If A is anti-hermitian, show that $|det(1+A)|^2 geqslant 1$
$begingroup$
Given that A is anti-hermitian, i.e. $A^dagger = -A$, show, by diagonalising $iA$, that $$|det(1+A)|^2 geqslant 1$$
$hspace{3mm}$
So what I thought was that I need to show that $det(1+A)[det(1+A)]^* geqslant 1$, i.e. $det(1+A)det(1+A^dagger) geqslant 1$.
Since I know the determinant is equal to the product of the eigenvalues, I have found the eigenvalues of $(1+A)$ by $$(1+A)textbf{x} = ktextbf{x}$$
$$Atextbf{x}=(k-1)textbf{x}$$ $$Atextbf{x}=lambda textbf{x}$$ $$k=lambda +1$$
where $lambda$ is purely imaginary because A is anti-hermitian.
Similarly, I can show that the eigenvalue of $(1+A^dagger)=(1-A)$ to be $(1-lambda)$. Therefore, $$det(1+A)det(1+A^dagger) = prod_{i=1}^{n} (1+lambda_i)(1-lambda_i)=prod_{i=1}^{n} (1+|lambda_i|^2) geqslant 1$$
However, I am not sure if this actually proves it, and especially I don't know why the question asks to diagonalize $iA$.
Thank you.
matrices determinant diagonalization
edited Jan 1 at 13:09
Botond
5,6982732
asked Jan 1 at 11:47
Student 1Student 1
668
$endgroup$
add a comment |
$begingroup$
Given that A is anti-hermitian, i.e. $A^dagger = -A$, show, by diagonalising $iA$, that $$|det(1+A)|^2 geqslant 1$$
$hspace{3mm}$
So what I thought was that I need to show that $det(1+A)[det(1+A)]^* geqslant 1$, i.e. $det(1+A)det(1+A^dagger) geqslant 1$.
Since I know the determinant is equal to the product of the eigenvalues, I have found the eigenvalues of $(1+A)$ by $$(1+A)textbf{x} = ktextbf{x}$$
$$Atextbf{x}=(k-1)textbf{x}$$ $$Atextbf{x}=lambda textbf{x}$$ $$k=lambda +1$$
where $lambda$ is purely imaginary because A is anti-hermitian.
Similarly, I can show that the eigenvalue of $(1+A^dagger)=(1-A)$ to be $(1-lambda)$. Therefore, $$det(1+A)det(1+A^dagger) = prod_{i=1}^{n} (1+lambda_i)(1-lambda_i)=prod_{i=1}^{n} (1+|lambda_i|^2) geqslant 1$$
However, I am not sure if this actually proves it, and especially I don't know why the question asks to diagonalize $iA$.
Thank you.
matrices determinant diagonalization
edited Jan 1 at 13:09
Botond
5,6982732
asked Jan 1 at 11:47
Student 1Student 1
668
$endgroup$
1
$begingroup$
Well, $iA$ is Hermitian, so its eigenvalues are real, so those of $A$ are purely imaginary.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:51
$begingroup$
@Lord Shark the Unknown So the questions asks us to diagonalise $iA$ so that we can show that eigenvalues of A are purely imaginary? Does it have no other purposes? What about the rest of my working, are they valid?
$endgroup$
– Student 1
Jan 1 at 11:58
add a comment |
$begingroup$
Given that A is anti-hermitian, i.e. $A^dagger = -A$, show, by diagonalising $iA$, that $$|det(1+A)|^2 geqslant 1$$
$hspace{3mm}$
So what I thought was that I need to show that $det(1+A)[det(1+A)]^* geqslant 1$, i.e. $det(1+A)det(1+A^dagger) geqslant 1$.
Since I know the determinant is equal to the product of the eigenvalues, I have found the eigenvalues of $(1+A)$ by $$(1+A)textbf{x} = ktextbf{x}$$
$$Atextbf{x}=(k-1)textbf{x}$$ $$Atextbf{x}=lambda textbf{x}$$ $$k=lambda +1$$
where $lambda$ is purely imaginary because A is anti-hermitian.
Similarly, I can show that the eigenvalue of $(1+A^dagger)=(1-A)$ to be $(1-lambda)$. Therefore, $$det(1+A)det(1+A^dagger) = prod_{i=1}^{n} (1+lambda_i)(1-lambda_i)=prod_{i=1}^{n} (1+|lambda_i|^2) geqslant 1$$
However, I am not sure if this actually proves it, and especially I don't know why the question asks to diagonalize $iA$.
