Mixing
Is this representation completely reducible?
$begingroup$
Is these representations completely reducible?
Definition:
A linear representation is said to be completely reducible if every invariant subspace has an invariant complement.
But I have no idea how to apply the definition, and I know that the answer for 2 is no while that for 3 is yes, could you please clarify for me how to check the definition in each case?
representation-theory invariant-theory invariant-subspace
asked Jan 24 at 15:07
IdonotknowIdonotknow
1138
$endgroup$
|
show 5 more comments
$begingroup$
Is these representations completely reducible?
Definition:
A linear representation is said to be completely reducible if every invariant subspace has an invariant complement.
But I have no idea how to apply the definition, and I know that the answer for 2 is no while that for 3 is yes, could you please clarify for me how to check the definition in each case?
representation-theory invariant-theory invariant-subspace
asked Jan 24 at 15:07
IdonotknowIdonotknow
1138
$endgroup$
$begingroup$
Have you first found the invariant subspaces?
$endgroup$
– Max
Jan 24 at 15:11
$begingroup$
math.stackexchange.com/questions/3083310/… @Max
$endgroup$
– Idonotknow
Jan 24 at 15:12
1
$begingroup$
Knowing the invariant subspaces, all you have to do is look at the list and check whether every element on that list has a complement that's also on the list (I'm a different Max by the way)
$endgroup$
– Max
Jan 24 at 17:08
$begingroup$
For number 3 the invariant subspaces are "Any subspace spanned by some set of eigenvectors of the operator $alpha$" ...... This is the answer written at the back of the book @Max
$endgroup$
– hopefully
Jan 26 at 14:12
1
$begingroup$
@hopefully : ok well then ? Take a space $F$ in the set of invariant subspaces, and look at all other invariant subspaces : is there one of them that is a complement of $F$ ?
$endgroup$
– Max
Jan 26 at 22:52
|
show 5 more comments
$begingroup$
Is these representations completely reducible?
Definition:
A linear representation is said to be completely reducible if every invariant subspace has an invariant complement.
But I have no idea how to apply the definition, and I know that the answer for 2 is no while that for 3 is yes, could you please clarify for me how to check the definition in each case?
representation-theory invariant-theory invariant-subspace
asked Jan 24 at 15:07
IdonotknowIdonotknow
1138
$endgroup$
Is these representations completely reducible?
Definition:
A linear representation is said to be completely reducible if every invariant subspace has an invariant complement.
But I have no idea how to apply the definition, and I know that the answer for 2 is no while that for 3 is yes, could you please clarify for me how to check the definition in each case?
representation-theory invariant-theory invariant-subspace
representation-theory invariant-theory invariant-subspace
asked Jan 24 at 15:07
IdonotknowIdonotknow
1138
asked Jan 24 at 15:07
IdonotknowIdonotknow
1138
asked Jan 24 at 15:07
IdonotknowIdonotknow
1138
asked Jan 24 at 15:07
IdonotknowIdonotknow
1138
asked Jan 24 at 15:07
IdonotknowIdonotknow
1138
1138
$begingroup$
Have you first found the invariant subspaces?
$endgroup$
– Max
Jan 24 at 15:11
$begingroup$
math.stackexchange.com/questions/3083310/… @Max
$endgroup$
– Idonotknow
Jan 24 at 15:12
1
$begingroup$
Knowing the invariant subspaces, all you have to do is look at the list and check whether every element on that list has a complement that's also on the list (I'm a different Max by the way)
$endgroup$
– Max
Jan 24 at 17:08
$begingroup$
For number 3 the invariant subspaces are "Any subspace spanned by some set of eigenvectors of the operator $alpha$" ...... This is the answer written at the back of the book @Max
$endgroup$
– hopefully
Jan 26 at 14:12
1
$begingroup$
@hopefully : ok well then ? Take a space $F$ in the set of invariant subspaces, and look at all other invariant subspaces : is there one of them that is a complement of $F$ ?
