What is the dimension of the kernel of a linear transformation from infinite dimensional to finite...
0
$begingroup$
Let T:V→W be a linear transformation where V is an infinite-dimensional vector space and W is a finite-dimensional vector space. What is the dimension of the kernel?
Hello everyone. Sorry my English, it's not my first language.
I tried doing this:
T(v1)=w1
T(v2)=w2
...
T(vn)=wn
But V has infinite vectors, so remaining vectors are in the kernel.
I don't know if I can use dim V = dim KerT + dim ImT here.
I'm not sure of the things I said. If someone may help me I would appreciate a lot. Thank you very much for attention!
linear-algebra abstract-algebra linear-transformations
edited Jan 9 at 22:18
jgon
13.5k22041
asked Jan 9 at 21:26
HarryHarry
83
$endgroup$
|
show 2 more comments
0
$begingroup$
Let T:V→W be a linear transformation where V is an infinite-dimensional vector space and W is a finite-dimensional vector space. What is the dimension of the kernel?
Hello everyone. Sorry my English, it's not my first language.
I tried doing this:
T(v1)=w1
T(v2)=w2
...
T(vn)=wn
But V has infinite vectors, so remaining vectors are in the kernel.
I don't know if I can use dim V = dim KerT + dim ImT here.
I'm not sure of the things I said. If someone may help me I would appreciate a lot. Thank you very much for attention!
linear-algebra abstract-algebra linear-transformations
edited Jan 9 at 22:18
jgon
13.5k22041
asked Jan 9 at 21:26
HarryHarry
83
$endgroup$
$begingroup$
What is the point of writing down $T(v_k) = w_k$ where the $v_k,w_k$ are undefined???
$endgroup$
– copper.hat
Jan 9 at 21:27
$begingroup$
Sorry about this, i tried to say that v belongs to V and w belong to W.
$endgroup$
– Harry
Jan 9 at 21:36
$begingroup$
That is still meaningless.
$endgroup$
– copper.hat
Jan 9 at 21:36
1
$begingroup$
I understand that. But what is the relevance to the question other than noise? Without specifying $v_k,w_k$ it means nothing.
$endgroup$
– copper.hat
Jan 9 at 21:44
1
$begingroup$
Look, I have a passing familiarity with algebra, that is not the issue. The issue is that unless you specify what $v_k, w_k$ are the statement $T(v_k) = w_k$ is nothing but a bunch of symbols. Nowhere in your question do you specify what the $v_k,w_k$ are.
$endgroup$
– copper.hat
Jan 9 at 22:00
|
show 2 more comments
0
0
0
$begingroup$
Let T:V→W be a linear transformation where V is an infinite-dimensional vector space and W is a finite-dimensional vector space. What is the dimension of the kernel?
Hello everyone. Sorry my English, it's not my first language.
I tried doing this:
T(v1)=w1
T(v2)=w2
...
T(vn)=wn
But V has infinite vectors, so remaining vectors are in the kernel.
I don't know if I can use dim V = dim KerT + dim ImT here.
I'm not sure of the things I said. If someone may help me I would appreciate a lot. Thank you very much for attention!
linear-algebra abstract-algebra linear-transformations
edited Jan 9 at 22:18
jgon
13.5k22041
asked Jan 9 at 21:26
HarryHarry
83
$endgroup$
Let T:V→W be a linear transformation where V is an infinite-dimensional vector space and W is a finite-dimensional vector space. What is the dimension of the kernel?
Hello everyone. Sorry my English, it's not my first language.
I tried doing this:
T(v1)=w1
T(v2)=w2
...
T(vn)=wn
But V has infinite vectors, so remaining vectors are in the kernel.
I don't know if I can use dim V = dim KerT + dim ImT here.
I'm not sure of the things I said. If someone may help me I would appreciate a lot. Thank you very much for attention!
linear-algebra abstract-algebra linear-transformations
linear-algebra abstract-algebra linear-transformations
edited Jan 9 at 22:18
jgon
13.5k22041
asked Jan 9 at 21:26
HarryHarry
83
edited Jan 9 at 22:18
jgon
13.5k22041
asked Jan 9 at 21:26
HarryHarry
83
edited Jan 9 at 22:18
jgon
13.5k22041
edited Jan 9 at 22:18
jgon
13.5k22041
edited Jan 9 at 22:18
jgon
13.5k22041
13.5k22041
asked Jan 9 at 21:26
HarryHarry
83
asked Jan 9 at 21:26
HarryHarry
83
asked Jan 9 at 21:26
HarryHarry
83
83
$begingroup$
What is the point of writing down $T(v_k) = w_k$ where the $v_k,w_k$ are undefined???
