Determining eigenvalues of a linear transformation on a field extension via embedding into $BbbC$
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I have been working on the following question:
Suppose that $K$ is an extension of $BbbQ$ of degree $n$. Let $sigma_1,dots,sigma_n:KhookrightarrowBbbC$ be the distinct embeddings of $K$ into $BbbC$. Let $alphain K$. Regarding $K$ as a vector space over $BbbQ$, let $phi:Kto K$ be the linear transformation $phi(x) = alpha x$. Show that the eigenvalues of $phi$ are $sigma_1(alpha),dots,sigma_n(alpha)$.
I don't quite understand what exactly the question is asking. I know that $phi$ is an endomorphism of $K$, but $sigma_i$ maps $K$ into $BbbC$. How can eigenvalues of $phi$ be in $BbbC$?
Perhaps the question is considering maps $phi_i:sigma_i(K)tosigma_i(K)$ by $xmapsto sigma_i(alpha)x$, in which case the answer seems fairly straightforward? Of course $sigma_i(alpha)$ is the eigenvalue for this map, just by definition.
The written answer I received, however, is far lengthier than this, and unfortunately is not very clear. Thus the above does not seem likely.
Any clarification or hints would be appreciated.
vector-spaces field-theory linear-transformations
asked Aug 10 at 18:38
Santana Afton
1,9031425
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up vote
3
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I have been working on the following question:
Suppose that $K$ is an extension of $BbbQ$ of degree $n$. Let $sigma_1,dots,sigma_n:KhookrightarrowBbbC$ be the distinct embeddings of $K$ into $BbbC$. Let $alphain K$. Regarding $K$ as a vector space over $BbbQ$, let $phi:Kto K$ be the linear transformation $phi(x) = alpha x$. Show that the eigenvalues of $phi$ are $sigma_1(alpha),dots,sigma_n(alpha)$.
I don't quite understand what exactly the question is asking. I know that $phi$ is an endomorphism of $K$, but $sigma_i$ maps $K$ into $BbbC$. How can eigenvalues of $phi$ be in $BbbC$?
Perhaps the question is considering maps $phi_i:sigma_i(K)tosigma_i(K)$ by $xmapsto sigma_i(alpha)x$, in which case the answer seems fairly straightforward? Of course $sigma_i(alpha)$ is the eigenvalue for this map, just by definition.
The written answer I received, however, is far lengthier than this, and unfortunately is not very clear. Thus the above does not seem likely.
Any clarification or hints would be appreciated.
vector-spaces field-theory linear-transformations
asked Aug 10 at 18:38
Santana Afton
1,9031425
Recall that the eigenvalues of an operator are the roots of the characteristic polynomial. In your case you have an operator defined over $K$, whose characteristic polynomial will lie in $K[X]$. It is possible that its roots will lie in $mathbbC$ but not in $K$, if that is what you are asking. As an example, I suggest you compute the eigenvalues for multiplication by $i$ in the field $K=mathbbQ(i)$.
– Gal Porat
Aug 10 at 20:43
@GalPorat Are the eigenvalues of that map not $pm iin K$?
– Santana Afton
Aug 10 at 23:43
Sure, so now you may see that the claim you want to prove is true and makes sense in that particular case.
– Gal Porat
Aug 11 at 6:35
Sorry, I must be misunderstanding. I thought the example was a linear map of a finite field extension where the eigenvalues do not lie in that field. In the $K = BbbQ(i)$ case, both $pm i$ do live in $K$.
– Santana Afton
Aug 11 at 12:05
I am sorry for the confusion. For an example where the eigenvalues do lie in the field, you should choose a field $K$ that is not Galois over $mathbbQ$, such as $mathbbQ(sqrt[3]2)$. My first example was for you to see how the proposition holds in a specific case.
– Gal Porat
Aug 11 at 12:25
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have been working on the following question:
Suppose that $K$ is an extension of $BbbQ$ of degree $n$. Let $sigma_1,dots,sigma_n:KhookrightarrowBbbC$ be the distinct embeddings of $K$ into $BbbC$. Let $alphain K$. Regarding $K$ as a vector space over $BbbQ$, let $phi:Kto K$ be the linear transformation $phi(x) = alpha x$. Show that the eigenvalues of $phi$ are $sigma_1(alpha),dots,sigma_n(alpha)$.
