Let $f:Eto F$ be linear, then prove that if $E=ker foplus H$, then $H$ is isomorphic to $operatornameIm f$

Clash Royale CLAN TAG#URR8PPP

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Let $f:Eto F$ be linear where $E,F$ are vector spaces and $dim E<infty$. Let $H$ be such that $E=ker foplus H$, then I wish to prove that $H $ is isomorphic to $ operatornameIm f.$

Here is what I have done:

If $f:Eto F$ is linear and $dim E<infty$, then $$dim E=dim ker f+dim operatornameIm f. ;;;;;(1)$$ Since $E=ker foplus H$, then by formula of four-dimension $$dim E=dim ker f+dim H. ;;;;;(2)$$
From $(1)$ and $(2)$, we have that
$$dim H=dim operatornameIm f.$$
Then, $H $ is isomorphic to $ operatornameIm f.$ Please, I'm I right or wrong? If no, I would love to see better solutions.

linear-algebra abstract-algebra

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edited Jul 27 at 17:01

asked Jul 27 at 16:37

Mike

62112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hardmath: My apologies for that!
  – Mike
  Jul 27 at 16:54
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

Let $f:Eto F$ be linear where $E,F$ are vector spaces and $dim E<infty$. Let $H$ be such that $E=ker foplus H$, then I wish to prove that $H $ is isomorphic to $ operatornameIm f.$

Here is what I have done:

If $f:Eto F$ is linear and $dim E<infty$, then $$dim E=dim ker f+dim operatornameIm f. ;;;;;(1)$$ Since $E=ker foplus H$, then by formula of four-dimension $$dim E=dim ker f+dim H. ;;;;;(2)$$
From $(1)$ and $(2)$, we have that
$$dim H=dim operatornameIm f.$$
Then, $H $ is isomorphic to $ operatornameIm f.$ Please, I'm I right or wrong? If no, I would love to see better solutions.

linear-algebra abstract-algebra

share|cite|improve this question

edited Jul 27 at 17:01

asked Jul 27 at 16:37

Mike

62112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hardmath: My apologies for that!
  – Mike
  Jul 27 at 16:54
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $f:Eto F$ be linear where $E,F$ are vector spaces and $dim E<infty$. Let $H$ be such that $E=ker foplus H$, then I wish to prove that $H $ is isomorphic to $ operatornameIm f.$

Here is what I have done:

If $f:Eto F$ is linear and $dim E<infty$, then $$dim E=dim ker f+dim operatornameIm f. ;;;;;(1)$$ Since $E=ker foplus H$, then by formula of four-dimension $$dim E=dim ker f+dim H. ;;;;;(2)$$
From $(1)$ and $(2)$, we have that
$$dim H=dim operatornameIm f.$$
Then, $H $ is isomorphic to $ operatornameIm f.$ Please, I'm I right or wrong? If no, I would love to see better solutions.

linear-algebra abstract-algebra

share|cite|improve this question

edited Jul 27 at 17:01

asked Jul 27 at 16:37

Mike

62112

Let $f:Eto F$ be linear where $E,F$ are vector spaces and $dim E<infty$. Let $H$ be such that $E=ker foplus H$, then I wish to prove that $H $ is isomorphic to $ operatornameIm f.$

Here is what I have done:

If $f:Eto F$ is linear and $dim E<infty$, then $$dim E=dim ker f+dim operatornameIm f. ;;;;;(1)$$ Since $E=ker foplus H$, then by formula of four-dimension $$dim E=dim ker f+dim H. ;;;;;(2)$$
From $(1)$ and $(2)$, we have that
$$dim H=dim operatornameIm f.$$
Then, $H $ is isomorphic to $ operatornameIm f.$ Please, I'm I right or wrong? If no, I would love to see better solutions.

