aarjaneiro/planetmath huggingface dataset:
invertible linear transformation

invertible linear transformation

An

invertible linear transformation

is a

linear transformation

T

:

V

⟶

W

which is a

bijection

.

If

V

and

W

are

finite dimensional

, the linear transformation

T

is

invertible

if and only if the matrix of

T

is not

singular

.

Title

invertible linear transformation

Canonical name

InvertibleLinearTransformation

Date of creation

2013-03-22 12:19:16

Last modified on

2013-03-22 12:19:16

Owner

djao (24)

Last modified by

djao (24)

Numerical id

5

Author

djao (24)

Entry type

Definition

Classification

msc 15A04

Defines

invertible

Generated on Fri Feb 9 18:21:31 2018 by

LaTeXML
isospectral

isospectral

Two

linear operators

are said to be

isospectral

or

cospectral

if they have the same

spectrum

. For linear operators over

finite dimensional

spaces, this means that every

eigenvalue

of one

operator

is also an eigenvalue of the other operator.

Title

isospectral

Canonical name

Isospectral

Date of creation

2013-03-22 14:44:54

Last modified on

2013-03-22 14:44:54

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

6

Author

asteroid (17536)

Entry type

Definition

Classification

msc 15A04

Synonym

cospectral

Generated on Fri Feb 9 18:21:32 2018 by

LaTeXML
kernel of a linear mapping

kernel of a linear mapping

Let

T

:

V

→

W

be a

linear mapping

between

vector spaces

.

The set of all vectors in

V

that

T

maps to

0

is called the

kernel

(or

nullspace

) of

T

,
and is denoted

ker

⁡

T

. So

ker

⁡

T

=

{

x

∈

V

∣

T

⁢

(

x

)

=

0

}

.

The kernel is a

vector subspace

of

V

,
and its

dimension

(

http://planetmath.org/Dimension2

) is called the

nullity

of

T

.

The function

T

is injective if and only if

ker

⁡

T

=

{

0

}

(see the attached proof (

http://planetmath.org/OperatornamekerL0IfAndOnlyIfLIsInjective

)).
In particular, if the dimensions of

V

and

W

are equal and finite,
then

T

is

invertible

if and only if

ker

⁡

T

=

{

0

}

.

If

U

is a vector subspace of

V

, then we have

ker

⁡

T

|

U

=

U

∩

ker

⁡

T

,

where

T

|

U

is
the

restriction

(

http://planetmath.org/RestrictionOfAFunction

) of

T

to

U

.

When the linear mappings are given by means of matrices,
the kernel of the matrix

A

is

ker

⁡

A

=

{

x

∈

V

∣

A

⁢

x

=

0

}

.

Title

kernel of a linear mapping

Canonical

name

KernelOfALinearMapping

Date of creation

2013-03-22 11:58:22

Last modified on

2013-03-22 11:58:22

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

20

Author

yark (2760)

Entry type

Definition

Classification

msc 15A04

Synonym

nullspace

Synonym

null-space

Synonym

kernel

Related topic

LinearTransformation

Related topic

ImageOfALinearTransformation

Related topic

Nullity

Related topic

RankNullityTheorem

Generated on Fri Feb 9 18:21:34 2018 by

LaTeXML
linear isomorphism

linear isomorphism

Definition 1

.

Suppose

V

and

W

are

vector spaces

and

L

:

V

→

W

is a

linear map

. Then

L

is a

linear isomorphism

if

L

is

bijective

.

Properties

1.

Compositions

and of linear isomorphisms is a linear isomorphism.

2.

The

inverse

of a linear isomorphisms is a linear isomorphism.

3.

If either

V

or

W

if finite dimensional, then

dim

⁡

V

=

dim

⁡

W

.
(This is a

consequence

of the

rank-nullity theorem

.)

