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Does $Ax=x$ imply $A^* x=x$, if $A^*$ is the conjugate transpose of $A$?

5

1

$begingroup$

I have a fairly simple question. If $A$ is a matrix and $A^*$ denotes its conjugate transpose, is it true that if $Ax = x$, then $A^*x = x$?

The matrix $A^*$ will certainly have $1$ as an eigenvalue, but will it be with the same eigenvector? And if not, what is the relation between the eigenvector of $A$ and the one of $A^*$?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 30 at 19:46

glS

790521

asked Feb 10 '13 at 16:21

user61408user61408

894

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I take it you mean that $A(x) = x$ for some particular $x$ (that is, not for any $x$)?
  $endgroup$
  – Muphrid
  Feb 10 '13 at 16:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One simple thing that should be clarified here: if one distinguishes left and right eigenvectors then one CAN say something, ie Ax = kx is the same as xA = kx.
  $endgroup$
  – GaryMak
  Sep 22 '13 at 10:07
  

  
  
  
  

add a comment | 

5

1

$begingroup$

I have a fairly simple question. If $A$ is a matrix and $A^*$ denotes its conjugate transpose, is it true that if $Ax = x$, then $A^*x = x$?

The matrix $A^*$ will certainly have $1$ as an eigenvalue, but will it be with the same eigenvector? And if not, what is the relation between the eigenvector of $A$ and the one of $A^*$?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 30 at 19:46

glS

790521

asked Feb 10 '13 at 16:21

user61408user61408

894

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I take it you mean that $A(x) = x$ for some particular $x$ (that is, not for any $x$)?
  $endgroup$
  – Muphrid
  Feb 10 '13 at 16:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One simple thing that should be clarified here: if one distinguishes left and right eigenvectors then one CAN say something, ie Ax = kx is the same as xA = kx.
  $endgroup$
  – GaryMak
  Sep 22 '13 at 10:07
  

  
  
  
  

add a comment | 

$begingroup$

I have a fairly simple question. If $A$ is a matrix and $A^*$ denotes its conjugate transpose, is it true that if $Ax = x$, then $A^*x = x$?

The matrix $A^*$ will certainly have $1$ as an eigenvalue, but will it be with the same eigenvector? And if not, what is the relation between the eigenvector of $A$ and the one of $A^*$?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 30 at 19:46

glS

790521

asked Feb 10 '13 at 16:21

user61408user61408

894

$endgroup$

share|cite|improve this question

edited Mar 30 at 19:46

glS

790521

asked Feb 10 '13 at 16:21

user61408user61408

894

share|cite|improve this question

edited Mar 30 at 19:46

glS

790521

asked Feb 10 '13 at 16:21

user61408user61408

894

edited Mar 30 at 19:46

glS

790521

edited Mar 30 at 19:46

glS

790521

asked Feb 10 '13 at 16:21

user61408user61408

894

asked Feb 10 '13 at 16:21

user61408user61408

894

3 Answers
3

active

oldest

votes

7

$begingroup$

No, take the matrix

$$beginpmatrix 1 & 1\ 0 & -1endpmatrix$$

which has $x=(1,0)^T$ as an eigenvector with eigenvalue 1. Yet $A^*x=(1,1)^Tneq x$.

share|cite|improve this answer

answered Feb 10 '13 at 16:39

Alex R.Alex R.

25.2k12452

$endgroup$

add a comment | 

3

$begingroup$

The easiest example would be to consider the rank one matrix
$$A = xy^top$$
Then $$Ax = (xy^top) x= x(y^top x) = xlambda = lambda x$$
and
$$ A^* x = (bary barx^top) x =bary (barx^top x )= bary k = kbary$$

share|cite|improve this answer

answered Feb 10 '13 at 16:59

adam Wadam W

4,59521141

$endgroup$

add a comment | 

2

$begingroup$

The eigenvectors of $A$ and $A^ast$, in general, have no relationship with each other.

Let $e_i$ denotes the $i$-th vector in the canonical basis of $mathbbC^n$. Pick any two vectors $u,v$, such that $u$ is linearly independent of $e_n$ and $v$ is linearly independent of $e_1$. Then there always exists an invertible matrix $S$ such that $Se_1 = u$ and $Sv = e_n$. Now, consider $A=SJS^-1$, where $J$ is an $ntimes n$ Jordan block associated with the eigenvalue $1$. Then $u$ is the eigenvector of $A$ (up to scalar multiples) and $barv$ is the eigenvector of $A^ast$. Since $u$ and $v$ are picked arbitrarily, they have no relationship with each other whatsoever.

share|cite|improve this answer

edited Feb 10 '13 at 17:15

answered Feb 10 '13 at 17:09

user1551user1551

74.3k566129

$endgroup$

add a comment | 

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7

$begingroup$

No, take the matrix

$$beginpmatrix 1 & 1\ 0 & -1endpmatrix$$

which has $x=(1,0)^T$ as an eigenvector with eigenvalue 1. Yet $A^*x=(1,1)^Tneq x$.

share|cite|improve this answer

answered Feb 10 '13 at 16:39

Alex R.Alex R.

25.2k12452

$endgroup$

add a comment | 

7

$begingroup$

No, take the matrix

$$beginpmatrix 1 & 1\ 0 & -1endpmatrix$$

which has $x=(1,0)^T$ as an eigenvector with eigenvalue 1. Yet $A^*x=(1,1)^Tneq x$.

share|cite|improve this answer

answered Feb 10 '13 at 16:39

Alex R.Alex R.

