say that

V

is

finite-dimensional

if there exists a finite basis of

V

. Otherwise we
call

V

infinite-dimensional

.

It can be shown that every basis of

V

has the same

cardinality

. We call this cardinality the

dimension

of

V

. In particular, if

V

is finite-dimensional, then every basis of

V

will consist of a

finite set

v

1

,

…

,

v

n

. We then call the

natural number

n

the

dimension

of

V

.

Next, let

U

⊂

V

a

subspace

. The dimension of the

quotient
vector space

V

/

U

is called the

codimension

of

U

relative to

V

.

In circumstances where the choice of field is ambiguous, the
dimension of a vector space depends on the choice of field. For
example, every

complex vector space

is also a

real vector space

, and
therefore has a real dimension, double its complex dimension.

Title

dimension (vector space)

Canonical

name

DimensionvectorSpace

Date of creation

2013-03-22 12:42:31

Last modified on

2013-03-22 12:42:31

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

13

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A03

Related topic

dimension3

Defines

dimension

Defines

codimension

Defines

finite-dimensional

Defines

infinite-dimensional

Generated on Fri Feb 9 18:18:25 2018 by

LaTeXML
direct sum of Hermitian and skew-Hermitian matrices

direct sum of Hermitian and skew-Hermitian matrices

In this example, we show that any

square matrix

with

complex

entries can uniquely be decomposed into the sum of one

Hermitian matrix

and
one

skew-Hermitian matrix

. A fancy way to say this is that
complex square matrices is the

direct sum of Hermitian and skew-Hermitian
matrices

.

Let us denote the

vector space

(over

ℂ

) of
complex square

n

×

n

matrices by

M

.
Further, we denote by

M

+

respectively

M

-

the

vector
subspaces

of Hermitian and skew-Hermitian matrices.
We claim that

M

=

M

+

⊕

M

-

.

(1)

Since

M

+

and

M

-

are vector subspaces of

M

, it is clear
that

M

+

+

M

-

is a vector subspace of

M

. Conversely, suppose

A

∈

M

. We can then define

A

+

=

1

2

⁢

(

A

+

A

∗

)

,

A

-

=

1

2

⁢

(

A

-

A

∗

)

.

Here

A

∗

=

A

¯

T

, and

A

¯

is the

complex conjugate

of

A

,
and

A

T

is the

transpose

of

A

. It follows that

A

+

is Hermitian
and

A

-

is anti-Hermitian. Since

A

=

A

+

+

A

-

, any element
in

M

can be written as
the sum of one element in

M

+

and one element in

M

-

. Let us check
that this

decomposition

is unique. If

A

∈

M

+

∩

M

-

, then

A

=

A

∗

=

-

A

, so

A

=

0

.
We have established

equation

1

.

Special cases

•

In the special case of

1

×

1

matrices, we obtain the
decomposition of a

complex number

into its real and

imaginary

components

.

•

In the special case of

real matrices

, we obtain the decomposition of
a

n

×

n

matrix into a

symmetric matrix

and

anti-symmetric

matrix.

Title

direct sum of Hermitian and skew-Hermitian matrices

Canonical name

DirectSumOfHermitianAndSkewHermitianMatrices

Date of creation

2013-03-22 13:36:30

Last modified on

2013-03-22 13:36:30

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

5

Author

mathcam (2727)

Entry type

Example

Classification

msc 15A03

Classification

msc 15A57

Related topic

DirectSumOfEvenoddFunctionsExample

Generated on Fri Feb 9 18:18:27 2018 by

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Euclidean space

Euclidean space

1

Definition

Euclidean

n

-space

is a

metric space

(

E

,

d

)

with the

property

that the

group of isometries

is

transitive

and is

isomorphic

to an

n

-dimensional

Euclidean vector space

. To be more
precise, we are saying that there exists an

n

-dimensional Euclidean
vector space

V

with

inner product

⟨

⋅

,

⋅

⟩

and a

mapping

+

:

E

×

V

→

E

such that the following hold:

1.