I don't quite understand what exactly the question is asking. I know that $phi$ is an endomorphism of $K$, but $sigma_i$ maps $K$ into $BbbC$. How can eigenvalues of $phi$ be in $BbbC$?
Perhaps the question is considering maps $phi_i:sigma_i(K)tosigma_i(K)$ by $xmapsto sigma_i(alpha)x$, in which case the answer seems fairly straightforward? Of course $sigma_i(alpha)$ is the eigenvalue for this map, just by definition.
The written answer I received, however, is far lengthier than this, and unfortunately is not very clear. Thus the above does not seem likely.
Any clarification or hints would be appreciated.
vector-spaces field-theory linear-transformations
asked Aug 10 at 18:38
Santana Afton
1,9031425
I have been working on the following question:
Suppose that $K$ is an extension of $BbbQ$ of degree $n$. Let $sigma_1,dots,sigma_n:KhookrightarrowBbbC$ be the distinct embeddings of $K$ into $BbbC$. Let $alphain K$. Regarding $K$ as a vector space over $BbbQ$, let $phi:Kto K$ be the linear transformation $phi(x) = alpha x$. Show that the eigenvalues of $phi$ are $sigma_1(alpha),dots,sigma_n(alpha)$.
I don't quite understand what exactly the question is asking. I know that $phi$ is an endomorphism of $K$, but $sigma_i$ maps $K$ into $BbbC$. How can eigenvalues of $phi$ be in $BbbC$?
Perhaps the question is considering maps $phi_i:sigma_i(K)tosigma_i(K)$ by $xmapsto sigma_i(alpha)x$, in which case the answer seems fairly straightforward? Of course $sigma_i(alpha)$ is the eigenvalue for this map, just by definition.
The written answer I received, however, is far lengthier than this, and unfortunately is not very clear. Thus the above does not seem likely.
Any clarification or hints would be appreciated.
vector-spaces field-theory linear-transformations
asked Aug 10 at 18:38
Santana Afton
1,9031425
edited Aug 10 at 19:24
asked Aug 10 at 18:38
Santana Afton
1,9031425
asked Aug 10 at 18:38
Santana Afton
1,9031425
asked Aug 10 at 18:38
Santana Afton
1,9031425
1,9031425
Recall that the eigenvalues of an operator are the roots of the characteristic polynomial. In your case you have an operator defined over $K$, whose characteristic polynomial will lie in $K[X]$. It is possible that its roots will lie in $mathbbC$ but not in $K$, if that is what you are asking. As an example, I suggest you compute the eigenvalues for multiplication by $i$ in the field $K=mathbbQ(i)$.
– Gal Porat
Aug 10 at 20:43
@GalPorat Are the eigenvalues of that map not $pm iin K$?
– Santana Afton
Aug 10 at 23:43
Sure, so now you may see that the claim you want to prove is true and makes sense in that particular case.
– Gal Porat
Aug 11 at 6:35
Sorry, I must be misunderstanding. I thought the example was a linear map of a finite field extension where the eigenvalues do not lie in that field. In the $K = BbbQ(i)$ case, both $pm i$ do live in $K$.
– Santana Afton
Aug 11 at 12:05
I am sorry for the confusion. For an example where the eigenvalues do lie in the field, you should choose a field $K$ that is not Galois over $mathbbQ$, such as $mathbbQ(sqrt[3]2)$. My first example was for you to see how the proposition holds in a specific case.
– Gal Porat
Aug 11 at 12:25
add a comment |Â
Recall that the eigenvalues of an operator are the roots of the characteristic polynomial. In your case you have an operator defined over $K$, whose characteristic polynomial will lie in $K[X]$. It is possible that its roots will lie in $mathbbC$ but not in $K$, if that is what you are asking. As an example, I suggest you compute the eigenvalues for multiplication by $i$ in the field $K=mathbbQ(i)$.
– Gal Porat
Aug 10 at 20:43
@GalPorat Are the eigenvalues of that map not $pm iin K$?
– Santana Afton
Aug 10 at 23:43
Sure, so now you may see that the claim you want to prove is true and makes sense in that particular case.
