Symmetric matrix with n rows has n linearly independent eigenvectors?
I'm trying to better understand SPD matrices and have started with symmetric matrices. It's apparent to me that symmetric matrices have real eigenvalues, but how to show that the algebraic multiplicity of each eigenvalue is no greater than it's geometric multiplicity?
EDIT:
The comments led me to a nice answer from another user which I think I understand.
Essentially the argument is:
Take any eigenvector of the symmetric matrix $A$, call it $x$. Then $A$ maps the set ${v : langle v, x rangle=0}=:V$ to itself since take $v in V$ then $(Av)^Tx=v^TA^Tx=v^TAx=lambda v^T x=0$ but this space V has 1 dimension less than $V cup text{ span} (x)$. Then because all endomorphisms which are linear functions have eigenvalues and $A$ restricted to V is an endomorphism, A has an eigenvalue/eigenvector pair on $V$. You can repeat the process and therefore all the eigenvectors are distinct and orthonormal.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Jan 6 at 11:08
DanDan
827
add a comment |
I'm trying to better understand SPD matrices and have started with symmetric matrices. It's apparent to me that symmetric matrices have real eigenvalues, but how to show that the algebraic multiplicity of each eigenvalue is no greater than it's geometric multiplicity?
EDIT:
The comments led me to a nice answer from another user which I think I understand.
Essentially the argument is:
Take any eigenvector of the symmetric matrix $A$, call it $x$. Then $A$ maps the set ${v : langle v, x rangle=0}=:V$ to itself since take $v in V$ then $(Av)^Tx=v^TA^Tx=v^TAx=lambda v^T x=0$ but this space V has 1 dimension less than $V cup text{ span} (x)$. Then because all endomorphisms which are linear functions have eigenvalues and $A$ restricted to V is an endomorphism, A has an eigenvalue/eigenvector pair on $V$. You can repeat the process and therefore all the eigenvectors are distinct and orthonormal.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Jan 6 at 11:08
DanDan
827
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
– A.Γ.
Jan 6 at 11:44
Possible duplicate math.stackexchange.com/questions/2777494/…
– A.Γ.
Jan 6 at 12:00
add a comment |
I'm trying to better understand SPD matrices and have started with symmetric matrices. It's apparent to me that symmetric matrices have real eigenvalues, but how to show that the algebraic multiplicity of each eigenvalue is no greater than it's geometric multiplicity?
EDIT:
The comments led me to a nice answer from another user which I think I understand.
Essentially the argument is:
Take any eigenvector of the symmetric matrix $A$, call it $x$. Then $A$ maps the set ${v : langle v, x rangle=0}=:V$ to itself since take $v in V$ then $(Av)^Tx=v^TA^Tx=v^TAx=lambda v^T x=0$ but this space V has 1 dimension less than $V cup text{ span} (x)$. Then because all endomorphisms which are linear functions have eigenvalues and $A$ restricted to V is an endomorphism, A has an eigenvalue/eigenvector pair on $V$. You can repeat the process and therefore all the eigenvectors are distinct and orthonormal.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Jan 6 at 11:08
DanDan
827
I'm trying to better understand SPD matrices and have started with symmetric matrices. It's apparent to me that symmetric matrices have real eigenvalues, but how to show that the algebraic multiplicity of each eigenvalue is no greater than it's geometric multiplicity?
EDIT:
The comments led me to a nice answer from another user which I think I understand.
Essentially the argument is:
Take any eigenvector of the symmetric matrix $A$, call it $x$. Then $A$ maps the set ${v : langle v, x rangle=0}=:V$ to itself since take $v in V$ then $(Av)^Tx=v^TA^Tx=v^TAx=lambda v^T x=0$ but this space V has 1 dimension less than $V cup text{ span} (x)$. Then because all endomorphisms which are linear functions have eigenvalues and $A$ restricted to V is an endomorphism, A has an eigenvalue/eigenvector pair on $V$. You can repeat the process and therefore all the eigenvectors are distinct and orthonormal.
linear-algebra eigenvalues-eigenvectors diagonalization
linear-algebra eigenvalues-eigenvectors diagonalization
asked Jan 6 at 11:08
DanDan
827
asked Jan 6 at 11:08
DanDan
827
edited Jan 6 at 13:16
Dan
asked Jan 6 at 11:08
DanDan
827
asked Jan 6 at 11:08
DanDan
827
asked Jan 6 at 11:08
DanDan
827
827
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
– A.Γ.
Jan 6 at 11:44
Possible duplicate math.stackexchange.com/questions/2777494/…
– A.Γ.
Jan 6 at 12:00
add a comment |
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
– A.Γ.
Jan 6 at 11:44
Possible duplicate math.stackexchange.com/questions/2777494/…
– A.Γ.
Jan 6 at 12:00
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
– A.Γ.
Jan 6 at 11:44
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
– A.Γ.
Jan 6 at 11:44
Possible duplicate math.stackexchange.com/questions/2777494/…
– A.Γ.
Jan 6 at 12:00
Possible duplicate math.stackexchange.com/questions/2777494/…
– A.Γ.
Jan 6 at 12:00
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%2f3063725%2fsymmetric-matrix-with-n-rows-has-n-linearly-independent-eigenvectors%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.
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%2f3063725%2fsymmetric-matrix-with-n-rows-has-n-linearly-independent-eigenvectors%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
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
– A.Γ.
Jan 6 at 11:44
Possible duplicate math.stackexchange.com/questions/2777494/…
– A.Γ.
Jan 6 at 12:00