$endgroup$
– Max
Jan 26 at 22:52
|
show 5 more comments
$begingroup$
Have you first found the invariant subspaces?
$endgroup$
– Max
Jan 24 at 15:11
$begingroup$
math.stackexchange.com/questions/3083310/… @Max
$endgroup$
– Idonotknow
Jan 24 at 15:12
1
$begingroup$
Knowing the invariant subspaces, all you have to do is look at the list and check whether every element on that list has a complement that's also on the list (I'm a different Max by the way)
$endgroup$
– Max
Jan 24 at 17:08
$begingroup$
For number 3 the invariant subspaces are "Any subspace spanned by some set of eigenvectors of the operator $alpha$" ...... This is the answer written at the back of the book @Max
$endgroup$
– hopefully
Jan 26 at 14:12
1
$begingroup$
@hopefully : ok well then ? Take a space $F$ in the set of invariant subspaces, and look at all other invariant subspaces : is there one of them that is a complement of $F$ ?
$endgroup$
– Max
Jan 26 at 22:52
$begingroup$
Have you first found the invariant subspaces?
$endgroup$
– Max
Jan 24 at 15:11
$begingroup$
Have you first found the invariant subspaces?
$endgroup$
– Max
Jan 24 at 15:11
$begingroup$
math.stackexchange.com/questions/3083310/… @Max
$endgroup$
– Idonotknow
Jan 24 at 15:12
$begingroup$
math.stackexchange.com/questions/3083310/… @Max
$endgroup$
– Idonotknow
Jan 24 at 15:12
1
1
$begingroup$
Knowing the invariant subspaces, all you have to do is look at the list and check whether every element on that list has a complement that's also on the list (I'm a different Max by the way)
$endgroup$
– Max
Jan 24 at 17:08
$begingroup$
Knowing the invariant subspaces, all you have to do is look at the list and check whether every element on that list has a complement that's also on the list (I'm a different Max by the way)
$endgroup$
– Max
Jan 24 at 17:08
$begingroup$
For number 3 the invariant subspaces are "Any subspace spanned by some set of eigenvectors of the operator $alpha$" ...... This is the answer written at the back of the book @Max
$endgroup$
– hopefully
Jan 26 at 14:12
$begingroup$
For number 3 the invariant subspaces are "Any subspace spanned by some set of eigenvectors of the operator $alpha$" ...... This is the answer written at the back of the book @Max
$endgroup$
– hopefully
Jan 26 at 14:12
1
1
$begingroup$
@hopefully : ok well then ? Take a space $F$ in the set of invariant subspaces, and look at all other invariant subspaces : is there one of them that is a complement of $F$ ?
$endgroup$
– Max
Jan 26 at 22:52
$begingroup$
@hopefully : ok well then ? Take a space $F$ in the set of invariant subspaces, and look at all other invariant subspaces : is there one of them that is a complement of $F$ ?
$endgroup$
– Max
Jan 26 at 22:52
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Consider polynomials of degree less than or equal to 1
$$
V_1=mathbb Ccdot xoplusmathbb Ccdot 1,
$$
this is clearly invariant under $F(t)$ as $F(t)$ does not change degree. Similarly consider
$$
V_0=mathbb Ccdot 1,
$$
the space of constant functions. It is clear that $V_0subset V_1$ and that they are both invariant under $F$. However, if we consider the complement of $V_1$ in $V_0$, $W$, then there exists some $ainmathbb C$ such that $x+ain W$. If $W$ is invariant, then we have that $L(1-a)(x+a)=x+1in W$. $W$ is a vector space so $x+a-L(1-a)(x+a)=1in W$. This means $V_0subset W$, a contradiction. Hence the space is not completely reducible.$alphain$ End$(V)$, with $Vcongmathbb C^n$, so $alpha$ has Jordan Normal form, but its characteristic polynomial has no multiplicities, so it is diagonalisable. In the diagonal basis we have that $alpha=text{diag}(lambda_1,dots,lambda_n)$. So
$$
F(t)=text{diag}(e^{tlambda_1},dots,e^{tlambda_n}),
$$
and
$$
V=mathbb C_{lambda_1}oplusdotsoplusmathbb C_{lambda_n},
$$
where $mathbb C_{lambda_i}$ is the one dimensional representation $vmapsto e^{tlambda_i}v$. So $V$ is completely reducible.
answered Feb 1 at 11:48
Alec B-GAlec B-G
52019
$endgroup$
$begingroup$
what is the invariant complement in 3?