For all

x

,

y

∈

E

there exists a unique

u

∈

V

satisfying

y

=

x

+

u

,

d

⁢

(

x

,

y

)

2

=

⟨

u

,

u

⟩

,

2.

For all

x

,

y

∈

E

and all

u

∈

V

we have

d

⁢

(

x

+

u

,

y

+

u

)

=

d

⁢

(

x

,

y

)

.

3.

For all

x

∈

E

and all

u

,

v

∈

V

we have

(

x

+

u

)

+

v

=

x

+

(

u

+

v

)

.

Putting it more succinctly:

V

acts transitively and effectively on

E

by isometries.

Remarks.

•

The

difference

between

Euclidean space

and a Euclidean vector space is
one of loss of

structure

. Euclidean space is a Euclidean vector space
that has “forgotten” its origin.

•

A 2-dimensional Euclidean space is often called a

Euclidean plane

.

Title

Euclidean space

Canonical

name

EuclideanSpace

Date of creation

2013-03-22 14:17:19

Last modified on

2013-03-22 14:17:19

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

16

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A03

Classification

msc 51M05

Related topic

EuclideanVectorProperties

Related topic

InnerProduct

Related topic

PositiveDefinite

Related topic

EuclideanDistance

Related topic

Vector

Defines

Euclidean plane

Generated on Fri Feb 9 18:18:29 2018 by

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every vector space has a basis

every vector space has a basis

This result,

trivial

in the finite case, is in fact rather surprising
when one thinks of

infinite

dimensionial

vector spaces

, and the
definition of a basis: just try to imagine a basis of the vector space
of all

continuous mappings

f

:

ℝ

→

ℝ

. The

theorem

is

equivalent

to the

axiom of choice

family of axioms and theorems. Here
we will only prove that Zorn’s lemma implies that every vector space
has a basis.

Theorem

.

Let

X

be any vector space over any field

F

and assume Zorn’s
lemma. Then if

L

is a linearly independent subset of

X

, there
exists a basis of

X

containing

L

. In particular,

X

does have a
basis at all.

Proof.

Let

𝒜

be the set of linearly independent subsets of

X

containing

L

(in particular,

𝒜

is not empty), then

𝒜

is partially ordered by

inclusion

. For each chain

C

⊆

𝒜

,
define

C

^

=

⋃

C

. Clearly,

C

^

is an

upper bound

of

C

. Next we
show that

C

^

∈

𝒜

. Let

V

:=

{

v

1

,

…

,

v

n

}

⊆

C

^

be a finite

collection

of vectors. Then there exist sets

C

1

,

…

,

C

n

∈

C

such that

v

i

∈

C

i

for all

1

≤

i

≤

n

. Since

C

is
a chain, there is a number

k

with

1

≤

k

≤

n

such that

C

k

=

⋃

i

=

1

n

C

i

and thus

V

⊆

C

k

, that is

V

is

linearly independent

. Therefore,

C

^

is an element of

𝒜

.

According to Zorn’s lemma

𝒜

has a

maximal element

,

M

, which
is linearly independent. We show now that

M

is a basis. Let

⟨

M

⟩

be the span of

M

. Assume there exists an

x

∈

X

∖

⟨

M

⟩

. Let

{

x

1

,

…

,

x

n

}

⊆

M

be a finite collection of vectors and

a

1

,

…

,

a

n

+

1

∈

F

elements

such that

a

1

⁢

x

1

+

⋯

+

a

n

⁢

x

n

-

a

n

+

1

⁢

x

=

0

.

If

a

n

+

1

was necessarily zero, so would be the other

a

i

,

1

≤

i

≤

n

,
making

{

x

}

∪

M

linearly independent in

contradiction

to the
maximality of

M

. If

a

n

+

1

≠

0

, we would have

x

=

a

1

a

n

+

1

⁢

x

1

+

⋯

+

a

n

a

n

+

1

⁢

x

n

,

contradicting

x

∉

⟨

M

⟩

. Thus such an

x

does not exist and

X

=

⟨

M

⟩

, so

M

is a

generating set

and hence a basis.

