aarjaneiro/planetmath huggingface dataset:
identity matrix

identity matrix

The

n

×

n

identity matrix

I

(or

I

n

) over a ring

R

(with an

identity

1) is the

square matrix

with coefficients in

R

given by

I

=

[

1

0

⋯

0

0

1

⋯

0

0

0

⋱

0

0

0

⋯

1

]

,

where the numeral “1” and “0” respectively represent the multiplicative and

additive identities

in

R

.

0.0.1

Properties

The identity matrix

I

n

serves as the

multiplicative identity

in the ring of

n

×

n

matrices over

R

with

standard matrix multiplication

. For any

n

×

n

matrix

M

, we have

I

n

⁢

M

=

M

⁢

I

n

=

M

, and the identity matrix is uniquely defined by this

property

. In

addition

, for any

n

×

m

matrix

A

and

m

×

n

B

, we have

I

⁢

A

=

A

and

B

⁢

I

=

B

.

The

n

×

n

identity matrix

I

satisfy

the following properties

•

For the

determinant

, we have

det

⁡

I

=

1

, and for the trace, we have

tr

⁡

I

=

n

.

•

The identity matrix has only one

eigenvalue

λ

=

1

of

multiplicity

n

. The corresponding

eigenvectors

can be chosen to be

v

1

=

(

1

,

0

,

…

,

0

)

,

…

,

v

n

=

(

0

,

…

,

0

,

1

)

.

•

The

matrix exponential

of

I

gives

e

I

=

e

⁢

I

.

•

The identity matrix is a

diagonal matrix

.

Title

identity matrix

Canonical

name

IdentityMatrix

Date of creation

2013-03-22 12:06:29

Last modified on

2013-03-22 12:06:29

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

13

Author

mathcam (2727)

Entry type

Definition

Classification

msc 15-01

Classification

msc 15A57

Related topic

KroneckerDelta

Related topic

ZeroMatrix

Related topic

IdentityMap

Generated on Fri Feb 9 18:17:22 2018 by

LaTeXML
invertible matrix

invertible matrix

Let

R

be a ring and

M

an

m

×

n

matrix over

R

.

M

is said to be

left invertible

if there is an

n

×

m

matrix such that

N

⁢

M

=

I

n

, where

I

n

is the

n

×

n

identity matrix

. We call

N

a

left inverse

of

M

. Similarly,

M

is

right invertible

if there is an

n

×

m

matrix

P

, called a

right inverse

of

M

, such that

M

⁢

P

=

I

m

, where

I

m

is the

m

×

m

identity matrix. If

M

is both left invertible and right invertible, we say that

M

is

invertible

. If

R

is an

associative ring

, and

M

is invertible, then it has a unique left and a unique right inverse, and they are in fact equal, we call this matrix the

inverse

of

M

.

If

R

is a

division ring

, then it can be shown that for any matrix

M

over

R

,

M

is left invertible iff it is invertible iff it is right invertible. In

addition

, when

M

is invertible, it is a

square matrix

. Furthermore,

R

is a field iff for any square matrix

M

(over

R

),

M

is invertible implies that

M

T

, its

transpose

, is invertible as well. Invertibility of matrices over a division ring can also be determined by quantities known as ranks and

determinants

. It can be shown that a matrix over a division ring is invertible iff its

left row rank

(or right column rank) is full iff its determinant is non-zero. For example, the

2

×

2

matrix

(

1

j

i

k

)

over the

Hamiltonian quaternions

is not invertible, as its determinant

k

-

j

⁢

i

=

0

. It is interesting to note that, however, its transpose

(

1

i

j

k

)

is invertible, whose determinant is

2

⁢

k

≠

0

. The relationship between determinants and matrix invertibility can also be used to prove the following: preservation of matrix invertibility upon matrix

transposition

implies

commutativity

of division ring

D

. This can be done as follows: given any

a

,

b

∈

D

, the

2

×

2

matrix

(

a

⁢

b

b

a

1

)

is not invertible because its determinant is

0

. Therefore, its transpose

(

a

⁢

b

a

b

1

)

is also not invertible, and its determinant is

0

=

a

⁢

b

-

b

⁢

a

, whence

D

is a field.