$endgroup$
– Idonotknow
Feb 1 at 16:34
$begingroup$
you mean that the complement of $V_{1}$ in $V_{0}$ is the constant term?
$endgroup$
– Idonotknow
Feb 1 at 16:36
$begingroup$
Is F in 2. L and t is 1-a and f is x +a I am confused about the notation .... could you clarify it please?
$endgroup$
– Idonotknow
Feb 1 at 16:39
1
$begingroup$
If $V$ is a subspace of $W$, both with an action by a group $G$, an invariant complement is a vector space $U$ such that $Gcdot U=U$ and $W=Voplus U$. A complement of $V_1$ in $V_0$ would be any one dimensional space spanned by $x-a$ for $ainmathbb C$. I showed that such a space cannot be invariant under $F(t)$ without also containing all of $V_1$.
$endgroup$
– Alec B-G
Feb 2 at 9:31
1
$begingroup$
$V_1$ is two dimensional and $V_0$ is one dimensional, so the complement, $W$, is a subspace of $V_1$ such that $V_0oplus W=V_1$, and so has to be one dimensional. Write $mathbf 1$ for the constant function 1, a vector in both $V_0$ and $V_1$, and $mathbf x$ for the function $x$, a vector in $V_1$. $W$ is a one dimensional vector space that sits inside $(mathbb Cmathbf 1oplus mathbb Cmathbf x)backslashmathbb Cmathbf 1$. This means that arbitrary elements of $W$ are of the form $lambdacdot(amathbf 1+mathbf x)$ for $a,lambdainmathbb C$.
$endgroup$
– Alec B-G
Feb 3 at 9:20
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3085966%2fis-this-representation-completely-reducible%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
$begingroup$
Consider polynomials of degree less than or equal to 1
$$
V_1=mathbb Ccdot xoplusmathbb Ccdot 1,
$$
this is clearly invariant under $F(t)$ as $F(t)$ does not change degree. Similarly consider
$$
V_0=mathbb Ccdot 1,
$$
the space of constant functions. It is clear that $V_0subset V_1$ and that they are both invariant under $F$. However, if we consider the complement of $V_1$ in $V_0$, $W$, then there exists some $ainmathbb C$ such that $x+ain W$. If $W$ is invariant, then we have that $L(1-a)(x+a)=x+1in W$. $W$ is a vector space so $x+a-L(1-a)(x+a)=1in W$. This means $V_0subset W$, a contradiction. Hence the space is not completely reducible.$alphain$ End$(V)$, with $Vcongmathbb C^n$, so $alpha$ has Jordan Normal form, but its characteristic polynomial has no multiplicities, so it is diagonalisable. In the diagonal basis we have that $alpha=text{diag}(lambda_1,dots,lambda_n)$. So
$$
F(t)=text{diag}(e^{tlambda_1},dots,e^{tlambda_n}),
$$
and
$$
V=mathbb C_{lambda_1}oplusdotsoplusmathbb C_{lambda_n},
$$
where $mathbb C_{lambda_i}$ is the one dimensional representation $vmapsto e^{tlambda_i}v$. So $V$ is completely reducible.
answered Feb 1 at 11:48
Alec B-GAlec B-G
52019
$endgroup$
$begingroup$
what is the invariant complement in 3?
$endgroup$
– Idonotknow
Feb 1 at 16:34
$begingroup$
you mean that the complement of $V_{1}$ in $V_{0}$ is the constant term?
