Linear transformation on infinite dimensional vector spaces
up vote
0
down vote
favorite
Let $E$ and $F$ be two infinite dimentional vector spaces.
And $Lin mathbb{L}(E,F)$, A continuous linear tranformation.
If $L$ is surjective implies that $L$ is bijective ?
general-topology functional-analysis linear-transformations
asked Nov 20 at 10:02
Anas BOUALII
253
add a comment |
up vote
0
down vote
favorite
Let $E$ and $F$ be two infinite dimentional vector spaces.
And $Lin mathbb{L}(E,F)$, A continuous linear tranformation.
If $L$ is surjective implies that $L$ is bijective ?
general-topology functional-analysis linear-transformations
asked Nov 20 at 10:02
Anas BOUALII
253
This is not true even in finite dimensional spaces...what is true is that surjectivity is equivalent to bijectivity for linear operators in $;mathcal L(V,V);$ , when $;dim V<infty;$ .
– DonAntonio
Nov 20 at 10:31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $E$ and $F$ be two infinite dimentional vector spaces.
And $Lin mathbb{L}(E,F)$, A continuous linear tranformation.
If $L$ is surjective implies that $L$ is bijective ?
general-topology functional-analysis linear-transformations
asked Nov 20 at 10:02
Anas BOUALII
253
Let $E$ and $F$ be two infinite dimentional vector spaces.
And $Lin mathbb{L}(E,F)$, A continuous linear tranformation.
If $L$ is surjective implies that $L$ is bijective ?
general-topology functional-analysis linear-transformations
general-topology functional-analysis linear-transformations
asked Nov 20 at 10:02
Anas BOUALII
253
asked Nov 20 at 10:02
Anas BOUALII
253
asked Nov 20 at 10:02
Anas BOUALII
253
asked Nov 20 at 10:02
Anas BOUALII
253
asked Nov 20 at 10:02
Anas BOUALII
253
253
This is not true even in finite dimensional spaces...what is true is that surjectivity is equivalent to bijectivity for linear operators in $;mathcal L(V,V);$ , when $;dim V<infty;$ .
– DonAntonio
Nov 20 at 10:31
add a comment |
This is not true even in finite dimensional spaces...what is true is that surjectivity is equivalent to bijectivity for linear operators in $;mathcal L(V,V);$ , when $;dim V<infty;$ .
– DonAntonio
Nov 20 at 10:31
This is not true even in finite dimensional spaces...what is true is that surjectivity is equivalent to bijectivity for linear operators in $;mathcal L(V,V);$ , when $;dim V<infty;$ .
– DonAntonio
Nov 20 at 10:31
This is not true even in finite dimensional spaces...what is true is that surjectivity is equivalent to bijectivity for linear operators in $;mathcal L(V,V);$ , when $;dim V<infty;$ .
– DonAntonio
Nov 20 at 10:31
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Consider the vector space $V$ of all sequences of real numbers. Let $L(a_1,a_2,cdots)=(a_2,a_3,cdots)$. Then $L$ is surjective but $L(0,0,cdots)=L(1,0,cdots)$ it is not injective.
answered Nov 20 at 10:06
Kavi Rama Murthy
44.7k31852
thank you professor. It answers my question.
– Anas BOUALII
Nov 20 at 10:08
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Consider the vector space $V$ of all sequences of real numbers. Let $L(a_1,a_2,cdots)=(a_2,a_3,cdots)$. Then $L$ is surjective but $L(0,0,cdots)=L(1,0,cdots)$ it is not injective.
answered Nov 20 at 10:06
Kavi Rama Murthy
44.7k31852
thank you professor. It answers my question.
– Anas BOUALII
Nov 20 at 10:08
add a comment |
up vote
1
down vote
accepted
Consider the vector space $V$ of all sequences of real numbers. Let $L(a_1,a_2,cdots)=(a_2,a_3,cdots)$. Then $L$ is surjective but $L(0,0,cdots)=L(1,0,cdots)$ it is not injective.
answered Nov 20 at 10:06
Kavi Rama Murthy
44.7k31852
thank you professor. It answers my question.
– Anas BOUALII
Nov 20 at 10:08
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Consider the vector space $V$ of all sequences of real numbers. Let $L(a_1,a_2,cdots)=(a_2,a_3,cdots)$. Then $L$ is surjective but $L(0,0,cdots)=L(1,0,cdots)$ it is not injective.
answered Nov 20 at 10:06
Kavi Rama Murthy
44.7k31852
Consider the vector space $V$ of all sequences of real numbers. Let $L(a_1,a_2,cdots)=(a_2,a_3,cdots)$. Then $L$ is surjective but $L(0,0,cdots)=L(1,0,cdots)$ it is not injective.
answered Nov 20 at 10:06
Kavi Rama Murthy
44.7k31852
answered Nov 20 at 10:06
Kavi Rama Murthy
44.7k31852
answered Nov 20 at 10:06
Kavi Rama Murthy
44.7k31852
answered Nov 20 at 10:06
Kavi Rama Murthy
44.7k31852
44.7k31852
thank you professor. It answers my question.
– Anas BOUALII
Nov 20 at 10:08
add a comment |
thank you professor. It answers my question.
– Anas BOUALII
Nov 20 at 10:08
thank you professor. It answers my question.
– Anas BOUALII
Nov 20 at 10:08
thank you professor. It answers my question.
– Anas BOUALII
Nov 20 at 10:08
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006134%2flinear-transformation-on-infinite-dimensional-vector-spaces%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
This is not true even in finite dimensional spaces...what is true is that surjectivity is equivalent to bijectivity for linear operators in $;mathcal L(V,V);$ , when $;dim V<infty;$ .
– DonAntonio
Nov 20 at 10:31