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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 Mar 31 at 11:14

Yanko

8,4692830

asked Mar 31 at 5:00

math maniac.math maniac.

1647

$endgroup$

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 Mar 31 at 11:14

Yanko

8,4692830

asked Mar 31 at 5:00

math maniac.math maniac.

1647

$endgroup$

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 Mar 31 at 11:14

Yanko

8,4692830

asked Mar 31 at 5:00

math maniac.math maniac.

1647

$endgroup$

share|cite|improve this question

edited Mar 31 at 11:14

Yanko

8,4692830

asked Mar 31 at 5:00

math maniac.math maniac.

1647

share|cite|improve this question

edited Mar 31 at 11:14

Yanko

8,4692830

asked Mar 31 at 5:00

math maniac.math maniac.

1647

edited Mar 31 at 11:14

Yanko

8,4692830

edited Mar 31 at 11:14

Yanko

8,4692830

asked Mar 31 at 5:00

math maniac.math maniac.

1647

asked Mar 31 at 5:00

math maniac.math maniac.

1647

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 Mar 31 at 5:04

Lord Shark the UnknownLord Shark the Unknown

108k1162136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how can Euclidean norm of non zero vector be zero?
  $endgroup$
  – math maniac.
  Mar 31 at 5:08
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @mathmaniac. How can $1^2+i^2$ equal zero?
  $endgroup$
  – Lord Shark the Unknown
  Mar 31 at 5:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  Mar 31 at 5:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
  $endgroup$
  – math maniac.
  Mar 31 at 5:19
  
  

  
  
  
  

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 Mar 31 at 5:04

Lord Shark the UnknownLord Shark the Unknown

108k1162136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how can Euclidean norm of non zero vector be zero?
  $endgroup$
  – math maniac.
  Mar 31 at 5:08
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @mathmaniac. How can $1^2+i^2$ equal zero?
  $endgroup$
  – Lord Shark the Unknown
  Mar 31 at 5:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  Mar 31 at 5:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
  $endgroup$
  – math maniac.
  Mar 31 at 5:19
  
  

  
  
  
  

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 Mar 31 at 5:04

Lord Shark the UnknownLord Shark the Unknown

108k1162136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how can Euclidean norm of non zero vector be zero?
  $endgroup$
  – math maniac.
  Mar 31 at 5:08
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @mathmaniac. How can $1^2+i^2$ equal zero?
  $endgroup$
  – Lord Shark the Unknown
  Mar 31 at 5:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  Mar 31 at 5:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
  $endgroup$
  – math maniac.
  Mar 31 at 5:19
  
  

  
  
  
  

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 Mar 31 at 5:04

Lord Shark the UnknownLord Shark the Unknown

108k1162136

$endgroup$

share|cite|improve this answer

answered Mar 31 at 5:04

Lord Shark the UnknownLord Shark the Unknown

108k1162136

answered Mar 31 at 5:04

Lord Shark the UnknownLord Shark the Unknown

108k1162136

answered Mar 31 at 5:04

Lord Shark the UnknownLord Shark the Unknown

108k1162136