How to find the Kernel of $T: V rightarrow V$, where $T(f)=f'$? [closed]

1

$begingroup$

How do you find the kernel of $T: V rightarrow V$, where $T(f)=f'$, where $f$ is infinitely differentiable and $V$ is a vector space over R? Also, what is the dimension of the kernel?

linear-algebra

share|cite|improve this question

edited Jan 31 at 22:42

hoya2021

asked Jan 28 at 22:54

hoya2021hoya2021

365

$endgroup$

closed as off-topic by Surb, Kavi Rama Murthy, Gibbs, Theo Bendit, Lord Shark the Unknown Jan 29 at 5:16

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." – Kavi Rama Murthy, Gibbs, Theo Bendit

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $V$? This doesn't make any sense without further specification.
  $endgroup$
  – Thorgott
  Jan 28 at 22:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, if $f'=0$ then $f$ is constant... but I agree with @Thorgott that there is a lack of context in your question.
  $endgroup$
  – Surb
  Jan 28 at 23:06
  

  
  
  
  

add a comment | 

1

$begingroup$

How do you find the kernel of $T: V rightarrow V$, where $T(f)=f'$, where $f$ is infinitely differentiable and $V$ is a vector space over R? Also, what is the dimension of the kernel?

linear-algebra

share|cite|improve this question

edited Jan 31 at 22:42

hoya2021

asked Jan 28 at 22:54

hoya2021hoya2021

365

$endgroup$

closed as off-topic by Surb, Kavi Rama Murthy, Gibbs, Theo Bendit, Lord Shark the Unknown Jan 29 at 5:16

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." – Kavi Rama Murthy, Gibbs, Theo Bendit

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $V$? This doesn't make any sense without further specification.
  $endgroup$
  – Thorgott
  Jan 28 at 22:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, if $f'=0$ then $f$ is constant... but I agree with @Thorgott that there is a lack of context in your question.
  $endgroup$
  – Surb
  Jan 28 at 23:06
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

How do you find the kernel of $T: V rightarrow V$, where $T(f)=f'$, where $f$ is infinitely differentiable and $V$ is a vector space over R? Also, what is the dimension of the kernel?

linear-algebra

share|cite|improve this question

edited Jan 31 at 22:42

hoya2021

asked Jan 28 at 22:54

hoya2021hoya2021

365

$endgroup$

How do you find the kernel of $T: V rightarrow V$, where $T(f)=f'$, where $f$ is infinitely differentiable and $V$ is a vector space over R? Also, what is the dimension of the kernel?

linear-algebra

linear-algebra

share|cite|improve this question

edited Jan 31 at 22:42

hoya2021

asked Jan 28 at 22:54

hoya2021hoya2021

365

share|cite|improve this question

edited Jan 31 at 22:42

hoya2021

asked Jan 28 at 22:54

hoya2021hoya2021

365

share|cite|improve this question

share|cite|improve this question

edited Jan 31 at 22:42

hoya2021

edited Jan 31 at 22:42

hoya2021

edited Jan 31 at 22:42

hoya2021

asked Jan 28 at 22:54

hoya2021hoya2021

365

asked Jan 28 at 22:54

hoya2021hoya2021

365

asked Jan 28 at 22:54

hoya2021hoya2021

365

365

closed as off-topic by Surb, Kavi Rama Murthy, Gibbs, Theo Bendit, Lord Shark the Unknown Jan 29 at 5:16

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." – Kavi Rama Murthy, Gibbs, Theo Bendit

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

closed as off-topic by Surb, Kavi Rama Murthy, Gibbs, Theo Bendit, Lord Shark the Unknown Jan 29 at 5:16

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." – Kavi Rama Murthy, Gibbs, Theo Bendit

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $V$? This doesn't make any sense without further specification.
  $endgroup$
  – Thorgott
  Jan 28 at 22:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, if $f'=0$ then $f$ is constant... but I agree with @Thorgott that there is a lack of context in your question.
  $endgroup$
  – Surb
  Jan 28 at 23:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $V$? This doesn't make any sense without further specification.
  $endgroup$
  – Thorgott
  Jan 28 at 22:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, if $f'=0$ then $f$ is constant... but I agree with @Thorgott that there is a lack of context in your question.
  $endgroup$
  – Surb
  Jan 28 at 23:06
  

  
  
  
  

$begingroup$
What is $V$? This doesn't make any sense without further specification.
$endgroup$
– Thorgott
Jan 28 at 22:55