Taking

L

=

∅

, we see that

X

does have a basis at all.
∎

Title

every vector space has a basis

Canonical name

EveryVectorSpaceHasABasis

Date of creation

2013-03-22 13:04:48

Last modified on

2013-03-22 13:04:48

Owner

GrafZahl (9234)

Last modified by

GrafZahl (9234)

Numerical id

14

Author

GrafZahl (9234)

Entry type

Theorem

Classification

msc 15A03

Synonym

every vector space has a

Hamel basis

Related topic

ZornsLemma

Related topic

AxiomOfChoice

Related topic

ZermelosWellOrderingTheorem

Related topic

HaudorffsMaximumPrinciple

Related topic

KuratowskisLemma

Generated on Fri Feb 9 18:18:33 2018 by

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example of construction of a Schauder basis

example of construction of a Schauder basis

Consider an

uniformly continuous function

f

:

[

0

,

1

]

→

ℝ

. A

Schauder basis

{

f

n

⁢

(

x

)

}

0

∞

∈

C

⁢

[

0

,

1

]

is constructed. For this purpose we set

f

0

⁢

(

x

)

=

1

,

f

1

⁢

(

x

)

=

x

. Let us consider the sequence of semi-open

intervals

in

[

0

,

1

]

I

n

=

[

2

-

k

⁢

(

2

⁢

n

-

2

)

,

2

-

k

⁢

(

2

⁢

n

-

1

)

)

,

J

n

=

[

2

-

k

⁢

(

2

⁢

n

-

1

)

,

2

-

k

⁢

2

⁢

n

)

,

where

2

k

-

1

<

n

≤

2

k

,

k

≥

1

. Define now

f

n

⁢

(

x

)

=

{

2

k

⁢

[

x

-

(

2

-

k

⁢

(

2

⁢

n

-

2

)

-

1

)

]

if

⁢

x

∈

I

n

,

1

-

2

k

⁢

[

x

-

(

2

-

k

⁢

(

2

⁢

n

-

1

)

-

1

)

]

if

⁢

x

∈

J

n

,

0

otherwise.

Geometrically these functions form a sequence of triangular functions of height

one

and width

2

-

(

k

-

1

)

, sweeping

[

0

,

1

]

. So that if

f

∈

C

⁢

(

[

0

,

1

]

)

, it is

expressible

in Fourier series

f

⁢

(

x

)

∼

∑

n

=

0

∞

c

n

⁢

f

n

⁢

(

x

)

and computing the

coefficients

c

n

by equating the values of

f

⁢

(

x

)

and the series at the points

x

=

2

-

k

⁢

m

,

m

=

0

,

1

,

…

,

2

k

. The resulting series converges uniformly to

f

⁢

(

x

)

by the imposed premise.

Title

example of construction of a Schauder basis

Canonical name

ExampleOfConstructionOfASchauderBasis

Date of creation

2013-03-22 17:49:18

Last modified on

2013-03-22 17:49:18

Owner

perucho (2192)

Last modified by

perucho (2192)

Numerical id

5

Author

perucho (2192)

Entry type

Example

Classification

msc 15A03

Classification

msc 42-00

Generated on Fri Feb 9 18:18:37 2018 by

LaTeXML
finite dimensional proper subspaces of a normed space are nowhere dense

finite dimensional proper subspaces of a normed space are nowhere dense

-

Let

V

be a

normed space

. If

S

⊆

V

is a

finite dimensional

proper

subspace

, then

S

is nowhere dense.

Proof :

It is known that for any

topological vector space

(in particular, normed spaces) every proper subspace has empty

interior

(

http://planetmath.org/ProperSubspacesOfATopologicalVectorSpaceHaveEmptyInterior

).

From the entry (

http://planetmath.org/EveryFiniteDimensionalSubspaceOfANormedSpaceIsClosed

) we also know that finite dimensional subspaces of

V

are closed.

