Mixing
Linear Operators and Eigenvalue
$begingroup$
Let $V$ be an n-dimensional vector space and define the operators $T, space G:Vrightarrow V$ over an arbitrary field $mathbb{F}$, with char($mathbb{F})=0$. Suppose that $T^n=0$, $text{null}(T) = 1$, and $GT- TG = T.$
Prove that the eigenvalues of $G$ are of the form $alpha,$ $alpha -1, alpha-2, ..., alpha-n$, for some $alpha in mathbb{F}$.
My attempt:
By assumption, $text{null}(T)=1$, so there exists a nonzero vector $v in text{ker}(T)$. Clearly,
$0= T(v)=(GT- TG)(v) Rightarrow T(G(v))=0 Rightarrow G(v) in text{ker}(T)$ which implies that $text{ker}(T)$ is $G$-invariant.
Let $u$ be an eigenvector of $G$, then $G(u) = lambda u$ for some $lambda in mathbb{F}$ and we have that $(GT - TG)(u) = T(u) Rightarrow G(T(u)) = (lambda+1)T(u)$.
Therefore, if $u$ is eigenvector of $G$, with an eigenvalue $lambda$, then $T(u)$ is an eigenvector with eigenvalue $(lambda+1)$.
But I don't know what I will do with this..
linear-algebra
edited Feb 2 at 19:12
Victoria M
42618
asked Feb 2 at 18:57
user638057user638057
998
$endgroup$
add a comment |
$begingroup$
Let $V$ be an n-dimensional vector space and define the operators $T, space G:Vrightarrow V$ over an arbitrary field $mathbb{F}$, with char($mathbb{F})=0$. Suppose that $T^n=0$, $text{null}(T) = 1$, and $GT- TG = T.$
Prove that the eigenvalues of $G$ are of the form $alpha,$ $alpha -1, alpha-2, ..., alpha-n$, for some $alpha in mathbb{F}$.
My attempt:
By assumption, $text{null}(T)=1$, so there exists a nonzero vector $v in text{ker}(T)$. Clearly,
$0= T(v)=(GT- TG)(v) Rightarrow T(G(v))=0 Rightarrow G(v) in text{ker}(T)$ which implies that $text{ker}(T)$ is $G$-invariant.
Let $u$ be an eigenvector of $G$, then $G(u) = lambda u$ for some $lambda in mathbb{F}$ and we have that $(GT - TG)(u) = T(u) Rightarrow G(T(u)) = (lambda+1)T(u)$.
Therefore, if $u$ is eigenvector of $G$, with an eigenvalue $lambda$, then $T(u)$ is an eigenvector with eigenvalue $(lambda+1)$.
But I don't know what I will do with this..
linear-algebra
edited Feb 2 at 19:12
Victoria M
42618
asked Feb 2 at 18:57
user638057user638057
998
$endgroup$
1
$begingroup$
Almost an engaging problem--I say almost because something ain't quite right, to wit: $dim_F V = n$, but the alleged eigenvalues of $G$, $alpha$, $alpha - 1$, $alpha - 2$, $ldots$, $alpha - n$ are $n + 1$ in number; but an operator on an $n$ dimensional vector space has at most $n$ eigenvalues. Somethings gotta give, but what? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 19:31
add a comment |
$begingroup$
Let $V$ be an n-dimensional vector space and define the operators $T, space G:Vrightarrow V$ over an arbitrary field $mathbb{F}$, with char($mathbb{F})=0$. Suppose that $T^n=0$, $text{null}(T) = 1$, and $GT- TG = T.$
Prove that the eigenvalues of $G$ are of the form $alpha,$ $alpha -1, alpha-2, ..., alpha-n$, for some $alpha in mathbb{F}$.
My attempt:
By assumption, $text{null}(T)=1$, so there exists a nonzero vector $v in text{ker}(T)$. Clearly,
$0= T(v)=(GT- TG)(v) Rightarrow T(G(v))=0 Rightarrow G(v) in text{ker}(T)$ which implies that $text{ker}(T)$ is $G$-invariant.
Let $u$ be an eigenvector of $G$, then $G(u) = lambda u$ for some $lambda in mathbb{F}$ and we have that $(GT - TG)(u) = T(u) Rightarrow G(T(u)) = (lambda+1)T(u)$.
