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Show that if $T$ is surjective and spans $V$, then $T(S)$ spans $W$.

2

$begingroup$

Given that $T: V to W$ is a linear transformation from $V$ to $W$. Show that if $T$ is surjective and $Ssubset V$ spans $V$, then $T(S)$ spans $W$.

I think the main thing stumping me right now is how to use spanning to show anything or what I would need to prove in order to show that $W$ has been successfully spanned. I understand how to use the definition of linear transformation and surjective, but I can't even get started (OR CAN I???) without a clear understanding of how spanning works.

I understand span in concrete numbers, so I need help getting a more abstract concept of $operatorname{span} (S)$, if possible. And I tried YouTube and reading definitions already, I am truly stuck.

linear-algebra linear-transformations transformation

share|cite|improve this question

edited Mar 23 at 22:05

Rócherz

3,0263823

asked Jan 22 '17 at 1:54

SnackbreakSnackbreak

314

$endgroup$

• 
  
  
  

  
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  $begingroup$
  How would a linear map span a subspace? What is S in all this?
  $endgroup$
  – Bernard
  Jan 22 '17 at 1:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $S$? It seems to have just shown up out of nowhere in this question.
  $endgroup$
  – The Count
  Jan 22 '17 at 1:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited your question. Please check that this is what you meant.
  $endgroup$
  – user378947
  Jan 22 '17 at 2:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, thank you for editing, that better says what I was trying to say.
  $endgroup$
  – Snackbreak
  Jan 22 '17 at 5:01
  

  
  
  
  

add a comment | 

2

$begingroup$

Given that $T: V to W$ is a linear transformation from $V$ to $W$. Show that if $T$ is surjective and $Ssubset V$ spans $V$, then $T(S)$ spans $W$.

I think the main thing stumping me right now is how to use spanning to show anything or what I would need to prove in order to show that $W$ has been successfully spanned. I understand how to use the definition of linear transformation and surjective, but I can't even get started (OR CAN I???) without a clear understanding of how spanning works.

I understand span in concrete numbers, so I need help getting a more abstract concept of $operatorname{span} (S)$, if possible. And I tried YouTube and reading definitions already, I am truly stuck.

linear-algebra linear-transformations transformation

share|cite|improve this question

edited Mar 23 at 22:05

Rócherz

3,0263823

asked Jan 22 '17 at 1:54

SnackbreakSnackbreak

314

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How would a linear map span a subspace? What is S in all this?
  $endgroup$
  – Bernard
  Jan 22 '17 at 1:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $S$? It seems to have just shown up out of nowhere in this question.
  $endgroup$
  – The Count
  Jan 22 '17 at 1:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited your question. Please check that this is what you meant.
  $endgroup$
  – user378947
  Jan 22 '17 at 2:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, thank you for editing, that better says what I was trying to say.
  $endgroup$
  – Snackbreak
  Jan 22 '17 at 5:01
  

  
  
  
  

add a comment | 

$begingroup$

Given that $T: V to W$ is a linear transformation from $V$ to $W$. Show that if $T$ is surjective and $Ssubset V$ spans $V$, then $T(S)$ spans $W$.

I think the main thing stumping me right now is how to use spanning to show anything or what I would need to prove in order to show that $W$ has been successfully spanned. I understand how to use the definition of linear transformation and surjective, but I can't even get started (OR CAN I???) without a clear understanding of how spanning works.

I understand span in concrete numbers, so I need help getting a more abstract concept of $operatorname{span} (S)$, if possible. And I tried YouTube and reading definitions already, I am truly stuck.

linear-algebra linear-transformations transformation

share|cite|improve this question

edited Mar 23 at 22:05

Rócherz

3,0263823

asked Jan 22 '17 at 1:54

SnackbreakSnackbreak

314

$endgroup$

share|cite|improve this question

edited Mar 23 at 22:05

Rócherz

3,0263823

asked Jan 22 '17 at 1:54

SnackbreakSnackbreak

314

share|cite|improve this question

edited Mar 23 at 22:05

Rócherz

3,0263823

asked Jan 22 '17 at 1:54

SnackbreakSnackbreak

314

edited Mar 23 at 22:05

Rócherz

3,0263823

edited Mar 23 at 22:05

Rócherz

3,0263823

asked Jan 22 '17 at 1:54

SnackbreakSnackbreak

314

asked Jan 22 '17 at 1:54

SnackbreakSnackbreak

314

2 Answers
2

active

oldest

votes

1

$begingroup$

$S$ spans $V$ simply means that every element $v in V$ can be written (not necessarily uniquely) as a finite linear combination of elements of $S$.

so it is required to prove that every element $w in W$ can be written as a finite linear combination of elements of $T(S)$

now, the fact that $T$ is surjective means that any $w in W$ is the image (under the mapping $T$) of some (not necessarily unique) element of $V$. i.e. we can find $u in V$ such that:

$$
w = T(u)
$$

write $u$ as a finite linear combination of elements of $S$:

$$
u = sum_k lambda_k s_k
$$
where the $lambda_k$ are scalars and each $s_k in S$

since the mapping $T$ is linear we have:

$$
w = T(u) = Tbigg(sum_k lambda_k s_k bigg) = sum_k lambda_k T(s_k)
$$
showing that $w$ is a linear combination of elements of $T(S)$ as required.

