Eigenvalues of a real orthogonal matrix.Do real matrices always have real eigenvalues?Generalized eigenvalue problem; why do real eigenvalues exist?If $A$ is a real symmetric matrix, then $A$ has real eigenvalues.Block diagonal form of elements of SO(n)Eigenvectors and eigenvalues of Hessian matrixproperties of, 3x3 matrix, determinant 1, real eigenvaluesWhy eigenvalues of an orthogonal matrix made with QR decomposition include -1?Determine the matrix of the orthogonal projectionLet $A in mathbbC^n times n$ be hermitian. Prove all eigenvalues of $A$ are real…Existence condition of Real Eigenvalues for Non-Symmetric Real Matrix
Why do I get negative height?
Sums of two squares in arithmetic progressions
Standard deduction V. mortgage interest deduction - is it basically only for the rich?
How to remove border from elements in the last row?
Bullying boss launched a smear campaign and made me unemployable
Is it a bad idea to plug the other end of ESD strap to wall ground?
What is a Samsaran Word™?
Pact of Blade Warlock with Dancing Blade
How to prevent "they're falling in love" trope
GFCI outlets - can they be repaired? Are they really needed at the end of a circuit?
How could indestructible materials be used in power generation?
Did 'Cinema Songs' exist during Hiranyakshipu's time?
Which ISO should I use for the cleanest image?
Processor speed limited at 0.4 Ghz
How dangerous is XSS
How do I exit BASH while loop using modulus operator?
Is it "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function"? If so, why?
When handwriting 黄 (huáng; yellow) is it incorrect to have a disconnected 草 (cǎo; grass) radical on top?
Why was the shrink from 8″ made only to 5.25″ and not smaller (4″ or less)
What historical events would have to change in order to make 19th century "steampunk" technology possible?
Finitely generated matrix groups whose eigenvalues are all algebraic
Could the museum Saturn V's be refitted for one more flight?
How can saying a song's name be a copyright violation?
How exploitable/balanced is this homebrew spell: Spell Permanency?
Eigenvalues of a real orthogonal matrix.
Do real matrices always have real eigenvalues?Generalized eigenvalue problem; why do real eigenvalues exist?If $A$ is a real symmetric matrix, then $A$ has real eigenvalues.Block diagonal form of elements of SO(n)Eigenvectors and eigenvalues of Hessian matrixproperties of, 3x3 matrix, determinant 1, real eigenvaluesWhy eigenvalues of an orthogonal matrix made with QR decomposition include -1?Determine the matrix of the orthogonal projectionLet $A in mathbbC^n times n$ be hermitian. Prove all eigenvalues of $A$ are real…Existence condition of Real Eigenvalues for Non-Symmetric Real Matrix
$begingroup$
Let $A$ be a real orthogonal matrix. Then $A^text T A = I.$ Let $lambda in Bbb C$ be an eigenvalue of $A$ corresponding to the eigenvector $X in Bbb C^n.$ Then we have
$$beginalign* X^text T A^text T A X = X^text T X. \ implies (AX)^text T AX & = X^text T X. \ implies (lambda X)^text T lambda X & = X^text T X. \ implies lambda^2 X^text T X & = X^text T X. \ implies (lambda^2 - 1) X^text T X & = 0. endalign*$$
Since $X$ is an eigenvector $X neq 0.$ Therefore $^2 = X^text T X neq 0.$ Hence we must have $lambda^2 - 1 = 0$ i.e. $lambda^2 = 1.$ So $lambda = pm 1.$
So according to my argument above it follows that eigenvalues of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigenvalues of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.
What's going wrong in my argument above. Please help me in this regard.
Thank you very much for your valuable time.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
edited 2 days ago
Yanko
8,2942830
asked 2 days ago
math maniac.math maniac.