$endgroup$
– Idonotknow
Feb 1 at 16:36
$begingroup$
Is F in 2. L and t is 1-a and f is x +a I am confused about the notation .... could you clarify it please?
$endgroup$
– Idonotknow
Feb 1 at 16:39
1
$begingroup$
If $V$ is a subspace of $W$, both with an action by a group $G$, an invariant complement is a vector space $U$ such that $Gcdot U=U$ and $W=Voplus U$. A complement of $V_1$ in $V_0$ would be any one dimensional space spanned by $x-a$ for $ainmathbb C$. I showed that such a space cannot be invariant under $F(t)$ without also containing all of $V_1$.
$endgroup$
– Alec B-G
Feb 2 at 9:31
1
$begingroup$
$V_1$ is two dimensional and $V_0$ is one dimensional, so the complement, $W$, is a subspace of $V_1$ such that $V_0oplus W=V_1$, and so has to be one dimensional. Write $mathbf 1$ for the constant function 1, a vector in both $V_0$ and $V_1$, and $mathbf x$ for the function $x$, a vector in $V_1$. $W$ is a one dimensional vector space that sits inside $(mathbb Cmathbf 1oplus mathbb Cmathbf x)backslashmathbb Cmathbf 1$. This means that arbitrary elements of $W$ are of the form $lambdacdot(amathbf 1+mathbf x)$ for $a,lambdainmathbb C$.
$endgroup$
– Alec B-G
Feb 3 at 9:20
|
show 1 more comment
$begingroup$
Consider polynomials of degree less than or equal to 1
$$
V_1=mathbb Ccdot xoplusmathbb Ccdot 1,
$$
this is clearly invariant under $F(t)$ as $F(t)$ does not change degree. Similarly consider
$$
V_0=mathbb Ccdot 1,
$$
the space of constant functions. It is clear that $V_0subset V_1$ and that they are both invariant under $F$. However, if we consider the complement of $V_1$ in $V_0$, $W$, then there exists some $ainmathbb C$ such that $x+ain W$. If $W$ is invariant, then we have that $L(1-a)(x+a)=x+1in W$. $W$ is a vector space so $x+a-L(1-a)(x+a)=1in W$. This means $V_0subset W$, a contradiction. Hence the space is not completely reducible.$alphain$ End$(V)$, with $Vcongmathbb C^n$, so $alpha$ has Jordan Normal form, but its characteristic polynomial has no multiplicities, so it is diagonalisable. In the diagonal basis we have that $alpha=text{diag}(lambda_1,dots,lambda_n)$. So
$$
F(t)=text{diag}(e^{tlambda_1},dots,e^{tlambda_n}),
$$
and
$$
V=mathbb C_{lambda_1}oplusdotsoplusmathbb C_{lambda_n},
$$
where $mathbb C_{lambda_i}$ is the one dimensional representation $vmapsto e^{tlambda_i}v$. So $V$ is completely reducible.
answered Feb 1 at 11:48
Alec B-GAlec B-G
52019
$endgroup$
$begingroup$
what is the invariant complement in 3?
$endgroup$
– Idonotknow
Feb 1 at 16:34
$begingroup$
you mean that the complement of $V_{1}$ in $V_{0}$ is the constant term?
$endgroup$
– Idonotknow
Feb 1 at 16:36
$begingroup$
Is F in 2. L and t is 1-a and f is x +a I am confused about the notation .... could you clarify it please?
$endgroup$
– Idonotknow
Feb 1 at 16:39
1
$begingroup$
If $V$ is a subspace of $W$, both with an action by a group $G$, an invariant complement is a vector space $U$ such that $Gcdot U=U$ and $W=Voplus U$. A complement of $V_1$ in $V_0$ would be any one dimensional space spanned by $x-a$ for $ainmathbb C$. I showed that such a space cannot be invariant under $F(t)$ without also containing all of $V_1$.
