Show that $A = sum_i lambda_i x_i x_i^T$
$begingroup$
I came across this formula:
$A = sum_i lambda_i x_i x_i^T$
where $lambda_i$ is the i'th eigenvalue and $x_i$ the i'th eigenvector of $A$. As I have no idea how it is called, I was not able to find a proof.
Could someone give a pointer where to find one?
Edit: I have looked at the eigendecomposition, but only found the decomposition $A = Q^T Lambda Q$.
linear-algebra eigenvalues-eigenvectors matrix-decomposition
asked Jan 25 at 10:24
user2316602user2316602
286
$endgroup$
add a comment |
$begingroup$
I came across this formula:
$A = sum_i lambda_i x_i x_i^T$
where $lambda_i$ is the i'th eigenvalue and $x_i$ the i'th eigenvector of $A$. As I have no idea how it is called, I was not able to find a proof.
Could someone give a pointer where to find one?
Edit: I have looked at the eigendecomposition, but only found the decomposition $A = Q^T Lambda Q$.
linear-algebra eigenvalues-eigenvectors matrix-decomposition
asked Jan 25 at 10:24
user2316602user2316602
286
$endgroup$
$begingroup$
I think this requires the matrix $A$ to be symmetric.
$endgroup$
– Mefitico
Jan 25 at 11:07
add a comment |
$begingroup$
I came across this formula:
$A = sum_i lambda_i x_i x_i^T$
where $lambda_i$ is the i'th eigenvalue and $x_i$ the i'th eigenvector of $A$. As I have no idea how it is called, I was not able to find a proof.
Could someone give a pointer where to find one?
Edit: I have looked at the eigendecomposition, but only found the decomposition $A = Q^T Lambda Q$.
linear-algebra eigenvalues-eigenvectors matrix-decomposition
asked Jan 25 at 10:24
user2316602user2316602
286
$endgroup$
I came across this formula:
$A = sum_i lambda_i x_i x_i^T$
where $lambda_i$ is the i'th eigenvalue and $x_i$ the i'th eigenvector of $A$. As I have no idea how it is called, I was not able to find a proof.
Could someone give a pointer where to find one?
Edit: I have looked at the eigendecomposition, but only found the decomposition $A = Q^T Lambda Q$.
linear-algebra eigenvalues-eigenvectors matrix-decomposition
linear-algebra eigenvalues-eigenvectors matrix-decomposition
asked Jan 25 at 10:24
user2316602user2316602
286
asked Jan 25 at 10:24
user2316602user2316602
286
edited Jan 25 at 10:31
user2316602
asked Jan 25 at 10:24
user2316602user2316602
286
asked Jan 25 at 10:24
user2316602user2316602
286
asked Jan 25 at 10:24
user2316602user2316602
286
286
$begingroup$
I think this requires the matrix $A$ to be symmetric.
$endgroup$
– Mefitico
Jan 25 at 11:07
add a comment |
$begingroup$
I think this requires the matrix $A$ to be symmetric.
$endgroup$
– Mefitico
Jan 25 at 11:07
$begingroup$
I think this requires the matrix $A$ to be symmetric.
$endgroup$
– Mefitico
Jan 25 at 11:07
$begingroup$
I think this requires the matrix $A$ to be symmetric.
$endgroup$
– Mefitico
Jan 25 at 11:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Indeed a one line calculus shows that $A = Q^T Lambda Q = sum_i lambda_i Q_i Q_i^T$ with $Q_i$ denoting the $i^{th}$ column of $Q$.
answered Jan 25 at 10:57
BertrandBertrand
45815
$endgroup$
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%2f3086944%2fshow-that-a-sum-i-lambda-i-x-i-x-it%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
$begingroup$
Indeed a one line calculus shows that $A = Q^T Lambda Q = sum_i lambda_i Q_i Q_i^T$ with $Q_i$ denoting the $i^{th}$ column of $Q$.
answered Jan 25 at 10:57
BertrandBertrand
45815
$endgroup$
add a comment |
$begingroup$
Indeed a one line calculus shows that $A = Q^T Lambda Q = sum_i lambda_i Q_i Q_i^T$ with $Q_i$ denoting the $i^{th}$ column of $Q$.
answered Jan 25 at 10:57
BertrandBertrand
45815
$endgroup$
add a comment |
$begingroup$
Indeed a one line calculus shows that $A = Q^T Lambda Q = sum_i lambda_i Q_i Q_i^T$ with $Q_i$ denoting the $i^{th}$ column of $Q$.
answered Jan 25 at 10:57
BertrandBertrand
45815
$endgroup$
Indeed a one line calculus shows that $A = Q^T Lambda Q = sum_i lambda_i Q_i Q_i^T$ with $Q_i$ denoting the $i^{th}$ column of $Q$.
answered Jan 25 at 10:57
BertrandBertrand
45815
edited Jan 25 at 14:45
answered Jan 25 at 10:57
BertrandBertrand
45815
answered Jan 25 at 10:57
BertrandBertrand
45815
answered Jan 25 at 10:57
BertrandBertrand
45815
45815
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%2f3086944%2fshow-that-a-sum-i-lambda-i-x-i-x-it%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
$begingroup$
I think this requires the matrix $A$ to be symmetric.
$endgroup$
– Mefitico
Jan 25 at 11:07