1417
$endgroup$
add a comment |
$begingroup$
Let $A$ be a real orthogonal matrix. Then $A^text T A = I.$ Let $lambda in Bbb C$ be an eigenvalue of $A$ corresponding to the eigenvector $X in Bbb C^n.$ Then we have
$$beginalign* X^text T A^text T A X = X^text T X. \ implies (AX)^text T AX & = X^text T X. \ implies (lambda X)^text T lambda X & = X^text T X. \ implies lambda^2 X^text T X & = X^text T X. \ implies (lambda^2 - 1) X^text T X & = 0. endalign*$$
Since $X$ is an eigenvector $X neq 0.$ Therefore $^2 = X^text T X neq 0.$ Hence we must have $lambda^2 - 1 = 0$ i.e. $lambda^2 = 1.$ So $lambda = pm 1.$
So according to my argument above it follows that eigenvalues of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigenvalues of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.
What's going wrong in my argument above. Please help me in this regard.
Thank you very much for your valuable time.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
edited 2 days ago
Yanko
8,2942830
asked 2 days ago
math maniac.math maniac.
1417
$endgroup$
add a comment |
$begingroup$
Let $A$ be a real orthogonal matrix. Then $A^text T A = I.$ Let $lambda in Bbb C$ be an eigenvalue of $A$ corresponding to the eigenvector $X in Bbb C^n.$ Then we have
$$beginalign* X^text T A^text T A X = X^text T X. \ implies (AX)^text T AX & = X^text T X. \ implies (lambda X)^text T lambda X & = X^text T X. \ implies lambda^2 X^text T X & = X^text T X. \ implies (lambda^2 - 1) X^text T X & = 0. endalign*$$
Since $X$ is an eigenvector $X neq 0.$ Therefore $^2 = X^text T X neq 0.$ Hence we must have $lambda^2 - 1 = 0$ i.e. $lambda^2 = 1.$ So $lambda = pm 1.$
So according to my argument above it follows that eigenvalues of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigenvalues of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.
What's going wrong in my argument above. Please help me in this regard.
Thank you very much for your valuable time.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
edited 2 days ago
Yanko
8,2942830
asked 2 days ago
math maniac.math maniac.
1417
$endgroup$
Let $A$ be a real orthogonal matrix. Then $A^text T A = I.$ Let $lambda in Bbb C$ be an eigenvalue of $A$ corresponding to the eigenvector $X in Bbb C^n.$ Then we have
$$beginalign* X^text T A^text T A X = X^text T X. \ implies (AX)^text T AX & = X^text T X. \ implies (lambda X)^text T lambda X & = X^text T X. \ implies lambda^2 X^text T X & = X^text T X. \ implies (lambda^2 - 1) X^text T X & = 0. endalign*$$
Since $X$ is an eigenvector $X neq 0.$ Therefore $^2 = X^text T X neq 0.$ Hence we must have $lambda^2 - 1 = 0$ i.e. $lambda^2 = 1.$ So $lambda = pm 1.$
So according to my argument above it follows that eigenvalues of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigenvalues of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.
What's going wrong in my argument above. Please help me in this regard.
Thank you very much for your valuable time.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
edited 2 days ago
Yanko
8,2942830
asked 2 days ago
math maniac.math maniac.
1417
edited 2 days ago
Yanko
8,2942830
asked 2 days ago
math maniac.math maniac.
1417
edited 2 days ago
Yanko
8,2942830
edited 2 days ago
Yanko
8,2942830
edited 2 days ago
Yanko
8,2942830
8,2942830
asked 2 days ago
math maniac.math maniac.
1417
asked 2 days ago
math maniac.math maniac.
1417
asked 2 days ago
math maniac.math maniac.
1417
1417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 2 days ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
2 days ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
2 days ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
2 days ago
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%2f3169070%2feigenvalues-of-a-real-orthogonal-matrix%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
$begingroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 2 days ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
2 days ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
2 days ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
2 days ago
add a comment |
$begingroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 2 days ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
2 days ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
2 days ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
2 days ago
add a comment |
$begingroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 2 days ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 2 days ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 2 days ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 2 days ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 2 days ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
2 days ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
2 days ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
2 days ago
add a comment |
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
2 days ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
2 days ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
2 days ago
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
2 days ago
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
2 days ago
2
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
2 days ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
2 days ago
1
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
2 days ago
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
2 days ago
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%2f3169070%2feigenvalues-of-a-real-orthogonal-matrix%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