Title

linear isomorphism

Canonical

name

LinearIsomorphism

Date of creation

2013-03-22 14:36:42

Last modified on

2013-03-22 14:36:42

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

7

Author

matte (1858)

Entry type

Definition

Classification

msc 15A04

Synonym

invertible linear map

Synonym

bijective linear map

Synonym

non-singular linear map

Generated on Fri Feb 9 18:21:36 2018 by

LaTeXML
linear transformation

linear transformation

Let

V

and

W

be

vector spaces

over the same field

F

. A

linear transformation

is a

function

T

:

V

→

W

such that:

•

T

⁢

(

v

+

w

)

=

T

⁢

(

v

)

+

T

⁢

(

w

)

for all

v

,

w

∈

V

•

T

⁢

(

λ

⁢

v

)

=

λ

⁢

T

⁢

(

v

)

for all

v

∈

V

, and

λ

∈

F

The set of all linear maps

V

→

W

is denoted by

Hom

F

⁡

(

V

,

W

)

or

ℒ

⁢

(

V

,

W

)

.

Examples:

•

Let

V

=

ℝ

n

and

W

=

ℝ

m

and

A

is any

m

×

n

matrix. Then the function

L

A

:

V

→

W

defined by

L

A

⁢

(

v

)

=

A

⁢

v

, the

multiplication

of matrix

A

and the vector

v

(considered as an

n

×

1

matrix), is a linear transformation.

•

Let

V

be the space of all

differentiable functions

over

ℝ

and

W

the space of all

continuous functions

over

ℝ

. Then

D

:

V

→

W

defined by

D

⁢

(

f

)

=

f

′

, the

derivative

of

f

, is a linear transformation.

Properties:

•

T

⁢

(

0

)

=

0

.

•

If

S

and

T

are linear transformations from

V

to

W

, and

k

∈

F

, then so are

S

+

T

and

k

⁢

T

. As a result,

Hom

F

⁡

(

V

,

W

)

is a vector space over F.

•

If

G

:

W

→

U

is a linear transformations then

G

∘

T

:

V

→

U

is also a linear transformation.

•

The kernel (

http://

planetmath

.org/KernelOfALinearTransformation

)

Ker

⁡

(

T

)

=

{

v

∈

V

∣

T

⁢

(

v

)

=

0

}

is a

subspace

of

V

.

•

The image (

http://planetmath.org/ImageOfALinearTransformation

)

Im

⁡

(

T

)

=

{

T

⁢

(

v

)

∣

v

∈

V

}

is a subspace of

W

.

•

The

inverse image

T

-

1

⁢

(

w

)

is a subspace if and only if

w

=

0

.

•

A linear transformation is

injective

if and only if

Ker

⁡

(

T

)

=

{

0

}

.

•

If

v

∈

V

then

T

-

1

⁢

(

T

⁢

(

v

)

)

=

v

+

Ker

⁡

(

T

)

.

•

If

w

∈

Im

⁡

(

T

)

then

T

⁢

(

T

-

1

⁢

(

w

)

)

=

{

w

}

.

Remark

. A linear transformation

T

:

V

→

W

such that

W

=

V

is called a

linear operator

, and a

linear functional

when

W

=

F

.

See also:

•

Wikipedia

,

http://www.wikipedia.org/wiki/Linear_transformation

linear transformation

Title

linear transformation

Canonical name

LinearTransformation

Date of creation

2013-03-22 11:56:41

Last modified on

2013-03-22 11:56:41

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

24

Author

CWoo (3771)

Entry type

Definition

Classification

msc 15A04

Synonym

linear map

Synonym

vector space homomorphism

Synonym

linear mapping

Related topic

Matrix

Related topic

InvariantSubspace

Related topic

DualHomomorphism

Related topic

KernelOfALinearTransformation

Related topic

EigenvalueOfALinearOperator

Related topic

NilpotentTransformation

Related topic

AffineTransformation

Related topic

SubLinear

Related topic

MatrixRepresentationOfALinearTransformation

Defines

linear operator

Generated on Fri Feb 9 18:21:38 2018 by

LaTeXML
linear transformation is continuous if its domain is finite dimensional

linear transformation is continuous if its domain is finite dimensional

Theorem 1

.

A

linear transformation

is

continuous

if the domain is finite dimensional.

Proof.