$begingroup$
What is $V$? This doesn't make any sense without further specification.
$endgroup$
– Thorgott
Jan 28 at 22:55

2

2

$begingroup$
Well, if $f'=0$ then $f$ is constant... but I agree with @Thorgott that there is a lack of context in your question.
$endgroup$
– Surb
Jan 28 at 23:06

$begingroup$
Well, if $f'=0$ then $f$ is constant... but I agree with @Thorgott that there is a lack of context in your question.
$endgroup$
– Surb
Jan 28 at 23:06

add a comment | 

2 Answers
2

active

oldest

votes

1

$begingroup$

Supposing that $V$ is some suitable subset of differentiable functions,
e.g. $C^1$-functions on the interval $[0,1]$:

You can show that a function's derivative vanishes on an interval if and only it is constant, using the mean value theorem.
Thus the functions that are mapped to the zero function by $T$ are precisely the constant functions.

share|cite|improve this answer

answered Jan 28 at 23:41

ffffforallffffforall

36028

$endgroup$

add a comment | 

0

$begingroup$

Let $V$ denote the space of all constant functions $f : mathbb{R} rightarrow mathbb{R}$ over $mathbb{R}$ with regular function addition and scalar multiplication. Then,

$$text{ker}(T) = {f in V mid T(f) = f' = 0} = V$$

and since the constant function $1:mathbb{R} rightarrow mathbb{R}$ forms a basis for $V$, we have

$$ text{dim }text{ker}(T) = 1$$

Now let $W$ denote the space of containing only the constant function $0:mathbb{R} rightarrow mathbb{R}$ with the same field and operations as $V$ but restricted to $W$. Then,

$$text{ker}(T) = {f in W mid T(f) = f' = 0} = W$$

and since $W$ contains only the zero function, we have

$$text{dim }text{ker}(T) = 0$$

Do you see now that you needed more information in your question to receive a more appropriate/specific answer? It was too broad, and I am pointing out that the answer depends on your space $V$. If you're looking for the general procedure, then I recommend looking up the relevant definitions, and then refining your question afterwards.

share|cite|improve this answer

answered Jan 28 at 23:39

MetricMetric

1,23659

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Supposing that $V$ is some suitable subset of differentiable functions,
e.g. $C^1$-functions on the interval $[0,1]$:

You can show that a function's derivative vanishes on an interval if and only it is constant, using the mean value theorem.
Thus the functions that are mapped to the zero function by $T$ are precisely the constant functions.

share|cite|improve this answer

answered Jan 28 at 23:41

ffffforallffffforall

36028

$endgroup$

add a comment | 

1

$begingroup$

Supposing that $V$ is some suitable subset of differentiable functions,
e.g. $C^1$-functions on the interval $[0,1]$:

You can show that a function's derivative vanishes on an interval if and only it is constant, using the mean value theorem.
Thus the functions that are mapped to the zero function by $T$ are precisely the constant functions.

share|cite|improve this answer

answered Jan 28 at 23:41

ffffforallffffforall

36028

$endgroup$

add a comment | 

1

1

1

$begingroup$

Supposing that $V$ is some suitable subset of differentiable functions,
e.g. $C^1$-functions on the interval $[0,1]$:

You can show that a function's derivative vanishes on an interval if and only it is constant, using the mean value theorem.
Thus the functions that are mapped to the zero function by $T$ are precisely the constant functions.

share|cite|improve this answer

answered Jan 28 at 23:41

ffffforallffffforall

36028

$endgroup$

Supposing that $V$ is some suitable subset of differentiable functions,
e.g. $C^1$-functions on the interval $[0,1]$:

You can show that a function's derivative vanishes on an interval if and only it is constant, using the mean value theorem.
Thus the functions that are mapped to the zero function by $T$ are precisely the constant functions.

share|cite|improve this answer

answered Jan 28 at 23:41

ffffforallffffforall

36028

share|cite|improve this answer

share|cite|improve this answer

answered Jan 28 at 23:41

ffffforallffffforall

36028

answered Jan 28 at 23:41

ffffforallffffforall

36028

answered Jan 28 at 23:41

ffffforallffffforall

36028

36028

add a comment | 

add a comment | 

0

$begingroup$

Let $V$ denote the space of all constant functions $f : mathbb{R} rightarrow mathbb{R}$ over $mathbb{R}$ with regular function addition and scalar multiplication. Then,

$$text{ker}(T) = {f in V mid T(f) = f' = 0} = V$$

and since the constant function $1:mathbb{R} rightarrow mathbb{R}$ forms a basis for $V$, we have

$$ text{dim }text{ker}(T) = 1$$

Now let $W$ denote the space of containing only the constant function $0:mathbb{R} rightarrow mathbb{R}$ with the same field and operations as $V$ but restricted to $W$. Then,

$$text{ker}(T) = {f in W mid T(f) = f' = 0} = W$$

and since $W$ contains only the zero function, we have

$$text{dim }text{ker}(T) = 0$$

Do you see now that you needed more information in your question to receive a more appropriate/specific answer? It was too broad, and I am pointing out that the answer depends on your space $V$. If you're looking for the general procedure, then I recommend looking up the relevant definitions, and then refining your question afterwards.