Then,

int

⁡

(

S

¯

)

=

int

⁡

(

S

)

=

∅

, which shows that

S

is nowhere dense.

□

Title

finite dimensional proper subspaces of a normed space are nowhere dense

Canonical name

FiniteDimensionalProperSubspacesOfANormedSpaceAreNowhereDense

Date of creation

2013-03-22 14:58:59

Last modified on

2013-03-22 14:58:59

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

9

Author

asteroid (17536)

Entry type

Result

Classification

msc 15A03

Classification

msc 46B99

Classification

msc 54E52

Generated on Fri Feb 9 18:18:39 2018 by

LaTeXML
Hamel function

Hamel function

A function

h

:

ℝ

n

→

ℝ

is said to be a

Hamel function

if

h

, considered as a subset

{

(

x

,

h

(

x

)

}

⊂

ℝ

n

+

1

, is a

Hamel basis

for

ℝ

n

+

1

over

ℚ

. We denote the set of

n

-dimensional Hamel function by

H

⁢

F

⁢

(

ℝ

n

)

.

References

•

Poltka, K.

On Functions Whose Graph is a Hamel Basis

. Unpublised Ph.D. work. Online at

http://academic.scranton.edu/faculty/PLOTKAK2/publications/ham_0911.pdf

Title

Hamel function

Canonical name

HamelFunction

Date of creation

2013-03-22 14:15:19

Last modified on

2013-03-22 14:15:19

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Definition

Classification

msc 15A03

Classification

msc 54C40

Generated on Fri Feb 9 18:18:46 2018 by

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linear combination

linear combination

A

linear combination

of vectors

v

1

,

…

,

v

n

of a

vector space

V

over a field

F

is a vector of the form

∑

i

=

1

n

a

i

⁢

v

i

,

where the

a

i

are elements of

F

.

Title

linear combination

Canonical name

LinearCombination

Date of creation

2013-03-22 11:58:15

Last modified on

2013-03-22 11:58:15

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

16

Author

yark (2760)

Entry type

Definition

Classification

msc 15A03

Related topic

Span

Generated on Fri Feb 9 18:18:48 2018 by

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linear manifold

linear manifold

Definition

Suppose

V

is a

vector space

and suppose that

L

is a
non-empty subset of

V

. If there exists a

v

∈

V

such that

L

+

v

=

{

v

+

l

∣

l

∈

L

}

is a

vector subspace

of

V

, then

L

is a

linear manifold

of

V

. Then we
say that the

dimension

of

L

is the dimension of

L

+

v

and write

dim

⁡

L

=

dim

⁡

(

L

+

v

)

.
In the important case

dim

⁡

L

=

dim

⁡

V

-

1

,

L

is called a

hyperplane

.

A linear manifold is, in other words, a linear subspace that has possibly been
shifted away from the origin.
For instance, in

ℝ

2

examples of linear
manifolds are points, lines (which are hyperplanes), and

ℝ

2

itself.
In

ℝ

n

hyperplanes naturally describe tangent planes to a smooth
hyper surface.

References

1

R. Cristescu,

Topological vector spaces

,
Noordhoff International Publishing, 1977.

Title

linear manifold

Canonical

name

LinearManifold

Date of creation

2013-03-22 14:04:32

Last modified on

2013-03-22 14:04:32

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

6

Author

matte (1858)

Entry type

Definition

Classification

msc 15A03

Classification

msc 15-00

Related topic

VectorSubspace

Related topic

LineSegment

Defines

hyperplane

Generated on Fri Feb 9 18:18:50 2018 by

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linearly independent

linearly independent

Let

V

be a

vector space

over a
field

F

. We say that

v

1

,

…

,

v

k

∈

V

are

linearly dependent

if there exist scalars

λ

1

,

…

,

λ

k

∈

F

, not all zero, such that

λ

1

⁢

v

1

+

⋯

+

λ

k

⁢

v

k

=

0

.