25.2k12452

$endgroup$

add a comment | 

$begingroup$

No, take the matrix

$$beginpmatrix 1 & 1\ 0 & -1endpmatrix$$

which has $x=(1,0)^T$ as an eigenvector with eigenvalue 1. Yet $A^*x=(1,1)^Tneq x$.

share|cite|improve this answer

answered Feb 10 '13 at 16:39

Alex R.Alex R.

25.2k12452

$endgroup$

share|cite|improve this answer

answered Feb 10 '13 at 16:39

Alex R.Alex R.

25.2k12452

answered Feb 10 '13 at 16:39

Alex R.Alex R.

25.2k12452

answered Feb 10 '13 at 16:39

Alex R.Alex R.

25.2k12452

3

$begingroup$

The easiest example would be to consider the rank one matrix
$$A = xy^top$$
Then $$Ax = (xy^top) x= x(y^top x) = xlambda = lambda x$$
and
$$ A^* x = (bary barx^top) x =bary (barx^top x )= bary k = kbary$$

share|cite|improve this answer

answered Feb 10 '13 at 16:59

adam Wadam W

4,59521141

$endgroup$

add a comment | 

3

$begingroup$

The easiest example would be to consider the rank one matrix
$$A = xy^top$$
Then $$Ax = (xy^top) x= x(y^top x) = xlambda = lambda x$$
and
$$ A^* x = (bary barx^top) x =bary (barx^top x )= bary k = kbary$$

share|cite|improve this answer

answered Feb 10 '13 at 16:59

adam Wadam W

4,59521141

$endgroup$

add a comment | 

$begingroup$

The easiest example would be to consider the rank one matrix
$$A = xy^top$$
Then $$Ax = (xy^top) x= x(y^top x) = xlambda = lambda x$$
and
$$ A^* x = (bary barx^top) x =bary (barx^top x )= bary k = kbary$$

share|cite|improve this answer

answered Feb 10 '13 at 16:59

adam Wadam W

4,59521141

$endgroup$

share|cite|improve this answer

answered Feb 10 '13 at 16:59

adam Wadam W

4,59521141

answered Feb 10 '13 at 16:59

adam Wadam W

4,59521141

answered Feb 10 '13 at 16:59

adam Wadam W

4,59521141

2

$begingroup$

The eigenvectors of $A$ and $A^ast$, in general, have no relationship with each other.

Let $e_i$ denotes the $i$-th vector in the canonical basis of $mathbbC^n$. Pick any two vectors $u,v$, such that $u$ is linearly independent of $e_n$ and $v$ is linearly independent of $e_1$. Then there always exists an invertible matrix $S$ such that $Se_1 = u$ and $Sv = e_n$. Now, consider $A=SJS^-1$, where $J$ is an $ntimes n$ Jordan block associated with the eigenvalue $1$. Then $u$ is the eigenvector of $A$ (up to scalar multiples) and $barv$ is the eigenvector of $A^ast$. Since $u$ and $v$ are picked arbitrarily, they have no relationship with each other whatsoever.

share|cite|improve this answer

edited Feb 10 '13 at 17:15

answered Feb 10 '13 at 17:09

user1551user1551

74.3k566129

$endgroup$

add a comment | 

2

$begingroup$

The eigenvectors of $A$ and $A^ast$, in general, have no relationship with each other.

Let $e_i$ denotes the $i$-th vector in the canonical basis of $mathbbC^n$. Pick any two vectors $u,v$, such that $u$ is linearly independent of $e_n$ and $v$ is linearly independent of $e_1$. Then there always exists an invertible matrix $S$ such that $Se_1 = u$ and $Sv = e_n$. Now, consider $A=SJS^-1$, where $J$ is an $ntimes n$ Jordan block associated with the eigenvalue $1$. Then $u$ is the eigenvector of $A$ (up to scalar multiples) and $barv$ is the eigenvector of $A^ast$. Since $u$ and $v$ are picked arbitrarily, they have no relationship with each other whatsoever.

share|cite|improve this answer

edited Feb 10 '13 at 17:15

answered Feb 10 '13 at 17:09

user1551user1551

74.3k566129

$endgroup$

add a comment | 

$begingroup$

The eigenvectors of $A$ and $A^ast$, in general, have no relationship with each other.

Let $e_i$ denotes the $i$-th vector in the canonical basis of $mathbbC^n$. Pick any two vectors $u,v$, such that $u$ is linearly independent of $e_n$ and $v$ is linearly independent of $e_1$. Then there always exists an invertible matrix $S$ such that $Se_1 = u$ and $Sv = e_n$. Now, consider $A=SJS^-1$, where $J$ is an $ntimes n$ Jordan block associated with the eigenvalue $1$. Then $u$ is the eigenvector of $A$ (up to scalar multiples) and $barv$ is the eigenvector of $A^ast$. Since $u$ and $v$ are picked arbitrarily, they have no relationship with each other whatsoever.

share|cite|improve this answer

edited Feb 10 '13 at 17:15

answered Feb 10 '13 at 17:09

user1551user1551

74.3k566129

$endgroup$

share|cite|improve this answer

edited Feb 10 '13 at 17:15

answered Feb 10 '13 at 17:09

user1551user1551

74.3k566129

answered Feb 10 '13 at 17:09

user1551user1551

74.3k566129

answered Feb 10 '13 at 17:09

user1551user1551

74.3k566129