linear-algebra abstract-algebra

share|cite|improve this question

edited Jul 27 at 17:01

asked Jul 27 at 16:37

Mike

62112

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 17:01

edited Jul 27 at 17:01

edited Jul 27 at 17:01

asked Jul 27 at 16:37

Mike

62112

asked Jul 27 at 16:37

Mike

62112

asked Jul 27 at 16:37

Mike

62112

62112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hardmath: My apologies for that!
  – Mike
  Jul 27 at 16:54
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hardmath: My apologies for that!
  – Mike
  Jul 27 at 16:54
  

  
  
  
  

@hardmath: My apologies for that!
– Mike
Jul 27 at 16:54

@hardmath: My apologies for that!
– Mike
Jul 27 at 16:54

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote



accepted

Yes, this looks correct.

Since you have an explicit assumption that $E$ is finite-dimensional, this is probably also what you're supposed to do.

But the conclusion is also true in the infinite-dimensional case -- in fact you should be able to prove that the restriction of $f$ to $H$ is the isomorphism you're looking for, without counting dimensions at all.

share|cite|improve this answer

edited Jul 27 at 17:04

answered Jul 27 at 16:50

Henning Makholm

225k16290516

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, I get it now!
  – Mike
  Jul 27 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, how do I do it without counting on dimensions?
  – Mike
  Jul 27 at 17:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mike: As I wrote: The restriction of $f$ to $H$ is an isomorphism $HtooperatornameIm f$. It should be simple to prove that it is (a) linear, (b) surjective, and (c) injective.
  – Henning Makholm
  Jul 27 at 18:45
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

Yes, your proof is right. It can be seen as a direct consequence of the rank–nullity theorem, the Grassmann formula and the fact that

Two finite dimensional vector spaces over a field $K$ are isomprphic iff they have the same dimension

The proof of this is straightforward because if they are isomorphic there is an isomorphism $phi$ that maps the base of one onto the base to the other so they have the same dimension. On the other hand if they have the same dimension they are both isomorphic to $K$ and so isomorphic to each other

share|cite|improve this answer

answered Jul 27 at 16:48

Davide Morgante

1,741220

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for your response!
  – Mike
  Jul 27 at 17:10
  

  
  
  
  

add a comment | 

up vote
1
down vote

The result is true regardless the dimension of $E$; that is, even when $E$ is infinite-dimensional:

$H$ is a complement of $textker f$, and all its complements are isomorphic to $V/textker f$. From this and applying the First Isomorphism Theorem, we get that
$$H simeq V/textker f simeq textIm f.$$

share|cite|improve this answer

answered Jul 27 at 18:19

Don

1263

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Yes, this looks correct.

Since you have an explicit assumption that $E$ is finite-dimensional, this is probably also what you're supposed to do.

But the conclusion is also true in the infinite-dimensional case -- in fact you should be able to prove that the restriction of $f$ to $H$ is the isomorphism you're looking for, without counting dimensions at all.

share|cite|improve this answer

edited Jul 27 at 17:04

answered Jul 27 at 16:50

Henning Makholm

225k16290516

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, I get it now!
  – Mike
  Jul 27 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, how do I do it without counting on dimensions?
  – Mike
  Jul 27 at 17:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mike: As I wrote: The restriction of $f$ to $H$ is an isomorphism $HtooperatornameIm f$. It should be simple to prove that it is (a) linear, (b) surjective, and (c) injective.
  – Henning Makholm
  Jul 27 at 18:45
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Yes, this looks correct.

Since you have an explicit assumption that $E$ is finite-dimensional, this is probably also what you're supposed to do.

But the conclusion is also true in the infinite-dimensional case -- in fact you should be able to prove that the restriction of $f$ to $H$ is the isomorphism you're looking for, without counting dimensions at all.

share|cite|improve this answer

edited Jul 27 at 17:04

answered Jul 27 at 16:50

Henning Makholm

225k16290516

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, I get it now!
  – Mike
  Jul 27 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, how do I do it without counting on dimensions?
  – Mike
  Jul 27 at 17:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mike: As I wrote: The restriction of $f$ to $H$ is an isomorphism $HtooperatornameIm f$. It should be simple to prove that it is (a) linear, (b) surjective, and (c) injective.
  – Henning Makholm
  Jul 27 at 18:45
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Yes, this looks correct.