$endgroup$
– copper.hat
Jan 9 at 21:27
$begingroup$
Sorry about this, i tried to say that v belongs to V and w belong to W.
$endgroup$
– Harry
Jan 9 at 21:36
$begingroup$
That is still meaningless.
$endgroup$
– copper.hat
Jan 9 at 21:36
1
$begingroup$
I understand that. But what is the relevance to the question other than noise? Without specifying $v_k,w_k$ it means nothing.
$endgroup$
– copper.hat
Jan 9 at 21:44
1
$begingroup$
Look, I have a passing familiarity with algebra, that is not the issue. The issue is that unless you specify what $v_k, w_k$ are the statement $T(v_k) = w_k$ is nothing but a bunch of symbols. Nowhere in your question do you specify what the $v_k,w_k$ are.
$endgroup$
– copper.hat
Jan 9 at 22:00
|
show 2 more comments
$begingroup$
What is the point of writing down $T(v_k) = w_k$ where the $v_k,w_k$ are undefined???
$endgroup$
– copper.hat
Jan 9 at 21:27
$begingroup$
Sorry about this, i tried to say that v belongs to V and w belong to W.
$endgroup$
– Harry
Jan 9 at 21:36
$begingroup$
That is still meaningless.
$endgroup$
– copper.hat
Jan 9 at 21:36
1
$begingroup$
I understand that. But what is the relevance to the question other than noise? Without specifying $v_k,w_k$ it means nothing.
$endgroup$
– copper.hat
Jan 9 at 21:44
1
$begingroup$
Look, I have a passing familiarity with algebra, that is not the issue. The issue is that unless you specify what $v_k, w_k$ are the statement $T(v_k) = w_k$ is nothing but a bunch of symbols. Nowhere in your question do you specify what the $v_k,w_k$ are.
$endgroup$
– copper.hat
Jan 9 at 22:00
$begingroup$
What is the point of writing down $T(v_k) = w_k$ where the $v_k,w_k$ are undefined???
$endgroup$
– copper.hat
Jan 9 at 21:27
$begingroup$
What is the point of writing down $T(v_k) = w_k$ where the $v_k,w_k$ are undefined???
$endgroup$
– copper.hat
Jan 9 at 21:27
$begingroup$
Sorry about this, i tried to say that v belongs to V and w belong to W.
$endgroup$
– Harry
Jan 9 at 21:36
$begingroup$
Sorry about this, i tried to say that v belongs to V and w belong to W.
$endgroup$
– Harry
Jan 9 at 21:36
$begingroup$
That is still meaningless.
$endgroup$
– copper.hat
Jan 9 at 21:36
$begingroup$
That is still meaningless.
$endgroup$
– copper.hat
Jan 9 at 21:36
1
1
$begingroup$
I understand that. But what is the relevance to the question other than noise? Without specifying $v_k,w_k$ it means nothing.
$endgroup$
– copper.hat
Jan 9 at 21:44
$begingroup$
I understand that. But what is the relevance to the question other than noise? Without specifying $v_k,w_k$ it means nothing.
$endgroup$
– copper.hat
Jan 9 at 21:44
1
1
$begingroup$
Look, I have a passing familiarity with algebra, that is not the issue. The issue is that unless you specify what $v_k, w_k$ are the statement $T(v_k) = w_k$ is nothing but a bunch of symbols. Nowhere in your question do you specify what the $v_k,w_k$ are.
$endgroup$
– copper.hat
Jan 9 at 22:00
$begingroup$
Look, I have a passing familiarity with algebra, that is not the issue. The issue is that unless you specify what $v_k, w_k$ are the statement $T(v_k) = w_k$ is nothing but a bunch of symbols. Nowhere in your question do you specify what the $v_k,w_k$ are.
$endgroup$
– copper.hat
Jan 9 at 22:00
|
show 2 more comments
1 Answer
1
active
oldest
votes
3
$begingroup$
Suppose the dimension of the kernel is finite, so $ker f$ has ${y_1,dots,y_n}$ as basis.
If ${f(x_1),dots,f(x_m)}$ is a basis of the image of $f$, prove that
$$
{x_1,dots,x_m,y_1,dots,y_n}
$$
is a spanning set for $V$ (actually a basis).
answered Jan 9 at 21:29
egregegreg
180k1485202
$endgroup$
$begingroup$
May I use this: math.stackexchange.com/questions/3070298/…?
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
If we have a linear transformation T:V→W and the dimension of V is greater than the dimension of W, then the kernel of T has to have a dimension at least as large as the difference. So, when V is infinite dimensional and W has a finite dimension, then the kernel has to be infinite dimensional.