$endgroup$
– Alec B-G
Feb 2 at 9:31
1
$begingroup$
$V_1$ is two dimensional and $V_0$ is one dimensional, so the complement, $W$, is a subspace of $V_1$ such that $V_0oplus W=V_1$, and so has to be one dimensional. Write $mathbf 1$ for the constant function 1, a vector in both $V_0$ and $V_1$, and $mathbf x$ for the function $x$, a vector in $V_1$. $W$ is a one dimensional vector space that sits inside $(mathbb Cmathbf 1oplus mathbb Cmathbf x)backslashmathbb Cmathbf 1$. This means that arbitrary elements of $W$ are of the form $lambdacdot(amathbf 1+mathbf x)$ for $a,lambdainmathbb C$.
$endgroup$
– Alec B-G
Feb 3 at 9:20
|
show 1 more comment
$begingroup$
Consider polynomials of degree less than or equal to 1
$$
V_1=mathbb Ccdot xoplusmathbb Ccdot 1,
$$
this is clearly invariant under $F(t)$ as $F(t)$ does not change degree. Similarly consider
$$
V_0=mathbb Ccdot 1,
$$
the space of constant functions. It is clear that $V_0subset V_1$ and that they are both invariant under $F$. However, if we consider the complement of $V_1$ in $V_0$, $W$, then there exists some $ainmathbb C$ such that $x+ain W$. If $W$ is invariant, then we have that $L(1-a)(x+a)=x+1in W$. $W$ is a vector space so $x+a-L(1-a)(x+a)=1in W$. This means $V_0subset W$, a contradiction. Hence the space is not completely reducible.$alphain$ End$(V)$, with $Vcongmathbb C^n$, so $alpha$ has Jordan Normal form, but its characteristic polynomial has no multiplicities, so it is diagonalisable. In the diagonal basis we have that $alpha=text{diag}(lambda_1,dots,lambda_n)$. So
$$
F(t)=text{diag}(e^{tlambda_1},dots,e^{tlambda_n}),
$$
and
$$
V=mathbb C_{lambda_1}oplusdotsoplusmathbb C_{lambda_n},
$$
where $mathbb C_{lambda_i}$ is the one dimensional representation $vmapsto e^{tlambda_i}v$. So $V$ is completely reducible.
answered Feb 1 at 11:48
Alec B-GAlec B-G
52019
$endgroup$
Consider polynomials of degree less than or equal to 1
$$
V_1=mathbb Ccdot xoplusmathbb Ccdot 1,
$$
this is clearly invariant under $F(t)$ as $F(t)$ does not change degree. Similarly consider
$$
V_0=mathbb Ccdot 1,
$$
the space of constant functions. It is clear that $V_0subset V_1$ and that they are both invariant under $F$. However, if we consider the complement of $V_1$ in $V_0$, $W$, then there exists some $ainmathbb C$ such that $x+ain W$. If $W$ is invariant, then we have that $L(1-a)(x+a)=x+1in W$. $W$ is a vector space so $x+a-L(1-a)(x+a)=1in W$. This means $V_0subset W$, a contradiction. Hence the space is not completely reducible.$alphain$ End$(V)$, with $Vcongmathbb C^n$, so $alpha$ has Jordan Normal form, but its characteristic polynomial has no multiplicities, so it is diagonalisable. In the diagonal basis we have that $alpha=text{diag}(lambda_1,dots,lambda_n)$. So
$$
F(t)=text{diag}(e^{tlambda_1},dots,e^{tlambda_n}),
$$
and
$$
V=mathbb C_{lambda_1}oplusdotsoplusmathbb C_{lambda_n},
$$
where $mathbb C_{lambda_i}$ is the one dimensional representation $vmapsto e^{tlambda_i}v$. So $V$ is completely reducible.
answered Feb 1 at 11:48
Alec B-GAlec B-G
52019
edited Feb 2 at 22:03
answered Feb 1 at 11:48
Alec B-GAlec B-G
52019
answered Feb 1 at 11:48
Alec B-GAlec B-G
52019
answered Feb 1 at 11:48
Alec B-GAlec B-G
52019
52019
$begingroup$
what is the invariant complement in 3?