Therefore, if $u$ is eigenvector of $G$, with an eigenvalue $lambda$, then $T(u)$ is an eigenvector with eigenvalue $(lambda+1)$.
But I don't know what I will do with this..
linear-algebra
edited Feb 2 at 19:12
Victoria M
42618
asked Feb 2 at 18:57
user638057user638057
998
$endgroup$
Let $V$ be an n-dimensional vector space and define the operators $T, space G:Vrightarrow V$ over an arbitrary field $mathbb{F}$, with char($mathbb{F})=0$. Suppose that $T^n=0$, $text{null}(T) = 1$, and $GT- TG = T.$
Prove that the eigenvalues of $G$ are of the form $alpha,$ $alpha -1, alpha-2, ..., alpha-n$, for some $alpha in mathbb{F}$.
My attempt:
By assumption, $text{null}(T)=1$, so there exists a nonzero vector $v in text{ker}(T)$. Clearly,
$0= T(v)=(GT- TG)(v) Rightarrow T(G(v))=0 Rightarrow G(v) in text{ker}(T)$ which implies that $text{ker}(T)$ is $G$-invariant.
Let $u$ be an eigenvector of $G$, then $G(u) = lambda u$ for some $lambda in mathbb{F}$ and we have that $(GT - TG)(u) = T(u) Rightarrow G(T(u)) = (lambda+1)T(u)$.
Therefore, if $u$ is eigenvector of $G$, with an eigenvalue $lambda$, then $T(u)$ is an eigenvector with eigenvalue $(lambda+1)$.
But I don't know what I will do with this..
linear-algebra
linear-algebra
edited Feb 2 at 19:12
Victoria M
42618
asked Feb 2 at 18:57
user638057user638057
998
edited Feb 2 at 19:12
Victoria M
42618
asked Feb 2 at 18:57
user638057user638057
998
edited Feb 2 at 19:12
Victoria M
42618
edited Feb 2 at 19:12
Victoria M
42618
edited Feb 2 at 19:12
Victoria M
42618
42618
asked Feb 2 at 18:57
user638057user638057
998
asked Feb 2 at 18:57
user638057user638057
998
asked Feb 2 at 18:57
user638057user638057
998
998
1
$begingroup$
Almost an engaging problem--I say almost because something ain't quite right, to wit: $dim_F V = n$, but the alleged eigenvalues of $G$, $alpha$, $alpha - 1$, $alpha - 2$, $ldots$, $alpha - n$ are $n + 1$ in number; but an operator on an $n$ dimensional vector space has at most $n$ eigenvalues. Somethings gotta give, but what? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 19:31
add a comment |
1
$begingroup$
Almost an engaging problem--I say almost because something ain't quite right, to wit: $dim_F V = n$, but the alleged eigenvalues of $G$, $alpha$, $alpha - 1$, $alpha - 2$, $ldots$, $alpha - n$ are $n + 1$ in number; but an operator on an $n$ dimensional vector space has at most $n$ eigenvalues. Somethings gotta give, but what? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 19:31
1
1
$begingroup$
Almost an engaging problem--I say almost because something ain't quite right, to wit: $dim_F V = n$, but the alleged eigenvalues of $G$, $alpha$, $alpha - 1$, $alpha - 2$, $ldots$, $alpha - n$ are $n + 1$ in number; but an operator on an $n$ dimensional vector space has at most $n$ eigenvalues. Somethings gotta give, but what? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 19:31
$begingroup$
Almost an engaging problem--I say almost because something ain't quite right, to wit: $dim_F V = n$, but the alleged eigenvalues of $G$, $alpha$, $alpha - 1$, $alpha - 2$, $ldots$, $alpha - n$ are $n + 1$ in number; but an operator on an $n$ dimensional vector space has at most $n$ eigenvalues. Somethings gotta give, but what? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 19:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: You're nearly there. Slight correction: suppose that $u$ is an eigenvector of $G$ with eigenvalue $lambda$. Then, as you calculated, we find that
$$
G(T(u)) = (lambda+1)T(u)
$$
That is, we have two possibilities: either $T(u)$ is an eigenvector of $G$ associated with $lambda+1$ or $T(u) = 0$, which is to say that $T(u) in ker G$. Because $operatorname{null}(T) = 1$, we can deduce that $T(u) = 0$ will happen for exactly one eigenvector of $G$.