Since you have an explicit assumption that $E$ is finite-dimensional, this is probably also what you're supposed to do.

But the conclusion is also true in the infinite-dimensional case -- in fact you should be able to prove that the restriction of $f$ to $H$ is the isomorphism you're looking for, without counting dimensions at all.

share|cite|improve this answer

edited Jul 27 at 17:04

answered Jul 27 at 16:50

Henning Makholm

225k16290516

Yes, this looks correct.

Since you have an explicit assumption that $E$ is finite-dimensional, this is probably also what you're supposed to do.

But the conclusion is also true in the infinite-dimensional case -- in fact you should be able to prove that the restriction of $f$ to $H$ is the isomorphism you're looking for, without counting dimensions at all.

share|cite|improve this answer

edited Jul 27 at 17:04

answered Jul 27 at 16:50

Henning Makholm

225k16290516

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 17:04

edited Jul 27 at 17:04

edited Jul 27 at 17:04

answered Jul 27 at 16:50

Henning Makholm

225k16290516

answered Jul 27 at 16:50

Henning Makholm

225k16290516

answered Jul 27 at 16:50

Henning Makholm

225k16290516

225k16290516

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, I get it now!
  – Mike
  Jul 27 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, how do I do it without counting on dimensions?
  – Mike
  Jul 27 at 17:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mike: As I wrote: The restriction of $f$ to $H$ is an isomorphism $HtooperatornameIm f$. It should be simple to prove that it is (a) linear, (b) surjective, and (c) injective.
  – Henning Makholm
  Jul 27 at 18:45
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, I get it now!
  – Mike
  Jul 27 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, how do I do it without counting on dimensions?
  – Mike
  Jul 27 at 17:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mike: As I wrote: The restriction of $f$ to $H$ is an isomorphism $HtooperatornameIm f$. It should be simple to prove that it is (a) linear, (b) surjective, and (c) injective.
  – Henning Makholm
  Jul 27 at 18:45
  
  

  
  
  
  

Okay, I get it now!
– Mike
Jul 27 at 16:59

Okay, I get it now!
– Mike
Jul 27 at 16:59

So, how do I do it without counting on dimensions?
– Mike
Jul 27 at 17:07

So, how do I do it without counting on dimensions?
– Mike
Jul 27 at 17:07

@Mike: As I wrote: The restriction of $f$ to $H$ is an isomorphism $HtooperatornameIm f$. It should be simple to prove that it is (a) linear, (b) surjective, and (c) injective.
– Henning Makholm
Jul 27 at 18:45

@Mike: As I wrote: The restriction of $f$ to $H$ is an isomorphism $HtooperatornameIm f$. It should be simple to prove that it is (a) linear, (b) surjective, and (c) injective.
– Henning Makholm
Jul 27 at 18:45

add a comment | 

up vote
1
down vote

Yes, your proof is right. It can be seen as a direct consequence of the rank–nullity theorem, the Grassmann formula and the fact that

Two finite dimensional vector spaces over a field $K$ are isomprphic iff they have the same dimension

The proof of this is straightforward because if they are isomorphic there is an isomorphism $phi$ that maps the base of one onto the base to the other so they have the same dimension. On the other hand if they have the same dimension they are both isomorphic to $K$ and so isomorphic to each other

share|cite|improve this answer

answered Jul 27 at 16:48

Davide Morgante

1,741220

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for your response!
  – Mike
  Jul 27 at 17:10
  

  
  
  
  

add a comment | 

up vote
1
down vote

Yes, your proof is right. It can be seen as a direct consequence of the rank–nullity theorem, the Grassmann formula and the fact that

