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Eigenvalues of a real orthogonal matrix.
The Next CEO of Stack OverflowDo 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 $X^2 = X^text T X neq 0.$ 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 58 mins ago
Yanko
8,3052830
asked 7 hours 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 $X^2 = X^text T X neq 0.$ 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 58 mins ago
Yanko
8,3052830
asked 7 hours 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 $X^2 = X^text T X neq 0.$ 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 58 mins ago
Yanko
8,3052830
asked 7 hours 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 $X^2 = X^text T X neq 0.$ 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 58 mins ago
Yanko
8,3052830
asked 7 hours ago
math maniac.math maniac.
1417
edited 58 mins ago
Yanko
8,3052830
asked 7 hours ago
math maniac.math maniac.
1417
edited 58 mins ago
Yanko
8,3052830
edited 58 mins ago
Yanko
8,3052830
edited 58 mins ago
Yanko
8,3052830
8,3052830
asked 7 hours ago
math maniac.math maniac.
1417
asked 7 hours ago
math maniac.math maniac.
1417
asked 7 hours 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 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
7 hours ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
6 hours 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.
6 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
6 hours ago
add a comment |
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1 Answer
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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 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
7 hours ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
6 hours 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.
6 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
6 hours 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 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
7 hours ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
6 hours 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.
6 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
6 hours 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 7 hours 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 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
7 hours ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
6 hours 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.
6 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
6 hours ago
add a comment |
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
7 hours ago
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
6 hours 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.
6 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
6 hours ago
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
7 hours ago
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
7 hours ago
2
2
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
6 hours ago
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
6 hours 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.
6 hours 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.
6 hours ago
1
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
6 hours ago
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
6 hours ago
add a comment |
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