Suppose

L

:

X

→

Y

is the

transformation

,

dim

⁡

X

=

n

,
and

∥

⋅

∥

X

,

∥

⋅

∥

Y

are the norms
on

X

,

Y

, respectively.
By this result (

http://planetmath.org/ContinuityIsPreservedWhenCodomainIsExtended

)
and this result (

http://planetmath.org/SubspaceTopologyInAMetricSpace

),
it suffices to prove that

L

:

X

→

L

⁢

(

X

)

is continuous
when

L

⁢

(

X

)

is equipped with the topology given by

∥

⋅

∥

Y

restricted onto

L

⁢

(

X

)

.
Also, since

continuity

and boundedness are

equivalent

, it suffices to
prove that

L

is bounded.
Let

e

1

,

…

,

e

n

be a basis for

X

such that

L

is

invertible

on

span

⁡

{

e

1

,

…

,

e

k

}

and

ker

⁡

L

=

span

⁡

{

e

k

+

1

,

…

,

e

n

}

for

k

=

1

,

…

,

n

. (The

zero map

is always continuous.)
Let

f

i

=

L

⁢

(

e

i

)

for

i

=

1

,

…

,

k

, so that

span

⁡

{

f

1

,

…

,

f

k

}

=

L

⁢

(

X

)

.
Let us define new norms on

X

and

L

⁢

(

X

)

,

∥

x

∥

X

′

=

∑

i

=

1

n

α

i

2

,

∥

y

∥

X

′

=

∑

i

=

1

k

β

i

2

,

for

x

=

∑

i

=

1

n

α

i

⁢

e

i

∈

X

and

y

=

∑

i

=

1

k

β

i

⁢

f

i

∈

Y

.
Since norms on finite dimensional

vector spaces

are equivalent, it follows
that

1

/

C

⁢

∥

x

∥

X

′

≤

∥

x

∥

X

≤

C

⁢

∥

x

∥

X

′

,

x

∈

X

1

/

D

⁢

∥

y

∥

Y

′

≤

∥

y

∥

Y

≤

D

⁢

∥

y

∥

Y

′

,

y

∈

L

⁢

(

X

)

for some

constants

C

,

D

>

0

.
For

x

=

∑

i

=

1

n

α

i

⁢

e

i

∈

X

,

∥

L

⁢

(

x

)

∥

Y

≤

D

⁢

∥

∑

i

=

1

k

α

i

⁢

f

i

∥

Y

′

=

D

⁢

∑

i

=

1

k

α

i

2

≤

D

⁢

∑

i

=

1

n

α

i

2

=

D

⁢

∥

x

∥

X

′

=

C

⁢

D

⁢

∥

x

∥

X

.

Thus

L

:

X

→

L

⁢

(

X

)

is bounded.
∎

Title

linear transformation is continuous if its domain is finite dimensional

Canonical name

LinearTransformationIsContinuousIfItsDomainIsFiniteDimensional

Date of creation

2013-03-22 15:17:59

Last modified on

2013-03-22 15:17:59

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

7

Author

matte (1858)

Entry type

Theorem

Classification

msc 15A04

Generated on Fri Feb 9 18:21:41 2018 by

LaTeXML
minimal polynomial (endomorphism)

minimal polynomial (endomorphism)

Let

T

be an

endomorphism

of an

n

-dimensional

vector space

V

.

Definitions

.

We define the

,

M

T

⁢

(

X

)

, to be the unique

monic polynomial

of such that

M

T

⁢

(

T

)

=

0

. We say that

P

⁢

(

X

)

is a

zero

for

T

if

P

⁢

(

T

)

is the zero endomorphism.

Note that the

minimal polynomial

exists by virtue of the

Cayley-Hamilton theorem

, which provides a

zero polynomial

for

T

.

.

Firstly,

End

⁡

(

V

)

is a vector space of

dimension

n

2

. Therefore the

n

2

+

1

vectors,

i

v

,

T

,

T

2

,

…

⁢

T

n

2

, are linearly dependant. So there are

coefficients

,

a

i

not all zero such that

∑

i

=

0

n

2

a

i

⁢

T

i

=

0

. We conclude that a non-trivial zero polynomial for

T

exists. We take

M

T

⁢

(

X

)

to be a zero polynomial for

T

of minimal degree with leading coefficient one.