Title

invertible matrix

Canonical

name

InvertibleMatrix

Date of creation

2013-03-22 19:23:09

Last modified on

2013-03-22 19:23:09

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

11

Author

CWoo (3771)

Entry type

Definition

Classification

msc 15-01

Classification

msc 15A09

Classification

msc 15A33

Generated on Fri Feb 9 18:17:32 2018 by

LaTeXML
isomorphism of rings of real and complex matrices

isomorphism of rings of real and complex matrices

Note that

submatrix notation

(

http://planetmath.org/

Submatrix

) will be used within this entry. Also, for any positive integer

n

,

M

n

×

n

⁢

(

R

)

will be used to denote the ring of

n

×

n

matrices with entries from the ring

R

, and

R

n

will be used to denote the following subring of

M

2

⁢

n

×

2

⁢

n

⁢

(

ℝ

)

:

R

n

=

{

P

∈

M

2

⁢

n

×

2

⁢

n

⁢

(

ℝ

)

:

P

=

(

A

B

-

B

A

)

⁢

for some

⁢

A

,

B

∈

M

n

×

n

⁢

(

ℝ

)

}

Theorem

.

For any positive integer

n

,

R

n

≅

M

n

×

n

⁢

(

C

)

.

Proof.

Define

φ

:

R

n

→

M

n

×

n

⁢

(

ℂ

)

by

φ

⁢

(

(

A

B

-

B

A

)

)

=

A

+

i

⁢

B

for

A

,

B

∈

M

n

×

n

⁢

(

ℝ

)

.

Let

A

,

B

,

C

,

D

∈

M

n

×

n

⁢

(

ℝ

)

such that

φ

⁢

(

(

A

B

-

B

A

)

)

=

φ

⁢

(

(

C

D

-

D

C

)

)

. Then

A

+

i

⁢

B

=

C

+

i

⁢

D

. Therefore,

A

=

C

and

B

=

D

. Hence,

(

A

B

-

B

A

)

=

(

C

D

-

D

C

)

. It follows that

φ

is

injective

.

Let

Z

∈

M

n

×

n

⁢

(

ℂ

)

. Then there exist

X

,

Y

∈

M

n

×

n

⁢

(

ℝ

)

such that

X

+

i

⁢

Y

=

Z

. Since

φ

⁢

(

(

X

Y

-

Y

X

)

)

=

X

+

i

⁢

Y

=

Z

, it follows that

φ

is

surjective

.

Let

A

,

B

,

C

,

D

∈

M

n

×

n

⁢

(

ℝ

)

. Then

φ

⁢

(

(

A

B

-

B

A

)

+

(

C

D

-

D

C

)

)

=

φ

⁢

(

(

A

+

C

B

+

D

-

B

-

D

A

+

C

)

)

=

A

+

C

+

i

⁢

(

B

+

D

)

=

A

+

i

⁢

B

+

C

+

i

⁢

D

=

φ

⁢

(

(

A

B

-

B

A

)

)

+

φ

⁢

(

(

C

D

-

D

C

)

)

and

φ

⁢

(

(

A

B

-

B

A

)

⁢

(

C

D

-

D

C

)

)

=

φ

⁢

(

(

A

⁢

C

-

B

⁢

D

A

⁢

D

+

B

⁢

C

-

A

⁢

D

-

B

⁢

C

A

⁢

C

-

B

⁢

D

)

)

=

A

⁢

C

-

B

⁢

D

+

i

⁢

(

A

⁢

D

+

B

⁢

C

)

=

(

A

+

i

⁢

B

)

⁢

(

C

+

i

⁢

D

)

=

φ

⁢

(

(

A

B

-

B

A

)

)

⁢

φ

⁢

(

(

C

D

-

D

C

)

)

.

It follows that

φ

is an

isomorphism

(

http://planetmath.org/RingIsomorphism

).
∎

Title

isomorphism of rings of real and complex matrices

Canonical name

IsomorphismOfRingsOfRealAndComplexMatrices

Date of creation

2013-03-22 16:17:15

Last modified on

2013-03-22 16:17:15

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 15-01

Classification

msc 15A33

Classification

msc 15A21

Generated on Fri Feb 9 18:17:36 2018 by

LaTeXML
Square of a generic sum of elements

Square of a generic sum of elements

The formula for computing the square of a generic sum of terms is as follows:

(

∑

k

=

1

K

a

k

)

2

=

∑

k

=

1

K

a

k

2

+

2

⁢

∑

j

=

1

K

∑

i

<

j

a

i

⁢

a

j

We can prove this

property

by induction, considering that it holds for K=2, since

(

a

+

b

)

