A short linear algebra proof [closed]

0

$begingroup$

I need help with the following question:

Let S and T be linear operators (V to V) such that for every v in V:

$S(T(v)) = S(v)$

$T(S(v)) = T(v)$

Prove: $ker(T) = ker(S)$

Thanks in advance!

linear-algebra

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edited Feb 2 at 16:47

jvdhooft

5,65961641

asked Feb 2 at 16:44

FrogfireFrogfire

111

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closed as off-topic by José Carlos Santos, Robert Z, Dietrich Burde, Pierre-Guy Plamondon, YiFan Feb 2 at 21:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Robert Z, Dietrich Burde, Pierre-Guy Plamondon, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MathSE. What have you tried?
  $endgroup$
  – jvdhooft
  Feb 2 at 16:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $xinker T$. Then, $T(x)=0$ implying that $S(x)=S(T(x))=S(0)=0$ since $F(0)=0$ for any linear operator $F$, so $xinker S$. This shows $ker Tsubseteqker S$. Can you show the other way around?
  $endgroup$
  – learner
  Feb 2 at 16:50
  

  
  
  
  

add a comment | 

0

$begingroup$

I need help with the following question:

Let S and T be linear operators (V to V) such that for every v in V:

$S(T(v)) = S(v)$

$T(S(v)) = T(v)$

Prove: $ker(T) = ker(S)$

Thanks in advance!

linear-algebra

share|cite|improve this question

edited Feb 2 at 16:47

jvdhooft

5,65961641

asked Feb 2 at 16:44

FrogfireFrogfire

111

$endgroup$

closed as off-topic by José Carlos Santos, Robert Z, Dietrich Burde, Pierre-Guy Plamondon, YiFan Feb 2 at 21:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Robert Z, Dietrich Burde, Pierre-Guy Plamondon, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MathSE. What have you tried?
  $endgroup$
  – jvdhooft
  Feb 2 at 16:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $xinker T$. Then, $T(x)=0$ implying that $S(x)=S(T(x))=S(0)=0$ since $F(0)=0$ for any linear operator $F$, so $xinker S$. This shows $ker Tsubseteqker S$. Can you show the other way around?
  $endgroup$
  – learner
  Feb 2 at 16:50
  

  
  
  
  

add a comment | 

0

0

0

0

$begingroup$

I need help with the following question:

Let S and T be linear operators (V to V) such that for every v in V:

$S(T(v)) = S(v)$

$T(S(v)) = T(v)$

Prove: $ker(T) = ker(S)$

Thanks in advance!

linear-algebra

share|cite|improve this question

edited Feb 2 at 16:47

jvdhooft

5,65961641

asked Feb 2 at 16:44

FrogfireFrogfire

111

$endgroup$

I need help with the following question:

Let S and T be linear operators (V to V) such that for every v in V:

$S(T(v)) = S(v)$

$T(S(v)) = T(v)$

Prove: $ker(T) = ker(S)$

Thanks in advance!

linear-algebra

linear-algebra

share|cite|improve this question

edited Feb 2 at 16:47

jvdhooft

5,65961641

asked Feb 2 at 16:44

FrogfireFrogfire

111

share|cite|improve this question

edited Feb 2 at 16:47

jvdhooft

5,65961641

asked Feb 2 at 16:44

FrogfireFrogfire

111

share|cite|improve this question

share|cite|improve this question

edited Feb 2 at 16:47

jvdhooft

5,65961641

edited Feb 2 at 16:47

jvdhooft

5,65961641

edited Feb 2 at 16:47

jvdhooft

5,65961641

5,65961641

asked Feb 2 at 16:44

FrogfireFrogfire

111

asked Feb 2 at 16:44

FrogfireFrogfire

111

asked Feb 2 at 16:44

FrogfireFrogfire

111

111

closed as off-topic by José Carlos Santos, Robert Z, Dietrich Burde, Pierre-Guy Plamondon, YiFan Feb 2 at 21:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Robert Z, Dietrich Burde, Pierre-Guy Plamondon, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, Robert Z, Dietrich Burde, Pierre-Guy Plamondon, YiFan Feb 2 at 21:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Robert Z, Dietrich Burde, Pierre-Guy Plamondon, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MathSE. What have you tried?
  $endgroup$
  – jvdhooft
  Feb 2 at 16:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $xinker T$. Then, $T(x)=0$ implying that $S(x)=S(T(x))=S(0)=0$ since $F(0)=0$ for any linear operator $F$, so $xinker S$. This shows $ker Tsubseteqker S$. Can you show the other way around?
  $endgroup$
  – learner
  Feb 2 at 16:50
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MathSE. What have you tried?
  $endgroup$
  – jvdhooft
  Feb 2 at 16:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let $xinker T$. Then, $T(x)=0$ implying that $S(x)=S(T(x))=S(0)=0$ since $F(0)=0$ for any linear operator $F$, so $xinker S$. This shows $ker Tsubseteqker S$. Can you show the other way around?
  $endgroup$
  – learner
  Feb 2 at 16:50
  

  
  
  
  

$begingroup$
Welcome to MathSE. What have you tried?
$endgroup$
– jvdhooft
Feb 2 at 16:47