Thank you.
matrices determinant diagonalization
edited Jan 1 at 13:09
Botond
5,6982732
asked Jan 1 at 11:47
Student 1Student 1
668
$endgroup$
Given that A is anti-hermitian, i.e. $A^dagger = -A$, show, by diagonalising $iA$, that $$|det(1+A)|^2 geqslant 1$$
$hspace{3mm}$
So what I thought was that I need to show that $det(1+A)[det(1+A)]^* geqslant 1$, i.e. $det(1+A)det(1+A^dagger) geqslant 1$.
Since I know the determinant is equal to the product of the eigenvalues, I have found the eigenvalues of $(1+A)$ by $$(1+A)textbf{x} = ktextbf{x}$$
$$Atextbf{x}=(k-1)textbf{x}$$ $$Atextbf{x}=lambda textbf{x}$$ $$k=lambda +1$$
where $lambda$ is purely imaginary because A is anti-hermitian.
Similarly, I can show that the eigenvalue of $(1+A^dagger)=(1-A)$ to be $(1-lambda)$. Therefore, $$det(1+A)det(1+A^dagger) = prod_{i=1}^{n} (1+lambda_i)(1-lambda_i)=prod_{i=1}^{n} (1+|lambda_i|^2) geqslant 1$$
However, I am not sure if this actually proves it, and especially I don't know why the question asks to diagonalize $iA$.
Thank you.
matrices determinant diagonalization
matrices determinant diagonalization
edited Jan 1 at 13:09
Botond
5,6982732
asked Jan 1 at 11:47
Student 1Student 1
668
edited Jan 1 at 13:09
Botond
5,6982732
asked Jan 1 at 11:47
Student 1Student 1
668
edited Jan 1 at 13:09
Botond
5,6982732
edited Jan 1 at 13:09
Botond
5,6982732
edited Jan 1 at 13:09
Botond
5,6982732
5,6982732
asked Jan 1 at 11:47
Student 1Student 1
668
asked Jan 1 at 11:47
Student 1Student 1
668
asked Jan 1 at 11:47
Student 1Student 1
668
668
1
$begingroup$
Well, $iA$ is Hermitian, so its eigenvalues are real, so those of $A$ are purely imaginary.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:51
$begingroup$
@Lord Shark the Unknown So the questions asks us to diagonalise $iA$ so that we can show that eigenvalues of A are purely imaginary? Does it have no other purposes? What about the rest of my working, are they valid?
$endgroup$
– Student 1
Jan 1 at 11:58
add a comment |
1
$begingroup$
Well, $iA$ is Hermitian, so its eigenvalues are real, so those of $A$ are purely imaginary.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:51
$begingroup$
@Lord Shark the Unknown So the questions asks us to diagonalise $iA$ so that we can show that eigenvalues of A are purely imaginary? Does it have no other purposes? What about the rest of my working, are they valid?
$endgroup$
– Student 1
Jan 1 at 11:58
1
1
$begingroup$
Well, $iA$ is Hermitian, so its eigenvalues are real, so those of $A$ are purely imaginary.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:51
$begingroup$
Well, $iA$ is Hermitian, so its eigenvalues are real, so those of $A$ are purely imaginary.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:51
$begingroup$
@Lord Shark the Unknown So the questions asks us to diagonalise $iA$ so that we can show that eigenvalues of A are purely imaginary? Does it have no other purposes? What about the rest of my working, are they valid?
$endgroup$
– Student 1
Jan 1 at 11:58
$begingroup$
@Lord Shark the Unknown So the questions asks us to diagonalise $iA$ so that we can show that eigenvalues of A are purely imaginary? Does it have no other purposes? What about the rest of my working, are they valid?
$endgroup$
– Student 1
Jan 1 at 11:58
add a comment |
0
active
oldest
votes
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%2f3058420%2fif-a-is-anti-hermitian-show-that-det1a2-geqslant-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3058420%2fif-a-is-anti-hermitian-show-that-det1a2-geqslant-1%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
$begingroup$
Well, $iA$ is Hermitian, so its eigenvalues are real, so those of $A$ are purely imaginary.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 11:51
$begingroup$
@Lord Shark the Unknown So the questions asks us to diagonalise $iA$ so that we can show that eigenvalues of A are purely imaginary? Does it have no other purposes? What about the rest of my working, are they valid?
$endgroup$
– Student 1
Jan 1 at 11:58