2

=

a

2

+

b

2

+

2

⁢

a

⁢

b

and that if the property holds for a generic K, then it holds also for K+1, as is proven in the following passages:

{aligned}

⁢

(

∑

k

=

1

K

+

1

a

k

)

2

⁢

&

=

(

∑

k

=

1

K

a

k

+

a

K

+

1

)

2

=

(

∑

k

=

1

K

+

1

a

k

)

2

+

a

K

+

1

2

+

2

⁢

∑

k

=

1

K

a

k

⁢

a

K

+

1

⁢

&

=

∑

k

=

1

K

a

k

2

+

2

⁢

∑

j

=

1

K

∑

i

<

j

a

i

⁢

a

j

+

a

K

+

1

2

+

2

⁢

∑

k

=

1

K

a

k

⁢

a

K

+

1

=

&

=

∑

k

=

1

K

+

1

a

k

2

+

2

⁢

∑

j

=

1

K

+

1

∑

i

<

j

a

i

⁢

a

j

(1)

Title

Square of a generic sum of elements

Canonical

name

SquareOfAGenericSumOfElements

Date of creation

2013-03-22 19:34:48

Last modified on

2013-03-22 19:34:48

Owner

mat (27197)

Last modified by

mat (27197)

Numerical id

8

Author

mat (27197)

Entry type

Definition

Classification

msc 15-01

Generated on Fri Feb 9 18:17:42 2018 by

LaTeXML
vectorization of matrix

vectorization of matrix

The

vectorization

of a

m

×

n

matrix

A

=

(

a

i

⁢

j

)

the matrix to a

column vector

vec

⁢

(

A

)

, which consists of all columns of

A

stacked in :

vec

⁢

(

A

)

:=

(

a

11

⁢

a

21

⁢

…

⁢

a

m

⁢

1

⁢

a

12

⁢

a

22

⁢

…

⁢

a

m

⁢

2

⁢

…

⁢

…

⁢

a

1

⁢

n

⁢

a

2

⁢

n

⁢

…

⁢

a

m

⁢

n

)

⊺

The mapping vec from the

vector space

formed by the

m

×

n

matrices to the vector space of the column vectors of the

m

⁢

n

is apparently a

linear transformation

.

Title

vectorization of matrix

Canonical

name

VectorizationOfMatrix

Date of creation

2013-03-22 18:12:20

Last modified on

2013-03-22 18:12:20

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

4

Author

pahio (2872)

Entry type

Definition

Classification

msc 15-01

Synonym

vectorization

Synonym

vectorisation

Related topic

transpose

Generated on Fri Feb 9 18:17:44 2018 by

LaTeXML
zero matrix

zero matrix

The

n

×

m

zero

O

over a ring

R

is the

n

×

m

matrix with
coefficients in

R

given by

O

=

[

0

⋯

0

⋮

⋱

⋮

0

⋯

0

]

,

where 0 is the

additive identity

(

http://planetmath.org/Ring

) in

R

.

0.0.1

Properties

The

zero matrix

is the additive identity in the ring of

n

×

n

matrices over

R

. This is an alternative definition of

O

(since there’s just one additive identity in any given ring (

http://planetmath.org/UniquenessOfAdditiveIdentityInARing2

)).

The

n

×

n

zero matrix

O

has the following properties:

•

The

determinant

of

O

is

det

⁡

O

=

0

, and its trace is

tr

⁡

O

=

0

.

•

O

has only one

eigenvalue

λ

=

0

of
multiplicity

n

. Any non-zero vector is an

eigenvector

of

O

, so if we’re looking for a basis of eigenvectors, we could pick the

standard basis

e

1

=

(

1

,

0

,

…

,

0

)

,

…

,

e

n

=

(

0

,

…

,

0

,

1

)

.

•

The

matrix exponential

of

O

is

e

O

=

I

, the

n

×

n

identity matrix

.

Title

zero matrix

Canonical name

ZeroMatrix

Date of creation

2013-03-22 14:19:19

Last modified on

2013-03-22 14:19:19

Owner

waj (4416)

Last modified by

waj (4416)

Numerical id

8

Author

waj (4416)

Entry type

Definition

Classification

msc 15-01

Related topic

Matrix

Related topic

IdentityMatrix

Generated on Fri Feb 9 18:17:46 2018 by

LaTeXML
annihilator of vector subspace

annihilator of vector subspace

If

V

is a

vector space

, and

S

is any subset of

V

,
the

annihilator

of

S

, denoted by

S

0

,
is the

subspace

of the

dual space

V

*

that kills every vector in

S

:

S

0

=

{

ϕ

∈

V

*

:

ϕ

⁢

(

v

)

=

0

⁢

for all

⁢

v

∈

S

}

.