– Gal Porat
Aug 11 at 6:35
Sorry, I must be misunderstanding. I thought the example was a linear map of a finite field extension where the eigenvalues do not lie in that field. In the $K = BbbQ(i)$ case, both $pm i$ do live in $K$.
– Santana Afton
Aug 11 at 12:05
I am sorry for the confusion. For an example where the eigenvalues do lie in the field, you should choose a field $K$ that is not Galois over $mathbbQ$, such as $mathbbQ(sqrt[3]2)$. My first example was for you to see how the proposition holds in a specific case.
– Gal Porat
Aug 11 at 12:25
Recall that the eigenvalues of an operator are the roots of the characteristic polynomial. In your case you have an operator defined over $K$, whose characteristic polynomial will lie in $K[X]$. It is possible that its roots will lie in $mathbbC$ but not in $K$, if that is what you are asking. As an example, I suggest you compute the eigenvalues for multiplication by $i$ in the field $K=mathbbQ(i)$.
– Gal Porat
Aug 10 at 20:43
Recall that the eigenvalues of an operator are the roots of the characteristic polynomial. In your case you have an operator defined over $K$, whose characteristic polynomial will lie in $K[X]$. It is possible that its roots will lie in $mathbbC$ but not in $K$, if that is what you are asking. As an example, I suggest you compute the eigenvalues for multiplication by $i$ in the field $K=mathbbQ(i)$.
– Gal Porat
Aug 10 at 20:43
@GalPorat Are the eigenvalues of that map not $pm iin K$?
– Santana Afton
Aug 10 at 23:43
@GalPorat Are the eigenvalues of that map not $pm iin K$?
– Santana Afton
Aug 10 at 23:43
Sure, so now you may see that the claim you want to prove is true and makes sense in that particular case.
– Gal Porat
Aug 11 at 6:35
Sure, so now you may see that the claim you want to prove is true and makes sense in that particular case.
– Gal Porat
Aug 11 at 6:35
Sorry, I must be misunderstanding. I thought the example was a linear map of a finite field extension where the eigenvalues do not lie in that field. In the $K = BbbQ(i)$ case, both $pm i$ do live in $K$.
– Santana Afton
Aug 11 at 12:05
Sorry, I must be misunderstanding. I thought the example was a linear map of a finite field extension where the eigenvalues do not lie in that field. In the $K = BbbQ(i)$ case, both $pm i$ do live in $K$.
– Santana Afton
Aug 11 at 12:05
I am sorry for the confusion. For an example where the eigenvalues do lie in the field, you should choose a field $K$ that is not Galois over $mathbbQ$, such as $mathbbQ(sqrt[3]2)$. My first example was for you to see how the proposition holds in a specific case.
– Gal Porat
Aug 11 at 12:25
I am sorry for the confusion. For an example where the eigenvalues do lie in the field, you should choose a field $K$ that is not Galois over $mathbbQ$, such as $mathbbQ(sqrt[3]2)$. My first example was for you to see how the proposition holds in a specific case.
– Gal Porat
Aug 11 at 12:25
add a comment |Â
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Recall that the eigenvalues of an operator are the roots of the characteristic polynomial. In your case you have an operator defined over $K$, whose characteristic polynomial will lie in $K[X]$. It is possible that its roots will lie in $mathbbC$ but not in $K$, if that is what you are asking. As an example, I suggest you compute the eigenvalues for multiplication by $i$ in the field $K=mathbbQ(i)$.
– Gal Porat
Aug 10 at 20:43
@GalPorat Are the eigenvalues of that map not $pm iin K$?
– Santana Afton
Aug 10 at 23:43
Sure, so now you may see that the claim you want to prove is true and makes sense in that particular case.
– Gal Porat
Aug 11 at 6:35
Sorry, I must be misunderstanding. I thought the example was a linear map of a finite field extension where the eigenvalues do not lie in that field. In the $K = BbbQ(i)$ case, both $pm i$ do live in $K$.
– Santana Afton
Aug 11 at 12:05
I am sorry for the confusion. For an example where the eigenvalues do lie in the field, you should choose a field $K$ that is not Galois over $mathbbQ$, such as $mathbbQ(sqrt[3]2)$. My first example was for you to see how the proposition holds in a specific case.
– Gal Porat
Aug 11 at 12:25