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
My question is, can I use the rank-nullity theorem in this case? When I have an infinite dimensional vector space?
$endgroup$
– Harry
Jan 11 at 23:29
$begingroup$
@Harry I didn't use the rank-nullity theorem. What I proved (in a way very similar to the theorem) that if the image is finite dimensional and the kernel is finite dimensional, then the domain is finite dimensional as well. Hence in your case the dimension of the kernel must be infinite.
$endgroup$
– egreg
Jan 11 at 23:45
$begingroup$
I understood! Thank you very much for your help. I'm very happy now.
$endgroup$
– Harry
Jan 11 at 23:51
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
Suppose the dimension of the kernel is finite, so $ker f$ has ${y_1,dots,y_n}$ as basis.
If ${f(x_1),dots,f(x_m)}$ is a basis of the image of $f$, prove that
$$
{x_1,dots,x_m,y_1,dots,y_n}
$$
is a spanning set for $V$ (actually a basis).
answered Jan 9 at 21:29
egregegreg
180k1485202
$endgroup$
$begingroup$
May I use this: math.stackexchange.com/questions/3070298/…?
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
If we have a linear transformation T:V→W and the dimension of V is greater than the dimension of W, then the kernel of T has to have a dimension at least as large as the difference. So, when V is infinite dimensional and W has a finite dimension, then the kernel has to be infinite dimensional.
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
My question is, can I use the rank-nullity theorem in this case? When I have an infinite dimensional vector space?
$endgroup$
– Harry
Jan 11 at 23:29
$begingroup$
@Harry I didn't use the rank-nullity theorem. What I proved (in a way very similar to the theorem) that if the image is finite dimensional and the kernel is finite dimensional, then the domain is finite dimensional as well. Hence in your case the dimension of the kernel must be infinite.
$endgroup$
– egreg
Jan 11 at 23:45
$begingroup$
I understood! Thank you very much for your help. I'm very happy now.
$endgroup$
– Harry
Jan 11 at 23:51
add a comment |
3
$begingroup$
Suppose the dimension of the kernel is finite, so $ker f$ has ${y_1,dots,y_n}$ as basis.
If ${f(x_1),dots,f(x_m)}$ is a basis of the image of $f$, prove that
$$
{x_1,dots,x_m,y_1,dots,y_n}
$$
is a spanning set for $V$ (actually a basis).
answered Jan 9 at 21:29
egregegreg
180k1485202
$endgroup$
$begingroup$
May I use this: math.stackexchange.com/questions/3070298/…?
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
If we have a linear transformation T:V→W and the dimension of V is greater than the dimension of W, then the kernel of T has to have a dimension at least as large as the difference. So, when V is infinite dimensional and W has a finite dimension, then the kernel has to be infinite dimensional.
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
My question is, can I use the rank-nullity theorem in this case? When I have an infinite dimensional vector space?
$endgroup$
– Harry
Jan 11 at 23:29
$begingroup$
@Harry I didn't use the rank-nullity theorem. What I proved (in a way very similar to the theorem) that if the image is finite dimensional and the kernel is finite dimensional, then the domain is finite dimensional as well. Hence in your case the dimension of the kernel must be infinite.
$endgroup$
– egreg
Jan 11 at 23:45
$begingroup$
I understood! Thank you very much for your help. I'm very happy now.
$endgroup$
– Harry
Jan 11 at 23:51
add a comment |
3
3
3
$begingroup$
Suppose the dimension of the kernel is finite, so $ker f$ has ${y_1,dots,y_n}$ as basis.
If ${f(x_1),dots,f(x_m)}$ is a basis of the image of $f$, prove that
$$
{x_1,dots,x_m,y_1,dots,y_n}
$$
is a spanning set for $V$ (actually a basis).
answered Jan 9 at 21:29
egregegreg
180k1485202
$endgroup$
Suppose the dimension of the kernel is finite, so $ker f$ has ${y_1,dots,y_n}$ as basis.
If ${f(x_1),dots,f(x_m)}$ is a basis of the image of $f$, prove that
$$
{x_1,dots,x_m,y_1,dots,y_n}
$$
is a spanning set for $V$ (actually a basis).
answered Jan 9 at 21:29
egregegreg
180k1485202
answered Jan 9 at 21:29
egregegreg
180k1485202
answered Jan 9 at 21:29
egregegreg
180k1485202
answered Jan 9 at 21:29
egregegreg
180k1485202
180k1485202
$begingroup$
May I use this: math.stackexchange.com/questions/3070298/…?
