$P,Q$ are Hermitian operators with only non-negative eigenvalues, then tr($PQ$) = 0 $implies$ that $PQ = 0$.
$begingroup$
So far I can prove:
suppose there is a basis set ${|i>}$ that diagonalize $P$ with eigenvalue $p_i$, then tr($PQ$) = $sum_i<i|PQ|i>$ = $sum_ip_i<i|Q|i>$.
Since $P, Q$ are Hermitian with non-negative eigenvalues, both $p_i$ and $<i|Q|i>$ is non-negative...
Now there are at least two possibilities to make tr() = $0$.
i) all $p_i$ = $0$, then of course $PQ = 0$
ii) all $<i|Q|i>$ = $0$, then I can't prove $PQ = 0$...
iii) $p_i$, $<i|Q|i>$ alternates to be $0$, $PQ$ = ???
I think they should be a very easy problem... But I am just stuck somehow..
linear-algebra operator-theory
edited Jan 19 at 6:36
Andrews
1,3062423
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
$endgroup$
add a comment |
$begingroup$
So far I can prove:
suppose there is a basis set ${|i>}$ that diagonalize $P$ with eigenvalue $p_i$, then tr($PQ$) = $sum_i<i|PQ|i>$ = $sum_ip_i<i|Q|i>$.
Since $P, Q$ are Hermitian with non-negative eigenvalues, both $p_i$ and $<i|Q|i>$ is non-negative...
Now there are at least two possibilities to make tr() = $0$.
i) all $p_i$ = $0$, then of course $PQ = 0$
ii) all $<i|Q|i>$ = $0$, then I can't prove $PQ = 0$...
iii) $p_i$, $<i|Q|i>$ alternates to be $0$, $PQ$ = ???
I think they should be a very easy problem... But I am just stuck somehow..
linear-algebra operator-theory
edited Jan 19 at 6:36
Andrews
1,3062423
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 5:30
1
$begingroup$
Try to use square roots $sqrt{P}, sqrt{Q}$.
$endgroup$
– Song
Jan 19 at 5:40
add a comment |
$begingroup$
So far I can prove:
suppose there is a basis set ${|i>}$ that diagonalize $P$ with eigenvalue $p_i$, then tr($PQ$) = $sum_i<i|PQ|i>$ = $sum_ip_i<i|Q|i>$.
Since $P, Q$ are Hermitian with non-negative eigenvalues, both $p_i$ and $<i|Q|i>$ is non-negative...
Now there are at least two possibilities to make tr() = $0$.
i) all $p_i$ = $0$, then of course $PQ = 0$
ii) all $<i|Q|i>$ = $0$, then I can't prove $PQ = 0$...
iii) $p_i$, $<i|Q|i>$ alternates to be $0$, $PQ$ = ???
I think they should be a very easy problem... But I am just stuck somehow..
linear-algebra operator-theory
edited Jan 19 at 6:36
Andrews
1,3062423
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
$endgroup$
So far I can prove:
suppose there is a basis set ${|i>}$ that diagonalize $P$ with eigenvalue $p_i$, then tr($PQ$) = $sum_i<i|PQ|i>$ = $sum_ip_i<i|Q|i>$.
Since $P, Q$ are Hermitian with non-negative eigenvalues, both $p_i$ and $<i|Q|i>$ is non-negative...
Now there are at least two possibilities to make tr() = $0$.
i) all $p_i$ = $0$, then of course $PQ = 0$
ii) all $<i|Q|i>$ = $0$, then I can't prove $PQ = 0$...
iii) $p_i$, $<i|Q|i>$ alternates to be $0$, $PQ$ = ???
I think they should be a very easy problem... But I am just stuck somehow..
linear-algebra operator-theory
linear-algebra operator-theory
edited Jan 19 at 6:36
Andrews
1,3062423
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
edited Jan 19 at 6:36
Andrews
1,3062423
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
edited Jan 19 at 6:36
Andrews
1,3062423
edited Jan 19 at 6:36
Andrews
1,3062423
edited Jan 19 at 6:36
Andrews
1,3062423
1,3062423
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
82
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 5:30
1
$begingroup$
Try to use square roots $sqrt{P}, sqrt{Q}$.
$endgroup$
– Song
Jan 19 at 5:40
add a comment |
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 5:30
1
$begingroup$
Try to use square roots $sqrt{P}, sqrt{Q}$.
$endgroup$
– Song
Jan 19 at 5:40
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 5:30
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 5:30
1
1
$begingroup$
Try to use square roots $sqrt{P}, sqrt{Q}$.
$endgroup$
– Song
Jan 19 at 5:40
$begingroup$
Try to use square roots $sqrt{P}, sqrt{Q}$.
$endgroup$
– Song
Jan 19 at 5:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a fairly direct argument.
You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
$begingroup$
Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
$endgroup$
– Jiashen Tang
Jan 19 at 19:18
$begingroup$
"Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
$endgroup$
– Martin Argerami
Jan 19 at 21:36
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%2f3079051%2fp-q-are-hermitian-operators-with-only-non-negative-eigenvalues-then-trpq%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
$begingroup$
Here is a fairly direct argument.
You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
$begingroup$
Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
$endgroup$
– Jiashen Tang
Jan 19 at 19:18
$begingroup$
"Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
$endgroup$
– Martin Argerami
Jan 19 at 21:36
add a comment |
$begingroup$
Here is a fairly direct argument.
You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
$begingroup$
Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
$endgroup$
– Jiashen Tang
Jan 19 at 19:18
$begingroup$
"Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
$endgroup$
– Martin Argerami
Jan 19 at 21:36
add a comment |
$begingroup$
Here is a fairly direct argument.
You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
Here is a fairly direct argument.
You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
130k1184185
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
130k1184185
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
130k1184185
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
130k1184185
130k1184185
$begingroup$
Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
$endgroup$
– Jiashen Tang
Jan 19 at 19:18
$begingroup$
"Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
$endgroup$
– Martin Argerami
Jan 19 at 21:36
add a comment |
$begingroup$
Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
$endgroup$
– Jiashen Tang
Jan 19 at 19:18
$begingroup$
"Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
$endgroup$
– Martin Argerami
Jan 19 at 21:36
$begingroup$
Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
$endgroup$
– Jiashen Tang
Jan 19 at 19:18
$begingroup$
Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
$endgroup$
– Jiashen Tang
Jan 19 at 19:18
$begingroup$
"Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
$endgroup$
– Martin Argerami
Jan 19 at 21:36
$begingroup$
"Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
$endgroup$
– Martin Argerami
Jan 19 at 21:36
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%2f3079051%2fp-q-are-hermitian-operators-with-only-non-negative-eigenvalues-then-trpq%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
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 5:30
1
$begingroup$
Try to use square roots $sqrt{P}, sqrt{Q}$.
$endgroup$
– Song
Jan 19 at 5:40