Show that if $T$ is surjective and spans $V$, then $T(S)$ spans $W$. Announcing the arrival of...
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Show that if $T$ is surjective and spans $V$, then $T(S)$ spans $W$.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to show that a linear map is surjective?Soft question: Why freshmen feel linear algebra is abstract?How to show a set spans a space?Invariant subspace - simplified definitionDifferences between finding linear independence and finding a spanning set.Linear independence, Spanning, Injectivity, and SurjectivityCompleteness, spanning and orthonormal basesBuilding a surjective functionSpanning set definition and theoremShow that $textrm{trace}: textrm{End}(V)rightarrow k$ defined by $textrm{trace}(phi ):=textrm{trace}(M(phi ))$ is a linear map.
$begingroup$
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
edited Mar 23 at 22:05
Rócherz
3,0263823
asked Jan 22 '17 at 1:54
SnackbreakSnackbreak
314
$endgroup$
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
edited Mar 23 at 22:05
Rócherz
3,0263823
asked Jan 22 '17 at 1:54
SnackbreakSnackbreak
314
$endgroup$
1
$begingroup$
How would a linearmapspan 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
edited Mar 23 at 22:05
Rócherz
3,0263823
asked Jan 22 '17 at 1:54
SnackbreakSnackbreak
314
$endgroup$
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
linear-algebra linear-transformations transformation
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
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
edited Mar 23 at 22:05
Rócherz
3,0263823
3,0263823
asked Jan 22 '17 at 1:54
SnackbreakSnackbreak
314
asked Jan 22 '17 at 1:54
SnackbreakSnackbreak
314
asked Jan 22 '17 at 1:54
SnackbreakSnackbreak
314
314
1
$begingroup$
How would a linearmapspan 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 |
1
$begingroup$
How would a linearmapspan 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
1
1
$begingroup$
How would a linear
map span a subspace? What is S in all this?$endgroup$
– Bernard
Jan 22 '17 at 1:59
$begingroup$
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
$begingroup$
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
$begingroup$
I've edited your question. Please check that this is what you meant.
$endgroup$
– user378947
Jan 22 '17 at 2:02
1
1
$begingroup$
Yes, thank you for editing, that better says what I was trying to say.
$endgroup$
– Snackbreak
Jan 22 '17 at 5:01
$begingroup$
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 Answers
2
active
oldest
votes
$begingroup$
$S$ 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.
answered Jan 22 '17 at 3:53
David HoldenDavid Holden
14.9k21226
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
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 |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$S$ 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.
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.
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.
answered Jan 22 '17 at 3:53
David HoldenDavid Holden
14.9k21226
$endgroup$
$S$ 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.
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
answered Jan 22 '17 at 3:53
David HoldenDavid Holden
14.9k21226
14.9k21226
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
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 |
$begingroup$
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.
$endgroup$
$begingroup$
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
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 |
$begingroup$
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.
$endgroup$
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.
answered Jan 22 '17 at 2:00
user378947
$begingroup$
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
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 |
$begingroup$
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
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
$begingroup$
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
$begingroup$
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
1
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
$begingroup$
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
$begingroup$
How would a linear
mapspan 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