$endgroup$
– Idonotknow
Feb 1 at 16:34
$begingroup$
you mean that the complement of $V_{1}$ in $V_{0}$ is the constant term?
$endgroup$
– Idonotknow
Feb 1 at 16:36
$begingroup$
Is F in 2. L and t is 1-a and f is x +a I am confused about the notation .... could you clarify it please?
$endgroup$
– Idonotknow
Feb 1 at 16:39
1
$begingroup$
If $V$ is a subspace of $W$, both with an action by a group $G$, an invariant complement is a vector space $U$ such that $Gcdot U=U$ and $W=Voplus U$. A complement of $V_1$ in $V_0$ would be any one dimensional space spanned by $x-a$ for $ainmathbb C$. I showed that such a space cannot be invariant under $F(t)$ without also containing all of $V_1$.
$endgroup$
– Alec B-G
Feb 2 at 9:31
1
$begingroup$
$V_1$ is two dimensional and $V_0$ is one dimensional, so the complement, $W$, is a subspace of $V_1$ such that $V_0oplus W=V_1$, and so has to be one dimensional. Write $mathbf 1$ for the constant function 1, a vector in both $V_0$ and $V_1$, and $mathbf x$ for the function $x$, a vector in $V_1$. $W$ is a one dimensional vector space that sits inside $(mathbb Cmathbf 1oplus mathbb Cmathbf x)backslashmathbb Cmathbf 1$. This means that arbitrary elements of $W$ are of the form $lambdacdot(amathbf 1+mathbf x)$ for $a,lambdainmathbb C$.
$endgroup$
– Alec B-G
Feb 3 at 9:20
|
show 1 more comment
$begingroup$
what is the invariant complement in 3?
$endgroup$
– Idonotknow
Feb 1 at 16:34
$begingroup$
you mean that the complement of $V_{1}$ in $V_{0}$ is the constant term?
$endgroup$
– Idonotknow
Feb 1 at 16:36
$begingroup$
Is F in 2. L and t is 1-a and f is x +a I am confused about the notation .... could you clarify it please?
$endgroup$
– Idonotknow
Feb 1 at 16:39
1
$begingroup$
If $V$ is a subspace of $W$, both with an action by a group $G$, an invariant complement is a vector space $U$ such that $Gcdot U=U$ and $W=Voplus U$. A complement of $V_1$ in $V_0$ would be any one dimensional space spanned by $x-a$ for $ainmathbb C$. I showed that such a space cannot be invariant under $F(t)$ without also containing all of $V_1$.
$endgroup$
– Alec B-G
Feb 2 at 9:31
1
$begingroup$
$V_1$ is two dimensional and $V_0$ is one dimensional, so the complement, $W$, is a subspace of $V_1$ such that $V_0oplus W=V_1$, and so has to be one dimensional. Write $mathbf 1$ for the constant function 1, a vector in both $V_0$ and $V_1$, and $mathbf x$ for the function $x$, a vector in $V_1$. $W$ is a one dimensional vector space that sits inside $(mathbb Cmathbf 1oplus mathbb Cmathbf x)backslashmathbb Cmathbf 1$. This means that arbitrary elements of $W$ are of the form $lambdacdot(amathbf 1+mathbf x)$ for $a,lambdainmathbb C$.
$endgroup$
– Alec B-G
Feb 3 at 9:20
$begingroup$
what is the invariant complement in 3?
$endgroup$
– Idonotknow
Feb 1 at 16:34
$begingroup$
what is the invariant complement in 3?
$endgroup$
– Idonotknow
Feb 1 at 16:34
$begingroup$
you mean that the complement of $V_{1}$ in $V_{0}$ is the constant term?
$endgroup$
– Idonotknow
Feb 1 at 16:36
$begingroup$
you mean that the complement of $V_{1}$ in $V_{0}$ is the constant term?