:

If

P

⁢

(

X

)

is a zero polynomial for

T

then

M

T

(

X

)

∣

P

(

X

)

.

Proof.

By the

division algorithm

for

polynomials

,

P

⁢

(

X

)

=

Q

⁢

(

X

)

⁢

M

T

⁢

(

X

)

+

R

⁢

(

X

)

with

d

⁢

e

⁢

g

⁢

R

<

d

⁢

e

⁢

g

⁢

M

T

. We note that

R

⁢

(

X

)

is also a zero polynomial for

T

and by minimality of

M

T

⁢

(

X

)

, must be just

0

. Thus we have shown

M

T

(

X

)

∣

P

(

X

)

.
∎

The minimal polynomial has a number of interesting

properties

:

1.

The roots are exactly the

eigenvalues

of the endomorphism

2.

If the minimal polynomial of

T

splits into linear factors then

T

is upper-triangular with respect to some basis

3.

The minimal polynomial of

T

splits into

distinct

linear factors (i.e. no repeated roots) if and only if

T

is

diagonal

with respect to some basis.

It is then a

corollary

of the

fundamental theorem of algebra

that every endomorphism of a finite dimensional vector space over

ℂ

may be upper-triangularized.

The minimal polynomial is intimately related to the

characteristic polynomial

for

T

. For let

χ

T

⁢

(

X

)

be the characteristc polynomial. Since

χ

T

⁢

(

T

)

=

0

, we have by the above lemma that

M

T

(

X

)

∣

χ

T

(

X

)

. It is also a fact that the eigenvalues of

T

are exactly the roots of

χ

T

. So when split into linear factors the only

difference

between

M

T

⁢

(

X

)

and

χ

T

⁢

(

X

)

is the

algebraic multiplicity

of the roots.

In general they may not be the same - for example any

diagonal matrix

with repeated eigenvalues.

Title

minimal polynomial (endomorphism)

Canonical

name

MinimalPolynomialendomorphism

Date of creation

2013-03-22 13:10:14

Last modified on

2013-03-22 13:10:14

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

12

Author

mathcam (2727)

Entry type

Definition

Classification

msc 15A04

Related topic

ZeroPolynomial2

Related topic

OppositePolynomial

Defines

zero polynomial

Defines

minimal polynomial

Generated on Fri Feb 9 18:21:55 2018 by

LaTeXML
necessary and sufficient condition for diagonalizability

necessary and sufficient condition for diagonalizability

Let

k

be a field,

V

a

vector space

over

k

of

dimension

n

,
and

T

∈

End

⁡

(

V

)

. Then

T

is

diagonalizable

if and only if its

minimal polynomial

(

http://planetmath.org/MinimalPolynomialEndomorphism

) has no

multiple roots

and all its roots lie in

k

.

Title

necessary and sufficient condition for diagonalizability

Canonical name

NecessaryAndSufficientConditionForDiagonalizability

Date of creation

2013-03-22 14:15:41

Last modified on

2013-03-22 14:15:41

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

11

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 15A04

Generated on Fri Feb 9 18:21:57 2018 by

LaTeXML
permutation operator

permutation operator

Let

V

be a

vector space

over a field. Let

σ

∈

S

n

, the symmetric group on

{

1

,

…

,

n

}

and define
a multilinear map

ϕ

:

V

×

⋯

×

V

→

V

⊗

n

=

V

⊗

⋯

⊗

V

⏞

n

⁢

times

by

ϕ

⁢

(

v

1

,

…

,

v

n

)

=

v

σ

-

1

⁢

(

1

)

⊗

⋯

⊗

v

σ

-

1

⁢

(

n

)

.

Then by the

universal

factorization

property

(

http://planetmath.org/TensorProduct

) for a

tensor product

(

http://planetmath.org/TensorProduct

) there is a
unique

linear map

P

⁢

(

σ

)

:

V

⊗

n

→

V

⊗

n

such that

P

(

σ

)

⊗

=

ϕ

. Then of course,

P

⁢

(

σ

)

⁢

v

1

⊗

⋯

⊗

v

n

=

v

σ

-

1

⁢

(

1

)

⊗

⋯

⊗

v

σ

-

1

⁢

(

n

)

.