If no such scalars exist, then we say that the vectors are

linearly independent

.
More generally, we say that a (possibly infinite) subset

S

⊂

V

is linearly independent if all finite subsets of

S

are linearly independent.

In the case of two vectors, linear dependence means that one of the
vectors is a

scalar multiple

of the other. As an alternate

characterization

of dependence, we also have the following.

Proposition 1

.

Let

S

⊂

V

be a subset of a vector space. Then,

S

is
linearly dependent if and only if there exists a

v

∈

S

such that

v

can be expressed as a

linear combination

of the vectors in the
set

S

\

{

v

}

(all the vectors in

S

other
than

v

(

http://planetmath.org/SetDifference

)).

Remark

.

Linear independence

can be defined more generally for modules over rings: if

M

is a (left) module over a ring

R

. A subset

S

of

M

is linearly independent if whenever

r

1

⁢

m

1

+

⋯

+

r

n

⁢

m

n

=

0

for

r

i

∈

R

and

m

i

∈

M

, then

r

1

=

⋯

=

r

n

=

0

.

Title

linearly independent

Canonical

name

LinearlyIndependent

Date of creation

2013-03-22 11:58:40

Last modified on

2013-03-22 11:58:40

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

30

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A03

Synonym

linear independence

Defines

linearly dependent

Defines

linear dependence

Generated on Fri Feb 9 18:18:53 2018 by

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list vector

list vector

Let

𝕂

be a field and

n

a positive natural number. We define

𝕂

n

to be the set of all

mappings

from the index list

(

1

,

2

,

…

,

n

)

to

𝕂

. Such a mapping

a

∈

𝕂

n

is
just a formal way of speaking of a list of field

elements

a

1

,

…

,

a

n

∈

𝕂

.

The above description is somewhat restrictive. A more flexible

definition

of a

list vector

is the following. Let

I

be a finite
list of indices

1

1

Distinct

index sets

are often used when
working with

multiple

frames of reference.

,

I

=

(

1

,

…

,

n

)

is one
such possibility, and let

𝕂

I

denote the set of all mappings
from

I

to

𝕂

. A list vector, an element of

𝕂

I

, is
just such a mapping. Conventionally,

superscripts

are used to denote
the values of a list vector, i.e. for

u

∈

𝕂

I

and

i

∈

I

,
we write

u

i

instead of

u

⁢

(

i

)

.

We add and scale list vectors point-wise, i.e. for

u

,

v

∈

𝕂

I

and

k

∈

𝕂

, we define

u

+

v

∈

𝕂

I

and

k

⁢

u

∈

𝕂

I

, respectively by

(

u

+

v

)

i

=

u

i

+

v

i

,

i

∈

I

,

(

k

⁢

u

)

i

=

k

⁢

u

i

,

i

∈

I

.

We also have the

zero vector

𝟎

∈

𝕂

I

, namely the

constant mapping

𝟎

i

=

0

,

i

∈

I

.

The above

operations

give

𝕂

I

the

structure

of an (abstract)

vector space

over

𝕂

.

Long-standing traditions of

linear algebra

hold that elements of

𝕂

I

be regarded as

column vectors

. For example, we write

a

∈

𝕂

n

as

a

=

(

a

1

a

2

⋮

a

n

)

.

Row vectors

are usually taken to represents

linear forms

on

𝕂

I

. In other words, row vectors are elements of the

dual
space

(

𝕂

I

)

*

. The

components

of a row vector are
customarily written with subscripts, rather than superscripts. Thus,
we express a row vector

α

∈

(

𝕂

n

)

*

as

α

=

(

α

1

,

…

,

α

n

)

.

Title

list vector

Canonical

name

ListVector

Date of creation

2013-03-22 12:51:50

Last modified on

2013-03-22 12:51:50

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

5

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A03

Classification

msc 15A90

Defines

column vector

Defines

row vector

Generated on Fri Feb 9 18:18:57 2018 by

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