$begingroup$
Welcome to MathSE. What have you tried?
$endgroup$
– jvdhooft
Feb 2 at 16:47

1

1

$begingroup$
Let $xinker T$. Then, $T(x)=0$ implying that $S(x)=S(T(x))=S(0)=0$ since $F(0)=0$ for any linear operator $F$, so $xinker S$. This shows $ker Tsubseteqker S$. Can you show the other way around?
$endgroup$
– learner
Feb 2 at 16:50

$begingroup$
Let $xinker T$. Then, $T(x)=0$ implying that $S(x)=S(T(x))=S(0)=0$ since $F(0)=0$ for any linear operator $F$, so $xinker S$. This shows $ker Tsubseteqker S$. Can you show the other way around?
$endgroup$
– learner
Feb 2 at 16:50

add a comment | 

2 Answers
2

active

oldest

votes

2

$begingroup$

Prove the double inclusion. Try it yourself before reading below.

---

Here a proof.

If $vin text{Ker } T$ then $T(v)=0$, and the first assumption implies that $0=S(0)=S(v)$, so that $vin text{Ker } S$. The other inclusion can be proved in the same way by using the second assumption.

share|cite|improve this answer

answered Feb 2 at 16:50

AlessioDVAlessioDV

1,208114

$endgroup$

add a comment | 

0

$begingroup$

Hint: If $v in ker{S}$, then by definition $T(S(v)) = T(0)$. Since $T$ is a linear transformation, what must $T(0)$ be?

share|cite|improve this answer

answered Feb 2 at 17:17

user458276user458276

7431315

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

Prove the double inclusion. Try it yourself before reading below.

---

Here a proof.

If $vin text{Ker } T$ then $T(v)=0$, and the first assumption implies that $0=S(0)=S(v)$, so that $vin text{Ker } S$. The other inclusion can be proved in the same way by using the second assumption.

share|cite|improve this answer

answered Feb 2 at 16:50

AlessioDVAlessioDV

1,208114

$endgroup$

add a comment | 

2

$begingroup$

Prove the double inclusion. Try it yourself before reading below.

---

Here a proof.

If $vin text{Ker } T$ then $T(v)=0$, and the first assumption implies that $0=S(0)=S(v)$, so that $vin text{Ker } S$. The other inclusion can be proved in the same way by using the second assumption.

share|cite|improve this answer

answered Feb 2 at 16:50

AlessioDVAlessioDV

1,208114

$endgroup$

add a comment | 

2

2

2

$begingroup$

Prove the double inclusion. Try it yourself before reading below.

---

Here a proof.

If $vin text{Ker } T$ then $T(v)=0$, and the first assumption implies that $0=S(0)=S(v)$, so that $vin text{Ker } S$. The other inclusion can be proved in the same way by using the second assumption.

share|cite|improve this answer

answered Feb 2 at 16:50

AlessioDVAlessioDV

1,208114

$endgroup$

Prove the double inclusion. Try it yourself before reading below.

---

Here a proof.

If $vin text{Ker } T$ then $T(v)=0$, and the first assumption implies that $0=S(0)=S(v)$, so that $vin text{Ker } S$. The other inclusion can be proved in the same way by using the second assumption.

share|cite|improve this answer

answered Feb 2 at 16:50

AlessioDVAlessioDV

1,208114

share|cite|improve this answer

share|cite|improve this answer

answered Feb 2 at 16:50

AlessioDVAlessioDV

1,208114

answered Feb 2 at 16:50

AlessioDVAlessioDV

1,208114

answered Feb 2 at 16:50

AlessioDVAlessioDV

1,208114

1,208114

add a comment | 

add a comment | 

0

$begingroup$

Hint: If $v in ker{S}$, then by definition $T(S(v)) = T(0)$. Since $T$ is a linear transformation, what must $T(0)$ be?

share|cite|improve this answer

answered Feb 2 at 17:17

user458276user458276

7431315

$endgroup$

add a comment | 

0

$begingroup$

Hint: If $v in ker{S}$, then by definition $T(S(v)) = T(0)$. Since $T$ is a linear transformation, what must $T(0)$ be?

share|cite|improve this answer

answered Feb 2 at 17:17

user458276user458276

7431315

$endgroup$

add a comment | 

0

0

0

$begingroup$

Hint: If $v in ker{S}$, then by definition $T(S(v)) = T(0)$. Since $T$ is a linear transformation, what must $T(0)$ be?

share|cite|improve this answer

answered Feb 2 at 17:17

user458276user458276

7431315

$endgroup$

Hint: If $v in ker{S}$, then by definition $T(S(v)) = T(0)$. Since $T$ is a linear transformation, what must $T(0)$ be?

share|cite|improve this answer

answered Feb 2 at 17:17

user458276user458276

7431315

share|cite|improve this answer

share|cite|improve this answer

answered Feb 2 at 17:17

user458276user458276

7431315

answered Feb 2 at 17:17

user458276user458276

7431315

answered Feb 2 at 17:17

user458276user458276

7431315

7431315

add a comment | 

add a comment | 

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