P

⁢

(

σ

)

is called the

permutation operator

associated with

σ

.

1

Properties

1.

P

⁢

(

σ

⁢

τ

)

=

P

⁢

(

σ

)

⁢

P

⁢

(

τ

)

2.

P

⁢

(

e

)

=

I

, where

I

is the

identity mapping

on

V

⊗

n

3.

P

⁢

(

σ

)

is

nonsingular

and

P

⁢

(

σ

)

-

1

=

P

⁢

(

σ

-

1

)

Title

permutation operator

Canonical

name

PermutationOperator

Date of creation

2013-03-22 16:15:38

Last modified on

2013-03-22 16:15:38

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

7

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 15A04

Generated on Fri Feb 9 18:21:59 2018 by

LaTeXML
primary decomposition theorem

primary decomposition theorem

This is an important theorem in

linear algebra

. It states the following:
Let

k

be a field,

V

a

vector space

over

k

,

dim

⁡

V

=

n

, and

T

:

V

→

V

a

linear operator

, such that its

minimal polynomial

(or its

annihilator polynomial

) is

m

T

, which decomposes in

k

⁢

[

X

]

into

irreducible

factors as

m

T

=

p

1

α

1

⁢

…

⁢

p

r

α

r

. Then,

1.

V

=

⊕

i

=

1

r

ker

⁡

(

p

i

α

i

⁢

(

T

)

)

2.

ker

⁡

(

p

i

α

i

⁢

(

T

)

)

is

T

-

invariant

for every

i

3.

If

T

i

is the

restriction

of

T

to

ker

⁡

(

p

i

α

i

⁢

(

T

)

)

, then

m

T

i

=

p

i

α

i

This is a consequence of a more general theorem:
Let

V

,

T

be as above, and

f

∈

k

⁢

[

X

]

such that

f

⁢

(

T

)

=

0

, with

f

=

f

1

⁢

…

⁢

f

r

and

(

f

i

,

f

j

)

=

1

if

i

≠

j

, then

1.

V

=

⊕

i

=

1

r

ker

⁡

(

f

i

⁢

(

T

)

)

2.

ker

⁡

(

f

i

⁢

(

T

)

)

is

T

-invariant for every

i

To illustrate its importance, the

primary decomposition theorem

, together with the

cyclic decomposition theorem

, imply the existence and uniqueness of the

Jordan canonical form

.

Title

primary decomposition theorem

Canonical name

PrimaryDecompositionTheorem

Date of creation

2013-03-22 14:15:30

Last modified on

2013-03-22 14:15:30

Owner

gumau (3545)

Last modified by

gumau (3545)

Numerical id

8

Author

gumau (3545)

Entry type

Theorem

Classification

msc 15A04

Generated on Fri Feb 9 18:22:00 2018 by

LaTeXML
proof of necessary and sufficient condition for diagonalizability

proof of necessary and sufficient condition for diagonalizability

First, suppose that

T

is

diagonalizable

. Then

V

has a basis whose elements

{

v

1

,

…

,

v

n

}

are

eigenvectors

of

T

associated with the

eigenvalues

{

λ

1

,

…

,

λ

n

}

respectively. For each

i

=

1

,

…

,

n

, as

v

i

is an eigenvector, its

annihilator polynomial

is

m

v

i

=

X

-

λ

i

. As these vectors form a basis of

V

, we have that the

minimal polynomial

(

http://planetmath.org/MinimalPolynomialEndomorphism

) of

T

is

m

T

=

lcm

⁡

(

X

-

λ

1

,

…

,

X

-

λ

n

)

which is trivially a product of linear factors.

