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Eigenvalues of a real orthogonal matrix.

4

0

$begingroup$

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 $_2^2 = X^text T X neq 0.$ 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

share|cite|improve this question

edited yesterday

Yanko

8,2672830

asked 2 days ago

math maniac.math maniac.

1417

$endgroup$

add a comment | 

4

0

$begingroup$

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 $_2^2 = X^text T X neq 0.$ 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

share|cite|improve this question

edited yesterday

Yanko

8,2672830

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 $_2^2 = X^text T X neq 0.$ 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

share|cite|improve this question

edited yesterday

Yanko

8,2672830

asked 2 days ago

math maniac.math maniac.

1417

$endgroup$

share|cite|improve this question

edited yesterday

Yanko

8,2672830

asked 2 days ago

math maniac.math maniac.

1417

share|cite|improve this question

edited yesterday

Yanko

8,2672830

asked 2 days ago

math maniac.math maniac.

1417

edited yesterday

Yanko

8,2672830

edited yesterday

Yanko

8,2672830

asked 2 days ago

math maniac.math maniac.

1417

asked 2 days ago

math maniac.math maniac.

1417

1 Answer
1

active

oldest

votes

4

$begingroup$

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 $|lambda|^2=1$.

share|cite|improve this answer

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.
  yesterday
  
  

  
  
  
  

add a comment | 

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4

$begingroup$

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 $|lambda|^2=1$.

share|cite|improve this answer

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.
  yesterday
  
  

  
  
  
  

add a comment | 

4

$begingroup$

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 $|lambda|^2=1$.

share|cite|improve this answer

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.
  yesterday
  
  

  
  
  
  

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 $|lambda|^2=1$.

share|cite|improve this answer

answered 2 days ago

Lord Shark the UnknownLord Shark the Unknown

107k1162135

$endgroup$

share|cite|improve this answer

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