Similarly, if

Λ

is any subset of

V

*

, the

annihilated subspace

of

Λ

is

Λ

-

0

=

{

v

∈

V

:

ϕ

⁢

(

v

)

=

0

⁢

for all

⁢

ϕ

∈

Λ

}

=

⋂

ϕ

∈

Λ

ker

⁡

ϕ

.

(Note: this may not be the standard notation.)

1

Properties

Assume

V

is

finite-dimensional

.
Let

W

and

Φ

denote subspaces of

V

and

V

*

, respectively,
and let

^

denote the

natural isomorphism

from

V

to its double dual

V

*

*

.

i.

S

0

=

(

span

⁡

S

)

0

ii.

Λ

-

0

=

(

span

⁡

Λ

)

-

0

iii.

W

00

=

W

^

iv.

(

Φ

-

0

)

0

=

Φ

v.

(

W

0

)

-

0

=

W

vi.

dim

⁡

W

+

dim

⁡

W

0

=

dim

⁡

V

(a

dimension

theorem)

vii.

dim

⁡

Φ

+

dim

⁡

Φ

-

0

=

dim

⁡

V

*

=

dim

⁡

V

viii.

(

W

1

+

W

2

)

0

=

W

1

0

∩

W

2

0

, where

W

1

+

W

2

denotes
the sum of two subspaces of

V

.

ix.

If

T

:

V

→

V

is a

linear operator

, and

W

=

ker

⁡

T

,
then the image of the

pullback

T

*

:

V

*

→

V

*

is

W

0

.

References

1

Friedberg, Insel, Spence.

Linear Algebra

. Prentice-Hall, 1997.

Title

annihilator of vector subspace

Canonical name

AnnihilatorOfVectorSubspace

Date of creation

2013-03-22 15:25:59

Last modified on

2013-03-22 15:25:59

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

5

Author

stevecheng (10074)

Entry type

Definition

Classification

msc 15A03

Defines

annihilator

Defines

annihilated subspace

Generated on Fri Feb 9 18:17:48 2018 by

LaTeXML
another proof of rank-nullity theorem

another proof of rank-nullity theorem

Let

ϕ

:

V

→

W

be a

linear transformation

from

vector spaces

V

to

W

. Recall that the rank of

ϕ

is the

dimension

of the image of

ϕ

and the

nullity

of

ϕ

is the dimension of the kernel of

ϕ

.

Proposition 1

.

dim

⁡

(

V

)

=

rank

⁡

(

ϕ

)

+

nullity

⁡

(

ϕ

)

.

Proof.

Let

K

=

ker

⁡

(

ϕ

)

.

K

is a

subspace

of

V

so it has a unique

algebraic complement

L

such that

V

=

K

⊕

L

. It is evident that

dim

⁡

(

V

)

=

dim

⁡

(

K

)

+

dim

⁡

(

L

)

since

K

and

L

have

disjoint

bases and the union of their bases is a basis for

V

.

Define

ϕ

′

:

L

→

ϕ

⁢

(

V

)

by

restriction

of

ϕ

to the subspace

L

.

ϕ

′

is obviously a linear transformation. If

ϕ

′

⁢

(

v

)

=

0

, then

ϕ

⁢

(

v

)

=

ϕ

′

⁢

(

v

)

=

0

so that

v

∈

K

. Since

v

∈

L

as well, we have

v

∈

K

∩

L

=

{

0

}

, or

v

=

0

. This means that

ϕ

′

is

one-to-one

. Next, pick any

w

∈

ϕ

⁢

(

V

)

. So there is some

v

∈

V

with

ϕ

⁢

(

v

)

=

w

. Write

v

=

x

+

y

with

x

∈

K

and

y

∈

L

. So

ϕ

′

⁢

(

y

)

=

ϕ

⁢

(

y

)

=

0

+

ϕ

⁢

(

y

)

=

ϕ

⁢

(

x

)

+

ϕ

⁢

(

y

)

=

ϕ

⁢

(

v

)

=

w

, and therefore

ϕ

′

is onto. This means that

L

is

isomorphic

to

ϕ

⁢

(

V

)

, which is

equivalent

to saying that

dim

⁡

(

L

)

=

dim

⁡

(

ϕ

⁢

(

V

)

)

=

rank

⁡

(

ϕ

)

. Finally, we have

dim

⁡

(

V

)

=

dim

⁡

(

K

)

+

dim

⁡

(

L

)

=

nullity

⁡

(

ϕ

)

+

rank

⁡

(

ϕ

)

.

