Positive Definite Matrices and eigenvalues
$begingroup$
Problem:
Let $A$ ∈ ${C}^{n×n}$ be, such that for every x ∈ $C^n$ <$A$x$, x$> ≥ 0
Show that all eigenvalues of $A$ are positive or zero
I suppose that from the standart inner product in the problem we can say that $A$ is a positive definite matrix and therefore follows that the eigenvalues of $A$ are positive.
However I am not sure if that is right,could someone give a hint?
linear-algebra eigenvalues-eigenvectors positive-definite
asked Jan 10 at 18:45
KaiKai
636
$endgroup$
add a comment |
$begingroup$
Problem:
Let $A$ ∈ ${C}^{n×n}$ be, such that for every x ∈ $C^n$ <$A$x$, x$> ≥ 0
Show that all eigenvalues of $A$ are positive or zero
I suppose that from the standart inner product in the problem we can say that $A$ is a positive definite matrix and therefore follows that the eigenvalues of $A$ are positive.
However I am not sure if that is right,could someone give a hint?
linear-algebra eigenvalues-eigenvectors positive-definite
asked Jan 10 at 18:45
KaiKai
636
$endgroup$
add a comment |
$begingroup$
Problem:
Let $A$ ∈ ${C}^{n×n}$ be, such that for every x ∈ $C^n$ <$A$x$, x$> ≥ 0
Show that all eigenvalues of $A$ are positive or zero
I suppose that from the standart inner product in the problem we can say that $A$ is a positive definite matrix and therefore follows that the eigenvalues of $A$ are positive.
However I am not sure if that is right,could someone give a hint?
linear-algebra eigenvalues-eigenvectors positive-definite
asked Jan 10 at 18:45
KaiKai
636
$endgroup$
Problem:
Let $A$ ∈ ${C}^{n×n}$ be, such that for every x ∈ $C^n$ <$A$x$, x$> ≥ 0
Show that all eigenvalues of $A$ are positive or zero
I suppose that from the standart inner product in the problem we can say that $A$ is a positive definite matrix and therefore follows that the eigenvalues of $A$ are positive.
However I am not sure if that is right,could someone give a hint?
linear-algebra eigenvalues-eigenvectors positive-definite
linear-algebra eigenvalues-eigenvectors positive-definite
asked Jan 10 at 18:45
KaiKai
636
asked Jan 10 at 18:45
KaiKai
636
asked Jan 10 at 18:45
KaiKai
636
asked Jan 10 at 18:45
KaiKai
636
asked Jan 10 at 18:45
KaiKai
636
636
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
add a comment |
$begingroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
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%2f3069034%2fpositive-definite-matrices-and-eigenvalues%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
add a comment |
$begingroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
add a comment |
$begingroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
It's a straightforward computation. If $lambda$ is an eigenvalue of $A$, choose an eigenvector $v$ with $langle v,vrangle=1$. Then
$$
lambda=langle lambda v,vrangle=langle Av,vranglegeq0.
$$
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
answered Jan 10 at 18:50
Martin ArgeramiMartin Argerami
128k1182183
128k1182183
add a comment |
add a comment |
$begingroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
add a comment |
$begingroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
add a comment |
$begingroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
Using the definition that you wrote, pick your $x = v$, where $v$ is any eigenvector, then
$$<Av,v> = lambda geq 0$$
This tells you that no matter which eigenvector you pick to be your $x$, you will get that its corresponding eigenvalue is non-negative.
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
answered Jan 10 at 18:53
Ahmad BazziAhmad Bazzi
8,3622824
8,3622824
add a comment |
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%2f3069034%2fpositive-definite-matrices-and-eigenvalues%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