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
If we have a linear transformation T:V→W and the dimension of V is greater than the dimension of W, then the kernel of T has to have a dimension at least as large as the difference. So, when V is infinite dimensional and W has a finite dimension, then the kernel has to be infinite dimensional.
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
My question is, can I use the rank-nullity theorem in this case? When I have an infinite dimensional vector space?
$endgroup$
– Harry
Jan 11 at 23:29
$begingroup$
@Harry I didn't use the rank-nullity theorem. What I proved (in a way very similar to the theorem) that if the image is finite dimensional and the kernel is finite dimensional, then the domain is finite dimensional as well. Hence in your case the dimension of the kernel must be infinite.
$endgroup$
– egreg
Jan 11 at 23:45
$begingroup$
I understood! Thank you very much for your help. I'm very happy now.
$endgroup$
– Harry
Jan 11 at 23:51
add a comment |
$begingroup$
May I use this: math.stackexchange.com/questions/3070298/…?
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
If we have a linear transformation T:V→W and the dimension of V is greater than the dimension of W, then the kernel of T has to have a dimension at least as large as the difference. So, when V is infinite dimensional and W has a finite dimension, then the kernel has to be infinite dimensional.
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
My question is, can I use the rank-nullity theorem in this case? When I have an infinite dimensional vector space?
$endgroup$
– Harry
Jan 11 at 23:29
$begingroup$
@Harry I didn't use the rank-nullity theorem. What I proved (in a way very similar to the theorem) that if the image is finite dimensional and the kernel is finite dimensional, then the domain is finite dimensional as well. Hence in your case the dimension of the kernel must be infinite.
$endgroup$
– egreg
Jan 11 at 23:45
$begingroup$
I understood! Thank you very much for your help. I'm very happy now.
$endgroup$
– Harry
Jan 11 at 23:51
$begingroup$
May I use this: math.stackexchange.com/questions/3070298/…?
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
May I use this: math.stackexchange.com/questions/3070298/…?
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
If we have a linear transformation T:V→W and the dimension of V is greater than the dimension of W, then the kernel of T has to have a dimension at least as large as the difference. So, when V is infinite dimensional and W has a finite dimension, then the kernel has to be infinite dimensional.
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
If we have a linear transformation T:V→W and the dimension of V is greater than the dimension of W, then the kernel of T has to have a dimension at least as large as the difference. So, when V is infinite dimensional and W has a finite dimension, then the kernel has to be infinite dimensional.
$endgroup$
– Harry
Jan 11 at 23:26
$begingroup$
My question is, can I use the rank-nullity theorem in this case? When I have an infinite dimensional vector space?
$endgroup$
– Harry
Jan 11 at 23:29
$begingroup$
My question is, can I use the rank-nullity theorem in this case? When I have an infinite dimensional vector space?
$endgroup$
– Harry
Jan 11 at 23:29
$begingroup$
@Harry I didn't use the rank-nullity theorem. What I proved (in a way very similar to the theorem) that if the image is finite dimensional and the kernel is finite dimensional, then the domain is finite dimensional as well. Hence in your case the dimension of the kernel must be infinite.
$endgroup$
– egreg
Jan 11 at 23:45
$begingroup$
@Harry I didn't use the rank-nullity theorem. What I proved (in a way very similar to the theorem) that if the image is finite dimensional and the kernel is finite dimensional, then the domain is finite dimensional as well. Hence in your case the dimension of the kernel must be infinite.
$endgroup$
– egreg
Jan 11 at 23:45
$begingroup$
I understood! Thank you very much for your help. I'm very happy now.
$endgroup$
– Harry
Jan 11 at 23:51
$begingroup$
I understood! Thank you very much for your help. I'm very happy now.
$endgroup$
– Harry
Jan 11 at 23:51
add a comment |
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$begingroup$
What is the point of writing down $T(v_k) = w_k$ where the $v_k,w_k$ are undefined???
$endgroup$
– copper.hat
Jan 9 at 21:27
$begingroup$
Sorry about this, i tried to say that v belongs to V and w belong to W.
$endgroup$
– Harry
Jan 9 at 21:36
$begingroup$
That is still meaningless.
$endgroup$
– copper.hat
Jan 9 at 21:36
1
$begingroup$
I understand that. But what is the relevance to the question other than noise? Without specifying $v_k,w_k$ it means nothing.
$endgroup$
– copper.hat
Jan 9 at 21:44
1
$begingroup$
Look, I have a passing familiarity with algebra, that is not the issue. The issue is that unless you specify what $v_k, w_k$ are the statement $T(v_k) = w_k$ is nothing but a bunch of symbols. Nowhere in your question do you specify what the $v_k,w_k$ are.
$endgroup$
– copper.hat
Jan 9 at 22:00