Spectrum of a pair of commuting operators
Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
$(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
$$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$
Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
$$sigma_H(I,A)={(1,1);(1,-1)}.$$
functional-analysis operator-theory hilbert-spaces
asked Nov 28 '18 at 12:57
Student
2,5202524
add a comment |
Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
$(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
$$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$
Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
$$sigma_H(I,A)={(1,1);(1,-1)}.$$
functional-analysis operator-theory hilbert-spaces
asked Nov 28 '18 at 12:57
Student
2,5202524
add a comment |
Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
$(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
$$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$
Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
$$sigma_H(I,A)={(1,1);(1,-1)}.$$
functional-analysis operator-theory hilbert-spaces
asked Nov 28 '18 at 12:57
Student
2,5202524
Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
$(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
$$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$
Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
$$sigma_H(I,A)={(1,1);(1,-1)}.$$
functional-analysis operator-theory hilbert-spaces
functional-analysis operator-theory hilbert-spaces
asked Nov 28 '18 at 12:57
Student
2,5202524
asked Nov 28 '18 at 12:57
Student
2,5202524
edited Nov 28 '18 at 15:11
asked Nov 28 '18 at 12:57
Student
2,5202524
asked Nov 28 '18 at 12:57
Student
2,5202524
asked Nov 28 '18 at 12:57
Student
2,5202524
2,5202524
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 '18 at 16:59
s.harp
8,42312049
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 '18 at 11:12
add a 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%2f3017110%2fspectrum-of-a-pair-of-commuting-operators%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
First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 '18 at 16:59
s.harp
8,42312049
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 '18 at 11:12
add a comment |
First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 '18 at 16:59
s.harp
8,42312049
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 '18 at 11:12
add a comment |
First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 '18 at 16:59
s.harp
8,42312049
First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 '18 at 16:59
s.harp
8,42312049
answered Nov 28 '18 at 16:59
s.harp
8,42312049
answered Nov 28 '18 at 16:59
s.harp
8,42312049
answered Nov 28 '18 at 16:59
s.harp
8,42312049
8,42312049
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 '18 at 11:12
add a comment |
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 '18 at 11:12
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 '18 at 11:12
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 '18 at 11:12
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%2f3017110%2fspectrum-of-a-pair-of-commuting-operators%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