Now, suppose that

m

T

=

(

X

-

λ

1

)

⁢

…

⁢

(

X

-

λ

p

)

for some

p

.
Let

v

∈

V

. Consider the

T

-

cyclic subspace

generated by

v

,

Z

⁢

(

v

,

T

)

=

⟨

v

,

T

⁢

v

,

…

,

T

r

⁢

v

⟩

. Let

T

v

be the restriction of

T

to

Z

⁢

(

v

,

T

)

. Of course,

v

is a cyclic vector of

Z

⁢

(

v

,

T

v

)

, and then

m

v

=

m

T

v

=

χ

T

. This is really easy to see: the

dimension

of

Z

⁢

(

v

,

T

)

is

r

+

1

, and it’s also the degree of

m

v

. But as

m

v

divides

m

T

v

(because

m

T

v

⁢

v

=

0

), and

m

T

divides

χ

T

v

(

Cayley-Hamilton theorem

), we have that

m

v

divides

χ

T

v

. As these are two

monic polynomials

of degree

r

+

1

and one divides the other, they are equal. And then by the same reasoning

m

v

=

m

T

v

=

χ

T

.
But as

m

v

divides

m

T

, then as

m

v

=

m

T

v

, we have that

m

T

v

divides

m

T

, and then

m

T

v

has no

multiple roots

and they all lie in

k

. But then so does

χ

T

v

. Suppose that these roots are

λ

1

,

…

,

λ

r

+

1

. Then

Z

⁢

(

v

,

T

)

=

⊕

λ

i

E

λ

i

, where

E

λ

i

is the

eigenspace

associated to

λ

i

. Then

v

is a sum of eigenvectors. QED.

Title

proof of necessary and sufficient condition for diagonalizability

Canonical name

ProofOfNecessaryAndSufficientConditionForDiagonalizability

Date of creation

2013-03-22 14:15:45

Last modified on

2013-03-22 14:15:45

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

13

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 15A04

Generated on Fri Feb 9 18:22:07 2018 by

LaTeXML
proof of theorem about cyclic subspaces

proof of theorem about cyclic subspaces

We first prove the case

r

=

2

. The

⊆

inclusion is clear, since the right side is a

T

-

invariant subspace

that contains

v

1

+

v

2

.

For the other inclusion, it is sufficient to show that

v

1

,

v

2

∈

Z

⁢

(

v

1

+

v

2

,

T

)

. The idea is that the action of

T

on

v

1

+

v

2

can ”isolate” the two summands if their

annihilator polynomials

are

coprime

. Let’s write

m

i

for

m

v

i

.

Since

(

m

1

,

m

2

)

=

1

, there exist

polynomials

p

and

q

such that

p

⁢

m

1

+

q

⁢

m

2

=

1

(1)

this is BÃÂ©zout’s lemma (or the

Euclidean algorithm

, or the fact that

k

⁢

[

X

]

is a

principal ideal domain

).

Now

p

⁢

m

1

⁢

(

T

)

is the

projection

from

Z

⁢

(

v

1

,

T

)

⊕

Z

⁢

(

v

2

,

T

)

to

Z

⁢

(

v

2

,

T

)

:

(

p

⁢

m

1

)

⁢

(

T

)

⁢

v

1

=

p

⁢

(

T

)

⁢

m

1

⁢

(

T

)

⁢

v

1

=

p

⁢

(

T

)

⁢

0

=

0

(2)

(by assumption that

m

1

is the annihilator polynomial of

v

1

) and

(

p

⁢

m

1

)

⁢

(

T

)

=

1

-

(

q

⁢

m

2

)

⁢

(

T

)

(3)

(by choice of

p

and

q

), so

(

p

⁢

m

1

)

⁢

(

T

)

⁢

v

2

=

v

2

-

q

⁢

(

T

)

⁢

m

2

⁢

(

T

)

⁢

v

2

=

v

2

-

q

⁢

(

T

)

⁢

0

=

v

2

(4)

Any

subspace

that is

invariant

under

T

is also invariant under polynomials of

T

. Therefore, the preceding equations show that

v

2

=

(

p

⁢

m

1

)

⁢

(

T

)

⁢

(

v

1

+

v

2

)

∈

Z

⁢

(

v

1

+

v

2

,

T

)

. By

symmetry

, we also get that

v

1

∈

Z

⁢

(

v

1

+

v

2

,

T

)

.

For the last claim, we note that the annihilator polynomial

m

of

Z

⁢

(

v

1

,

T

)

⊕

Z

⁢

(

v

2

,

T

)

is the least common

multiple

of

m

1

and

m

2

(that

m

is a multiple of

m

1

follows from the fact that

m

must annihilate

v

1

, and the set of polynomials that annihilate

v

1

is the

ideal generated by

m

1

). Since

m

1

and

m

2

are coprime, the lcm is just their

product

.