∎

Remark

. The dimension of

V

is not assumed to be finite in this proof. For another approach (where finite dimensionality of

V

is assumed), please see this entry (

http://planetmath.org/ProofOfRankNullityTheorem

).

Title

another

proof of rank-nullity theorem

Canonical name

AnotherProofOfRanknullityTheorem

Date of creation

2013-03-22 18:06:14

Last modified on

2013-03-22 18:06:14

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

4

Author

CWoo (3771)

Entry type

Proof

Classification

msc 15A03

Related topic

ProofOfRankNullityTheorem

Generated on Fri Feb 9 18:17:57 2018 by

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axiomatization of dependence

axiomatization of dependence

As noted by van der Waerden, it is possible to define the notion of dependence axiomatically in such a way that one can deal with

linear dependence

,

algebraic dependence

, and other sorts of dependence via a general theory. In this general theoretical framework, one can prove results about bases,

dimension

, and the like.

Let

S

be a set. The basic object of this theory is a

relation

D

between

S

and the

power set

of

S

. This relation

satisfies

the following three axioms:

Axiom 1

If

Y

is a subset of

S

and

x

∈

Y

, then

D

⁢

(

x

,

Y

)

.

Axiom 2

If, for some set

X

⊆

S

and some

y

,

z

∈

S

, it happens that

D

⁢

(

y

,

X

∪

{

z

}

)

but not

D

⁢

(

y

,

X

)

, then

D

⁢

(

z

,

X

∪

{

y

}

)

.

Axiom 3

If, for some sets

Y

,

Z

⊆

S

and some

x

∈

S

, it happens that

D

⁢

(

x

,

Y

)

and, for every

y

∈

Y

, it is the case that

D

⁢

(

y

,

Z

)

, then

D

⁢

(

x

,

Z

)

.

Title

axiomatization of dependence

Canonical name

AxiomatizationOfDependence

Date of creation

2013-03-22 16:27:46

Last modified on

2013-03-22 16:27:46

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 15A03

Related topic

DependenceRelation

Generated on Fri Feb 9 18:17:59 2018 by

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basal units

basal units

The set of all

n

×

n

matrices over a skew field forms an

n

2

-dimensional

associative algebra

, for the basis of which one can take the

basal units

.  A basal unit of the

algebra

has all

components

zeroes except only one which is 1.

E.g.

(

0

0

0

0

0

1

0

0

0

)

is a basal unit of the algebra of

3

×

3

matrices.

References

1

L. E. Dickson

:

Algebras and their

arithmetics

.  Dover Publications, Inc.  New York (1923; second edition 1960).

Title

basal units

Canonical

name

BasalUnits

Date of creation

2013-03-22 19:17:13

Last modified on

2013-03-22 19:17:13

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Definition

Classification

msc 15A03

Related topic

MatrixRing

Related topic

StandardBasis

Generated on Fri Feb 9 18:18:01 2018 by

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basis

basis

A (Hamel) basis of a

vector space

is a

linearly independent

spanning set

.

It can be proved that any two bases of the same vector space must have the same

cardinality

. This introduces the notion of

dimension

of a vector space, which is precisely the cardinality of the basis, and is denoted by

dim

⁡

(

V

)

, where

V

is the vector space.

The fact that every vector space has a

Hamel basis

(

http://planetmath.org/EveryVectorSpaceHasABasis

) is an important

consequence

of the

axiom of choice

(in fact, that

proposition

is

equivalent

to the axiom of choice.)

Examples.

•

β

=

{

e

i

}

,

1

≤

i

≤

n

, is a basis for

ℝ

n

(the

n

-dimensional vector space over the reals). For

n

=

4

,

β

=

{

(

1

0

0

0

)

,

(

0

1

0

0

)

,

(

0

0

1

0

)

,

(

0

0

0

1

)

}

•

β

=

{

1

,

x

,

x

2

}

is a basis for the vector space of

polynomials

with degree at most 2, over a

division ring

.

