If $V$ is finite dimensional and $T:V to V , S:V to V$ are normal linear transformations and $TS=ST$ then they share a common basis of egienvectors

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If $V$ is over the field $mathbbR$ then this is not always true, because in $mathbbR$ if a linear transformation $T$ is normal , then it is not always true that $T$ diagonalizable and thus not having a basis for $V$ that contains eigenvectors of $T$

But if $V$ is over the field $mathbbC$ then I think this is true. I would like to get a clue please on how to proceed.

linear-algebra

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asked Jul 20 at 4:52

idan di

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  Any two commuting diagonal matrices are simultaneously diagonalisable.
  – Lord Shark the Unknown
  Jul 20 at 5:10
  

  
  
  
  

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up vote
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down vote

favorite

If $V$ is over the field $mathbbR$ then this is not always true, because in $mathbbR$ if a linear transformation $T$ is normal , then it is not always true that $T$ diagonalizable and thus not having a basis for $V$ that contains eigenvectors of $T$

But if $V$ is over the field $mathbbC$ then I think this is true. I would like to get a clue please on how to proceed.

linear-algebra

share|cite|improve this question

asked Jul 20 at 4:52

idan di

372110

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Any two commuting diagonal matrices are simultaneously diagonalisable.
  – Lord Shark the Unknown
  Jul 20 at 5:10
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

If $V$ is over the field $mathbbR$ then this is not always true, because in $mathbbR$ if a linear transformation $T$ is normal , then it is not always true that $T$ diagonalizable and thus not having a basis for $V$ that contains eigenvectors of $T$

But if $V$ is over the field $mathbbC$ then I think this is true. I would like to get a clue please on how to proceed.

linear-algebra

share|cite|improve this question

asked Jul 20 at 4:52

idan di

372110

If $V$ is over the field $mathbbR$ then this is not always true, because in $mathbbR$ if a linear transformation $T$ is normal , then it is not always true that $T$ diagonalizable and thus not having a basis for $V$ that contains eigenvectors of $T$

But if $V$ is over the field $mathbbC$ then I think this is true. I would like to get a clue please on how to proceed.

linear-algebra

share|cite|improve this question

asked Jul 20 at 4:52

idan di

372110

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 4:52

idan di

372110

asked Jul 20 at 4:52

idan di

372110

asked Jul 20 at 4:52

idan di

372110

372110

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Any two commuting diagonal matrices are simultaneously diagonalisable.
  – Lord Shark the Unknown
  Jul 20 at 5:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Any two commuting diagonal matrices are simultaneously diagonalisable.
  – Lord Shark the Unknown
  Jul 20 at 5:10
  

  
  
  
  

2

2

Any two commuting diagonal matrices are simultaneously diagonalisable.
– Lord Shark the Unknown
Jul 20 at 5:10

Any two commuting diagonal matrices are simultaneously diagonalisable.
– Lord Shark the Unknown
Jul 20 at 5:10

add a comment | 

1 Answer
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Hint: Take a basis of $V$ that consists of eigenvectors of $T$ and see what you get from $ST=TS$. Can this basis also consist of eigenvectors of $S$?

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edited Jul 20 at 6:57

answered Jul 20 at 6:49

Tengu

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
0
down vote

Hint: Take a basis of $V$ that consists of eigenvectors of $T$ and see what you get from $ST=TS$. Can this basis also consist of eigenvectors of $S$?

share|cite|improve this answer

edited Jul 20 at 6:57

answered Jul 20 at 6:49

Tengu

2,3391920

add a comment | 

up vote
0
down vote

Hint: Take a basis of $V$ that consists of eigenvectors of $T$ and see what you get from $ST=TS$. Can this basis also consist of eigenvectors of $S$?

share|cite|improve this answer

edited Jul 20 at 6:57

answered Jul 20 at 6:49

Tengu

2,3391920

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint: Take a basis of $V$ that consists of eigenvectors of $T$ and see what you get from $ST=TS$. Can this basis also consist of eigenvectors of $S$?

share|cite|improve this answer

edited Jul 20 at 6:57

answered Jul 20 at 6:49

Tengu

2,3391920

Hint: Take a basis of $V$ that consists of eigenvectors of $T$ and see what you get from $ST=TS$. Can this basis also consist of eigenvectors of $S$?

share|cite|improve this answer

edited Jul 20 at 6:57

answered Jul 20 at 6:49

Tengu

2,3391920

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 6:57

edited Jul 20 at 6:57

edited Jul 20 at 6:57

answered Jul 20 at 6:49

Tengu

2,3391920

answered Jul 20 at 6:49

Tengu

2,3391920

answered Jul 20 at 6:49

Tengu

2,3391920

2,3391920

add a comment | 

add a comment | 

 

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