share|cite|improve this answer

answered Jan 22 '17 at 3:53

David HoldenDavid Holden

14.9k21226

$endgroup$

add a comment | 

2

$begingroup$

Let $win W$. As $T$ is surjective we know that there is some $vin V$ such that $T(v)=w$. We also know that $S$ spans $V$, so there are $v_1,...,v_kin S$ and scalars $lambda_1,...,lambda_k$ such that $v=lambda_1v_1+cdots+lambda_kv_k.$ Applying $T$ to both sides of this equation yields the desired conclusion.

share|cite|improve this answer

answered Jan 22 '17 at 2:00

user378947

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, saying that a subset $S$ of a vector space $V$ spans $V$ means that for every $vin V$ there are a finite number of vectors $v_1,...,v_kin S$ and scalars $lambda_1,...,lambda_k$ such that $v=lambda_1v_1+cdots+lambda_kv_k$. In words, that every vector in $V$ can be expressed as a finite linear combination of elements in $S$.
  $endgroup$
  – user378947
  Jan 22 '17 at 2:07
  
  
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, I think the lambdas were throwing me off in the Wikipedia definition, I wasn't sure how to express anything useful with a sum of (λ sub i)(v sub i), but you breaking it down like that helps significantly. Thank you! @ mathbeing @David Holden
  $endgroup$
  – Snackbreak
  Jan 22 '17 at 5:05
  

  
  
  
  

add a comment | 

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1

$begingroup$

$S$ spans $V$ simply means that every element $v in V$ can be written (not necessarily uniquely) as a finite linear combination of elements of $S$.

so it is required to prove that every element $w in W$ can be written as a finite linear combination of elements of $T(S)$

now, the fact that $T$ is surjective means that any $w in W$ is the image (under the mapping $T$) of some (not necessarily unique) element of $V$. i.e. we can find $u in V$ such that:

$$
w = T(u)
$$

write $u$ as a finite linear combination of elements of $S$:

$$
u = sum_k lambda_k s_k
$$
where the $lambda_k$ are scalars and each $s_k in S$

since the mapping $T$ is linear we have:

$$
w = T(u) = Tbigg(sum_k lambda_k s_k bigg) = sum_k lambda_k T(s_k)
$$
showing that $w$ is a linear combination of elements of $T(S)$ as required.

share|cite|improve this answer

answered Jan 22 '17 at 3:53

David HoldenDavid Holden

14.9k21226

$endgroup$

add a comment | 

1

$begingroup$

$S$ spans $V$ simply means that every element $v in V$ can be written (not necessarily uniquely) as a finite linear combination of elements of $S$.

so it is required to prove that every element $w in W$ can be written as a finite linear combination of elements of $T(S)$

now, the fact that $T$ is surjective means that any $w in W$ is the image (under the mapping $T$) of some (not necessarily unique) element of $V$. i.e. we can find $u in V$ such that:

$$
w = T(u)
$$

write $u$ as a finite linear combination of elements of $S$:

$$
u = sum_k lambda_k s_k
$$
where the $lambda_k$ are scalars and each $s_k in S$

since the mapping $T$ is linear we have:

$$
w = T(u) = Tbigg(sum_k lambda_k s_k bigg) = sum_k lambda_k T(s_k)
$$
showing that $w$ is a linear combination of elements of $T(S)$ as required.

share|cite|improve this answer

answered Jan 22 '17 at 3:53

David HoldenDavid Holden

14.9k21226

$endgroup$

add a comment | 

$begingroup$

$S$ spans $V$ simply means that every element $v in V$ can be written (not necessarily uniquely) as a finite linear combination of elements of $S$.

so it is required to prove that every element $w in W$ can be written as a finite linear combination of elements of $T(S)$

now, the fact that $T$ is surjective means that any $w in W$ is the image (under the mapping $T$) of some (not necessarily unique) element of $V$. i.e. we can find $u in V$ such that:

$$
w = T(u)
$$

write $u$ as a finite linear combination of elements of $S$:

$$
u = sum_k lambda_k s_k
$$
where the $lambda_k$ are scalars and each $s_k in S$

since the mapping $T$ is linear we have:

$$
w = T(u) = Tbigg(sum_k lambda_k s_k bigg) = sum_k lambda_k T(s_k)
$$
showing that $w$ is a linear combination of elements of $T(S)$ as required.

share|cite|improve this answer

answered Jan 22 '17 at 3:53

David HoldenDavid Holden

14.9k21226

$endgroup$

share|cite|improve this answer

answered Jan 22 '17 at 3:53

David HoldenDavid Holden

14.9k21226

answered Jan 22 '17 at 3:53

David HoldenDavid Holden

14.9k21226

answered Jan 22 '17 at 3:53

David HoldenDavid Holden

14.9k21226