•

The set

β

=

{

[

1

0

0

0

]

,

[

0

1

0

0

]

,

[

0

0

0

1

]

,

[

0

0

1

0

]

}

is a basis for the vector space of

2

×

2

matrices over a division ring, and assuming that the

characteristic

of the ring is not 2, then so is

β

′

=

{

[

2

0

0

0

]

,

[

0

1

0

0

]

,

[

0

0

0

4

]

,

[

0

0

1

2

0

]

}

.

•

The

empty set

is a basis for the

trivial vector space

which consists of the unique

element

0

.

Remark

. More generally, for any (left)

right module

M

over a ring

R

, one may define a (left) right basis for

M

as a subset

B

of

M

such that

B

spans

M

and is linearly independent. However, unlike bases for a vector space, bases for a module may not have the same cardinality.

Title

basis

Canonical name

Basis

Date of creation

2013-03-22 12:01:57

Last modified on

2013-03-22 12:01:57

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

22

Author

mathcam (2727)

Entry type

Definition

Classification

msc 15A03

Synonym

Hamel basis

Related topic

Span

Related topic

IntegralBasis

Related topic

BasicTensor

Related topic

Aliasing

Related topic

Subbasis

Related topic

Blade

Related topic

ProofOfGramSchmidtOrthogonalizationProcedure

Related topic

LinearExtension

Generated on Fri Feb 9 18:18:04 2018 by

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blade

blade

A blade is a term often used to describe a basis entity in the space defined by a

geometric algebra

. Since a geometric algebra is a multi-graded space, the basis entities also have

multiple

grades. To distinguish the various graded entities, the blades are often prefixed by their grade. For example a grade-

k

basis entity would be called a

k

-blade.

The number of

linearly independent

k

-blades in a particular geometric algebra is dependent on the number of

dimensions

of the manifold on which the

algebra

is defined. For an

n

-dimensional manifold, the number of

k

-blades is given by the

binomial coefficient

.

N

k

=

(

n

k

)

The total number of basis blades of all grades in a geometric algebra defined on an

n

-manifold is then:

N

=

∑

k

=

0

n

N

k

=

2

n

Title

blade

Canonical

name

Blade

Date of creation

2013-03-22 15:58:40

Last modified on

2013-03-22 15:58:40

Owner

PhysBrain (974)

Last modified by

PhysBrain (974)

Numerical id

5

Author

PhysBrain (974)

Entry type

Definition

Classification

msc 15A03

Classification

msc 15A75

Classification

msc 15A66

Related topic

Basis

Related topic

UnitVector

Generated on Fri Feb 9 18:18:08 2018 by

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closure of a vector subspace in a normed space is a vector subspace

closure of a vector subspace in a normed space is a vector subspace

Let

(

V

,

∥

⋅

∥

)

be a

normed space

, and

S

⊂

V

a

vector subspace

. Then

S

¯

is a vector subspace in

V

.

Proof

First of all,

0

∈

S

¯

because

0

∈

S

. Now, let

x

,

y

∈

S

¯

, and

λ

∈

K

(where

K

is the ground field of the

vector space

V

). Then there are two sequences in

S

, say

(

x

n

)

n

∈

ℕ

and

(

y

n

)

n

∈

ℕ

which

converge

to

x

and

y

respectively.

Then, the sequence

(

x

n

+

λ

⋅

y

n

)

n

∈

ℕ

is a sequence in

S

(because

S

is a vector subspace), and it’s trivial (use properties of the norm) that this sequence converges to

x

+

λ

⋅

y

, and so this sum is a vector which lies in

S

¯

.

We have proved that

S

¯

is a vector subspace. QED.

Title

closure

of a vector subspace in a normed space is a vector subspace

Canonical name

ClosureOfAVectorSubspaceInANormedSpaceIsAVectorSubspace

Date of creation

2013-03-22 15:00:16

Last modified on

2013-03-22 15:00:16

Owner

gumau (3545)

Last modified by

gumau (3545)

Numerical id

7

Author

gumau (3545)

Entry type

Result

Classification

msc 15A03

Classification

msc 46B99

Classification

msc 54A05

Related topic

ClosureOfAVectorSubspaceIsAVectorSubspace2

Related topic

ClosureOfSetsClosedUnderAFinitaryOperation

Generated on Fri Feb 9 18:18:10 2018 by

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