answered Feb 2 at 21:50
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
Note: this result is important in the study of semisimple Lie algebras
$endgroup$
– Omnomnomnom
Feb 2 at 21:51
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3097651%2flinear-operators-and-eigenvalue%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: You're nearly there. Slight correction: suppose that $u$ is an eigenvector of $G$ with eigenvalue $lambda$. Then, as you calculated, we find that
$$
G(T(u)) = (lambda+1)T(u)
$$
That is, we have two possibilities: either $T(u)$ is an eigenvector of $G$ associated with $lambda+1$ or $T(u) = 0$, which is to say that $T(u) in ker G$. Because $operatorname{null}(T) = 1$, we can deduce that $T(u) = 0$ will happen for exactly one eigenvector of $G$.
answered Feb 2 at 21:50
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
Note: this result is important in the study of semisimple Lie algebras
$endgroup$
– Omnomnomnom
Feb 2 at 21:51
add a comment |
$begingroup$
Hint: You're nearly there. Slight correction: suppose that $u$ is an eigenvector of $G$ with eigenvalue $lambda$. Then, as you calculated, we find that
$$
G(T(u)) = (lambda+1)T(u)
$$
That is, we have two possibilities: either $T(u)$ is an eigenvector of $G$ associated with $lambda+1$ or $T(u) = 0$, which is to say that $T(u) in ker G$. Because $operatorname{null}(T) = 1$, we can deduce that $T(u) = 0$ will happen for exactly one eigenvector of $G$.
answered Feb 2 at 21:50
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
Note: this result is important in the study of semisimple Lie algebras
$endgroup$
– Omnomnomnom
Feb 2 at 21:51
add a comment |
$begingroup$
Hint: You're nearly there. Slight correction: suppose that $u$ is an eigenvector of $G$ with eigenvalue $lambda$. Then, as you calculated, we find that
$$
G(T(u)) = (lambda+1)T(u)
$$
That is, we have two possibilities: either $T(u)$ is an eigenvector of $G$ associated with $lambda+1$ or $T(u) = 0$, which is to say that $T(u) in ker G$. Because $operatorname{null}(T) = 1$, we can deduce that $T(u) = 0$ will happen for exactly one eigenvector of $G$.
answered Feb 2 at 21:50
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
Hint: You're nearly there. Slight correction: suppose that $u$ is an eigenvector of $G$ with eigenvalue $lambda$. Then, as you calculated, we find that
$$
G(T(u)) = (lambda+1)T(u)
$$
That is, we have two possibilities: either $T(u)$ is an eigenvector of $G$ associated with $lambda+1$ or $T(u) = 0$, which is to say that $T(u) in ker G$. Because $operatorname{null}(T) = 1$, we can deduce that $T(u) = 0$ will happen for exactly one eigenvector of $G$.
answered Feb 2 at 21:50
OmnomnomnomOmnomnomnom
129k794188
answered Feb 2 at 21:50
OmnomnomnomOmnomnomnom
129k794188
answered Feb 2 at 21:50
OmnomnomnomOmnomnomnom
129k794188
answered Feb 2 at 21:50
OmnomnomnomOmnomnomnom
129k794188
129k794188
$begingroup$
Note: this result is important in the study of semisimple Lie algebras
$endgroup$
– Omnomnomnom
Feb 2 at 21:51
add a comment |
$begingroup$
Note: this result is important in the study of semisimple Lie algebras
$endgroup$
– Omnomnomnom
Feb 2 at 21:51
$begingroup$
Note: this result is important in the study of semisimple Lie algebras
$endgroup$
– Omnomnomnom
Feb 2 at 21:51
$begingroup$
Note: this result is important in the study of semisimple Lie algebras
$endgroup$
– Omnomnomnom
Feb 2 at 21:51
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.
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%2f3097651%2flinear-operators-and-eigenvalue%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
1
$begingroup$
Almost an engaging problem--I say almost because something ain't quite right, to wit: $dim_F V = n$, but the alleged eigenvalues of $G$, $alpha$, $alpha - 1$, $alpha - 2$, $ldots$, $alpha - n$ are $n + 1$ in number; but an operator on an $n$ dimensional vector space has at most $n$ eigenvalues. Somethings gotta give, but what? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 19:31