Show that endomorphism is automorphism
up vote
0
down vote
favorite
Let dim V = n, $ain End(V)$, and for some $vin V$ set of $a(v)$, $a^2(v)$, $ldots$, $a^n(v)$ is linearly independent. (where $a^i(v) = acirc acirc ldotscirc a$ i times, $circ$ is a composition. Show that a has invertible. I need to show that a is injection, if I will prove it I can prove that this is surjetive, then I will show that this is bijection and this will mean that we have invertible $a^{-1}$. but the problem is that I dont know how to prove that this is injection. Can you give me a hint?
linear-algebra
add a comment |
up vote
0
down vote
favorite
Let dim V = n, $ain End(V)$, and for some $vin V$ set of $a(v)$, $a^2(v)$, $ldots$, $a^n(v)$ is linearly independent. (where $a^i(v) = acirc acirc ldotscirc a$ i times, $circ$ is a composition. Show that a has invertible. I need to show that a is injection, if I will prove it I can prove that this is surjetive, then I will show that this is bijection and this will mean that we have invertible $a^{-1}$. but the problem is that I dont know how to prove that this is injection. Can you give me a hint?
linear-algebra
It's probably easier to show directly that $a$ is surjective (from which it will follow that $a$ is also injective). For that, note that ${ a(v), a^2(v), ldots, a^n(v) }$ must be a basis of $V$...
– Daniel Schepler
Nov 20 at 22:31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let dim V = n, $ain End(V)$, and for some $vin V$ set of $a(v)$, $a^2(v)$, $ldots$, $a^n(v)$ is linearly independent. (where $a^i(v) = acirc acirc ldotscirc a$ i times, $circ$ is a composition. Show that a has invertible. I need to show that a is injection, if I will prove it I can prove that this is surjetive, then I will show that this is bijection and this will mean that we have invertible $a^{-1}$. but the problem is that I dont know how to prove that this is injection. Can you give me a hint?
linear-algebra
Let dim V = n, $ain End(V)$, and for some $vin V$ set of $a(v)$, $a^2(v)$, $ldots$, $a^n(v)$ is linearly independent. (where $a^i(v) = acirc acirc ldotscirc a$ i times, $circ$ is a composition. Show that a has invertible. I need to show that a is injection, if I will prove it I can prove that this is surjetive, then I will show that this is bijection and this will mean that we have invertible $a^{-1}$. but the problem is that I dont know how to prove that this is injection. Can you give me a hint?
linear-algebra
linear-algebra
asked Nov 20 at 22:23
user596269
It's probably easier to show directly that $a$ is surjective (from which it will follow that $a$ is also injective). For that, note that ${ a(v), a^2(v), ldots, a^n(v) }$ must be a basis of $V$...
– Daniel Schepler
Nov 20 at 22:31
add a comment |
It's probably easier to show directly that $a$ is surjective (from which it will follow that $a$ is also injective). For that, note that ${ a(v), a^2(v), ldots, a^n(v) }$ must be a basis of $V$...
– Daniel Schepler
Nov 20 at 22:31
It's probably easier to show directly that $a$ is surjective (from which it will follow that $a$ is also injective). For that, note that ${ a(v), a^2(v), ldots, a^n(v) }$ must be a basis of $V$...
– Daniel Schepler
Nov 20 at 22:31
It's probably easier to show directly that $a$ is surjective (from which it will follow that $a$ is also injective). For that, note that ${ a(v), a^2(v), ldots, a^n(v) }$ must be a basis of $V$...
– Daniel Schepler
Nov 20 at 22:31
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
The vectors $a(v),ldots,a^n(v)$ lies in the image of $a$ and they form a basis of $V$. So $a$ is suryective and, because the dimension is finite, it is an automorphism.
answered Nov 20 at 22:32
Dante Grevino
7787
Thank you! But I did not understood why this is basis leads to surjectivity?
– user596269
Nov 20 at 22:34
The image of a vector space under a linear function is always a linear subspace. Do you see why? So if you have a generating set in the image this is equivalent to surjectivity.
– Dante Grevino
Nov 21 at 0:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The vectors $a(v),ldots,a^n(v)$ lies in the image of $a$ and they form a basis of $V$. So $a$ is suryective and, because the dimension is finite, it is an automorphism.
answered Nov 20 at 22:32
Dante Grevino
7787
Thank you! But I did not understood why this is basis leads to surjectivity?
– user596269
Nov 20 at 22:34
The image of a vector space under a linear function is always a linear subspace. Do you see why? So if you have a generating set in the image this is equivalent to surjectivity.
– Dante Grevino
Nov 21 at 0:48
add a comment |
up vote
0
down vote
The vectors $a(v),ldots,a^n(v)$ lies in the image of $a$ and they form a basis of $V$. So $a$ is suryective and, because the dimension is finite, it is an automorphism.
answered Nov 20 at 22:32
Dante Grevino
7787
Thank you! But I did not understood why this is basis leads to surjectivity?
– user596269
Nov 20 at 22:34
The image of a vector space under a linear function is always a linear subspace. Do you see why? So if you have a generating set in the image this is equivalent to surjectivity.
– Dante Grevino
Nov 21 at 0:48
add a comment |
up vote
0
down vote
up vote
0
down vote
The vectors $a(v),ldots,a^n(v)$ lies in the image of $a$ and they form a basis of $V$. So $a$ is suryective and, because the dimension is finite, it is an automorphism.
answered Nov 20 at 22:32
Dante Grevino
7787
The vectors $a(v),ldots,a^n(v)$ lies in the image of $a$ and they form a basis of $V$. So $a$ is suryective and, because the dimension is finite, it is an automorphism.
answered Nov 20 at 22:32
Dante Grevino
7787
answered Nov 20 at 22:32
Dante Grevino
7787
answered Nov 20 at 22:32
Dante Grevino
7787
answered Nov 20 at 22:32
Dante Grevino
7787
7787
Thank you! But I did not understood why this is basis leads to surjectivity?
– user596269
Nov 20 at 22:34
The image of a vector space under a linear function is always a linear subspace. Do you see why? So if you have a generating set in the image this is equivalent to surjectivity.
– Dante Grevino
Nov 21 at 0:48
add a comment |
Thank you! But I did not understood why this is basis leads to surjectivity?
– user596269
Nov 20 at 22:34
The image of a vector space under a linear function is always a linear subspace. Do you see why? So if you have a generating set in the image this is equivalent to surjectivity.
– Dante Grevino
Nov 21 at 0:48
Thank you! But I did not understood why this is basis leads to surjectivity?
– user596269
Nov 20 at 22:34
Thank you! But I did not understood why this is basis leads to surjectivity?
– user596269
Nov 20 at 22:34
The image of a vector space under a linear function is always a linear subspace. Do you see why? So if you have a generating set in the image this is equivalent to surjectivity.
– Dante Grevino
Nov 21 at 0:48
The image of a vector space under a linear function is always a linear subspace. Do you see why? So if you have a generating set in the image this is equivalent to surjectivity.
– Dante Grevino
Nov 21 at 0:48
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%2f3006983%2fshow-that-endomorphism-is-automorphism%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
It's probably easier to show directly that $a$ is surjective (from which it will follow that $a$ is also injective). For that, note that ${ a(v), a^2(v), ldots, a^n(v) }$ must be a basis of $V$...
– Daniel Schepler
Nov 20 at 22:31