$endgroup$
– Idonotknow
Feb 1 at 16:36
$begingroup$
Is F in 2. L and t is 1-a and f is x +a I am confused about the notation .... could you clarify it please?
$endgroup$
– Idonotknow
Feb 1 at 16:39
$begingroup$
Is F in 2. L and t is 1-a and f is x +a I am confused about the notation .... could you clarify it please?
$endgroup$
– Idonotknow
Feb 1 at 16:39
1
1
$begingroup$
If $V$ is a subspace of $W$, both with an action by a group $G$, an invariant complement is a vector space $U$ such that $Gcdot U=U$ and $W=Voplus U$. A complement of $V_1$ in $V_0$ would be any one dimensional space spanned by $x-a$ for $ainmathbb C$. I showed that such a space cannot be invariant under $F(t)$ without also containing all of $V_1$.
$endgroup$
– Alec B-G
Feb 2 at 9:31
$begingroup$
If $V$ is a subspace of $W$, both with an action by a group $G$, an invariant complement is a vector space $U$ such that $Gcdot U=U$ and $W=Voplus U$. A complement of $V_1$ in $V_0$ would be any one dimensional space spanned by $x-a$ for $ainmathbb C$. I showed that such a space cannot be invariant under $F(t)$ without also containing all of $V_1$.
$endgroup$
– Alec B-G
Feb 2 at 9:31
1
1
$begingroup$
$V_1$ is two dimensional and $V_0$ is one dimensional, so the complement, $W$, is a subspace of $V_1$ such that $V_0oplus W=V_1$, and so has to be one dimensional. Write $mathbf 1$ for the constant function 1, a vector in both $V_0$ and $V_1$, and $mathbf x$ for the function $x$, a vector in $V_1$. $W$ is a one dimensional vector space that sits inside $(mathbb Cmathbf 1oplus mathbb Cmathbf x)backslashmathbb Cmathbf 1$. This means that arbitrary elements of $W$ are of the form $lambdacdot(amathbf 1+mathbf x)$ for $a,lambdainmathbb C$.
$endgroup$
– Alec B-G
Feb 3 at 9:20
$begingroup$
$V_1$ is two dimensional and $V_0$ is one dimensional, so the complement, $W$, is a subspace of $V_1$ such that $V_0oplus W=V_1$, and so has to be one dimensional. Write $mathbf 1$ for the constant function 1, a vector in both $V_0$ and $V_1$, and $mathbf x$ for the function $x$, a vector in $V_1$. $W$ is a one dimensional vector space that sits inside $(mathbb Cmathbf 1oplus mathbb Cmathbf x)backslashmathbb Cmathbf 1$. This means that arbitrary elements of $W$ are of the form $lambdacdot(amathbf 1+mathbf x)$ for $a,lambdainmathbb C$.
$endgroup$
– Alec B-G
Feb 3 at 9:20
|
show 1 more 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%2f3085966%2fis-this-representation-completely-reducible%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
$begingroup$
Have you first found the invariant subspaces?
$endgroup$
– Max
Jan 24 at 15:11
$begingroup$
math.stackexchange.com/questions/3083310/… @Max
$endgroup$
– Idonotknow
Jan 24 at 15:12
1
$begingroup$
Knowing the invariant subspaces, all you have to do is look at the list and check whether every element on that list has a complement that's also on the list (I'm a different Max by the way)
$endgroup$
– Max
Jan 24 at 17:08
$begingroup$
For number 3 the invariant subspaces are "Any subspace spanned by some set of eigenvectors of the operator $alpha$" ...... This is the answer written at the back of the book @Max
$endgroup$
– hopefully
Jan 26 at 14:12
1
$begingroup$
@hopefully : ok well then ? Take a space $F$ in the set of invariant subspaces, and look at all other invariant subspaces : is there one of them that is a complement of $F$ ?
$endgroup$
– Max
Jan 26 at 22:52