That concludes the proof for

r

=

2

. If

r

is arbitrary, we can simply apply the

r

=

2

case inductively. We only have to check that the coprimality condition is preserved under applying the

r

=

2

case to

i

=

1

,

2

. But it is well-known that if

p

,

q

,

r

(in

k

⁢

[

X

]

or in any principal ideal domain) are

pairwise coprime

, then

p

⁢

q

and

r

are also coprime.

Title

proof of theorem about cyclic subspaces

Canonical name

ProofOfTheoremAboutCyclicSubspaces

Date of creation

2013-03-22 17:32:53

Last modified on

2013-03-22 17:32:53

Owner

FunctorSalad (18100)

Last modified by

FunctorSalad (18100)

Numerical id

7

Author

FunctorSalad (18100)

Entry type

Proof

Classification

msc 15A04

Generated on Fri Feb 9 18:22:10 2018 by

LaTeXML
some facts about injective and surjective linear maps

some facts about injective and surjective linear maps

Let

k

be a field and

V

,

W

be

vector spaces

over

k

.

Proposition

. Let

f

:

V

→

W

be an

injective

linear map

. Then there exists a (

surjective

) linear map

g

:

W

→

V

such that

g

∘

f

=

id

V

.

Proof

. Of course

Im

⁢

(

f

)

is a

subspace

of

W

so

f

:

V

→

Im

⁢

(

f

)

is a

linear isomorphism

. Let

(

e

i

)

i

∈

I

be a basis of

Im

⁢

(

f

)

and

(

e

j

)

j

∈

J

be its completion to the basis of

W

, i.e.

(

e

i

)

i

∈

I

∪

J

is a basis of

W

. Define

g

:

W

→

V

on the basis as follows:

g

⁢

(

e

i

)

=

f

-

1

⁢

(

e

i

)

,

if

⁢

i

∈

I

;

g

⁢

(

e

j

)

=

0

,

if

⁢

j

∈

J

.

We will show that

g

∘

f

=

id

V

.

Let

v

∈

V

. Then

f

⁢

(

v

)

=

∑

i

∈

I

α

i

⁢

e

i

,

where

α

i

∈

k

(note that the

indexing set

is

I

). Thus we have

(

g

∘

f

)

⁢

(

v

)

=

g

⁢

(

∑

i

∈

I

α

i

⁢

e

i

)

=

∑

i

∈

I

α

i

⁢

g

⁢

(

e

i

)

=

∑

i

∈

I

α

i

⁢

f

-

1

⁢

(

e

i

)

=

=

f

-

1

⁢

(

∑

i

∈

I

α

i

⁢

e

i

)

=

f

-

1

⁢

(

f

⁢

(

v

)

)

=

v

.

It is clear that the equality

g

∘

f

=

id

V

implies that

g

is surjective.

□

Proposition

. Let

g

:

W

→

V

be a surjective linear map. Then there exists a (injective) linear map

f

:

V

→

W

such that

g

∘

f

=

id

V

.

Proof

. Let

(

e

i

)

i

∈

I

be a basis of

V

. Since

g

is onto, then for any

i

∈

I

there exist

w

i

∈

W

such that

g

⁢

(

w

i

)

=

e

i

.
Now define

f

:

V

→

W

by the

formula

f

⁢

(

e

i

)

=

w

i

.

It is clear that

g

∘

f

=

id

V

, which implies that

f

is injective.

□

If we combine these two propositions, we have the following corollary:

Corollary

. There exists an injective linear map

f

:

V

→

W

if and only if there exists a surjective linear map

g

:

W

→

V

.

Title

some facts about injective and surjective linear maps

Canonical name

SomeFactsAboutInjectiveAndSurjectiveLinearMaps

Date of creation

2013-03-22 18:32:22

Last modified on

2013-03-22 18:32:22

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

6

Author

joking (16130)

Entry type

Derivation

Classification

msc 15A04

Generated on Fri Feb 9 18:22:16 2018 by

LaTeXML