Two finite dimensional vector spaces over a field $K$ are isomprphic iff they have the same dimension

The proof of this is straightforward because if they are isomorphic there is an isomorphism $phi$ that maps the base of one onto the base to the other so they have the same dimension. On the other hand if they have the same dimension they are both isomorphic to $K$ and so isomorphic to each other

share|cite|improve this answer

answered Jul 27 at 16:48

Davide Morgante

1,741220

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for your response!
  – Mike
  Jul 27 at 17:10
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Yes, your proof is right. It can be seen as a direct consequence of the rank–nullity theorem, the Grassmann formula and the fact that

Two finite dimensional vector spaces over a field $K$ are isomprphic iff they have the same dimension

The proof of this is straightforward because if they are isomorphic there is an isomorphism $phi$ that maps the base of one onto the base to the other so they have the same dimension. On the other hand if they have the same dimension they are both isomorphic to $K$ and so isomorphic to each other

share|cite|improve this answer

answered Jul 27 at 16:48

Davide Morgante

1,741220

Yes, your proof is right. It can be seen as a direct consequence of the rank–nullity theorem, the Grassmann formula and the fact that

Two finite dimensional vector spaces over a field $K$ are isomprphic iff they have the same dimension

The proof of this is straightforward because if they are isomorphic there is an isomorphism $phi$ that maps the base of one onto the base to the other so they have the same dimension. On the other hand if they have the same dimension they are both isomorphic to $K$ and so isomorphic to each other

share|cite|improve this answer

answered Jul 27 at 16:48

Davide Morgante

1,741220

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 16:48

Davide Morgante

1,741220

answered Jul 27 at 16:48

Davide Morgante

1,741220

answered Jul 27 at 16:48

Davide Morgante

1,741220

1,741220

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for your response!
  – Mike
  Jul 27 at 17:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for your response!
  – Mike
  Jul 27 at 17:10
  

  
  
  
  

Thanks a lot for your response!
– Mike
Jul 27 at 17:10

Thanks a lot for your response!
– Mike
Jul 27 at 17:10

add a comment | 

up vote
1
down vote

The result is true regardless the dimension of $E$; that is, even when $E$ is infinite-dimensional:

$H$ is a complement of $textker f$, and all its complements are isomorphic to $V/textker f$. From this and applying the First Isomorphism Theorem, we get that
$$H simeq V/textker f simeq textIm f.$$

share|cite|improve this answer

answered Jul 27 at 18:19

Don

1263

add a comment | 

up vote
1
down vote

The result is true regardless the dimension of $E$; that is, even when $E$ is infinite-dimensional:

$H$ is a complement of $textker f$, and all its complements are isomorphic to $V/textker f$. From this and applying the First Isomorphism Theorem, we get that
$$H simeq V/textker f simeq textIm f.$$

share|cite|improve this answer

answered Jul 27 at 18:19

Don

1263

add a comment | 

up vote
1
down vote

up vote
1
down vote

The result is true regardless the dimension of $E$; that is, even when $E$ is infinite-dimensional:

$H$ is a complement of $textker f$, and all its complements are isomorphic to $V/textker f$. From this and applying the First Isomorphism Theorem, we get that
$$H simeq V/textker f simeq textIm f.$$

share|cite|improve this answer

answered Jul 27 at 18:19

Don

1263

The result is true regardless the dimension of $E$; that is, even when $E$ is infinite-dimensional:

$H$ is a complement of $textker f$, and all its complements are isomorphic to $V/textker f$. From this and applying the First Isomorphism Theorem, we get that
$$H simeq V/textker f simeq textIm f.$$

share|cite|improve this answer

answered Jul 27 at 18:19

Don

1263

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 18:19

Don

1263

answered Jul 27 at 18:19

Don

1263

answered Jul 27 at 18:19

Don

1263

1263

add a comment | 

add a comment | 

 

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