share|cite|improve this answer

answered Jan 28 at 23:39

MetricMetric

1,23659

$endgroup$

add a comment | 

0

$begingroup$

Let $V$ denote the space of all constant functions $f : mathbb{R} rightarrow mathbb{R}$ over $mathbb{R}$ with regular function addition and scalar multiplication. Then,

$$text{ker}(T) = {f in V mid T(f) = f' = 0} = V$$

and since the constant function $1:mathbb{R} rightarrow mathbb{R}$ forms a basis for $V$, we have

$$ text{dim }text{ker}(T) = 1$$

Now let $W$ denote the space of containing only the constant function $0:mathbb{R} rightarrow mathbb{R}$ with the same field and operations as $V$ but restricted to $W$. Then,

$$text{ker}(T) = {f in W mid T(f) = f' = 0} = W$$

and since $W$ contains only the zero function, we have

$$text{dim }text{ker}(T) = 0$$

Do you see now that you needed more information in your question to receive a more appropriate/specific answer? It was too broad, and I am pointing out that the answer depends on your space $V$. If you're looking for the general procedure, then I recommend looking up the relevant definitions, and then refining your question afterwards.

share|cite|improve this answer

answered Jan 28 at 23:39

MetricMetric

1,23659

$endgroup$

add a comment | 

0

0

0

$begingroup$

Let $V$ denote the space of all constant functions $f : mathbb{R} rightarrow mathbb{R}$ over $mathbb{R}$ with regular function addition and scalar multiplication. Then,

$$text{ker}(T) = {f in V mid T(f) = f' = 0} = V$$

and since the constant function $1:mathbb{R} rightarrow mathbb{R}$ forms a basis for $V$, we have

$$ text{dim }text{ker}(T) = 1$$

Now let $W$ denote the space of containing only the constant function $0:mathbb{R} rightarrow mathbb{R}$ with the same field and operations as $V$ but restricted to $W$. Then,

$$text{ker}(T) = {f in W mid T(f) = f' = 0} = W$$

and since $W$ contains only the zero function, we have

$$text{dim }text{ker}(T) = 0$$

Do you see now that you needed more information in your question to receive a more appropriate/specific answer? It was too broad, and I am pointing out that the answer depends on your space $V$. If you're looking for the general procedure, then I recommend looking up the relevant definitions, and then refining your question afterwards.

share|cite|improve this answer

answered Jan 28 at 23:39

MetricMetric

1,23659

$endgroup$

Let $V$ denote the space of all constant functions $f : mathbb{R} rightarrow mathbb{R}$ over $mathbb{R}$ with regular function addition and scalar multiplication. Then,

$$text{ker}(T) = {f in V mid T(f) = f' = 0} = V$$

and since the constant function $1:mathbb{R} rightarrow mathbb{R}$ forms a basis for $V$, we have

$$ text{dim }text{ker}(T) = 1$$

Now let $W$ denote the space of containing only the constant function $0:mathbb{R} rightarrow mathbb{R}$ with the same field and operations as $V$ but restricted to $W$. Then,

$$text{ker}(T) = {f in W mid T(f) = f' = 0} = W$$

and since $W$ contains only the zero function, we have

$$text{dim }text{ker}(T) = 0$$

Do you see now that you needed more information in your question to receive a more appropriate/specific answer? It was too broad, and I am pointing out that the answer depends on your space $V$. If you're looking for the general procedure, then I recommend looking up the relevant definitions, and then refining your question afterwards.

share|cite|improve this answer

answered Jan 28 at 23:39

MetricMetric

1,23659

share|cite|improve this answer

share|cite|improve this answer

answered Jan 28 at 23:39

MetricMetric

1,23659

answered Jan 28 at 23:39

MetricMetric

1,23659

answered Jan 28 at 23:39

MetricMetric

1,23659

1,23659

add a comment | 

add a comment | 

This page is only for reference, If you need detailed information, please check here