Similarity of two self-adjoint operators












3












$begingroup$


I am wondering whether the following is correct:



Let $A$ and $B$ be two bounded self-adjoint, positive and invertible linear operators such that $sigma(A)=sigma(B)$ and $AB=BA$. Can we say that $A$ is necessarily similar to $B$?



I just couldn't find an answer.



Thanks.



Math.






operator-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where did you get the question?
    $endgroup$
    – NicNic8
    Jan 8 at 20:07










  • $begingroup$
    In fact, I was trying to understand an example in some paper and if my question turns out to hold, then I am done.
    $endgroup$
    – Math
    Jan 8 at 20:11






  • 2




    $begingroup$
    Hint: Any two diagonal matrices commute with each other.
    $endgroup$
    – Daniel Schepler
    Jan 8 at 20:12










  • $begingroup$
    Daniel, are you saying that the result is not true?
    $endgroup$
    – Math
    Jan 8 at 20:25
















3












$begingroup$


I am wondering whether the following is correct:



Let $A$ and $B$ be two bounded self-adjoint, positive and invertible linear operators such that $sigma(A)=sigma(B)$ and $AB=BA$. Can we say that $A$ is necessarily similar to $B$?



I just couldn't find an answer.



Thanks.



Math.






operator-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where did you get the question?
    $endgroup$
    – NicNic8
    Jan 8 at 20:07










  • $begingroup$
    In fact, I was trying to understand an example in some paper and if my question turns out to hold, then I am done.
    $endgroup$
    – Math
    Jan 8 at 20:11






  • 2




    $begingroup$
    Hint: Any two diagonal matrices commute with each other.
    $endgroup$
    – Daniel Schepler
    Jan 8 at 20:12










  • $begingroup$
    Daniel, are you saying that the result is not true?
    $endgroup$
    – Math
    Jan 8 at 20:25














3












3








3





$begingroup$


I am wondering whether the following is correct:



Let $A$ and $B$ be two bounded self-adjoint, positive and invertible linear operators such that $sigma(A)=sigma(B)$ and $AB=BA$. Can we say that $A$ is necessarily similar to $B$?



I just couldn't find an answer.



Thanks.



Math.






operator-theory







share|cite|improve this question











$endgroup$




I am wondering whether the following is correct:



Let $A$ and $B$ be two bounded self-adjoint, positive and invertible linear operators such that $sigma(A)=sigma(B)$ and $AB=BA$. Can we say that $A$ is necessarily similar to $B$?



I just couldn't find an answer.



Thanks.



Math.






operator-theory




operator-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 20:43









Aweygan

14.3k21441




14.3k21441










asked Jan 8 at 19:56









MathMath

945




945












  • $begingroup$
    Where did you get the question?
    $endgroup$
    – NicNic8
    Jan 8 at 20:07










  • $begingroup$
    In fact, I was trying to understand an example in some paper and if my question turns out to hold, then I am done.
    $endgroup$
    – Math
    Jan 8 at 20:11






  • 2




    $begingroup$
    Hint: Any two diagonal matrices commute with each other.
    $endgroup$
    – Daniel Schepler
    Jan 8 at 20:12










  • $begingroup$
    Daniel, are you saying that the result is not true?
    $endgroup$
    – Math
    Jan 8 at 20:25


















  • $begingroup$
    Where did you get the question?
    $endgroup$
    – NicNic8
    Jan 8 at 20:07










  • $begingroup$
    In fact, I was trying to understand an example in some paper and if my question turns out to hold, then I am done.
    $endgroup$
    – Math
    Jan 8 at 20:11






  • 2




    $begingroup$
    Hint: Any two diagonal matrices commute with each other.
    $endgroup$
    – Daniel Schepler
    Jan 8 at 20:12










  • $begingroup$
    Daniel, are you saying that the result is not true?
    $endgroup$
    – Math
    Jan 8 at 20:25
















$begingroup$
Where did you get the question?
$endgroup$
– NicNic8
Jan 8 at 20:07




$begingroup$
Where did you get the question?
$endgroup$
– NicNic8
Jan 8 at 20:07












$begingroup$
In fact, I was trying to understand an example in some paper and if my question turns out to hold, then I am done.
$endgroup$
– Math
Jan 8 at 20:11




$begingroup$
In fact, I was trying to understand an example in some paper and if my question turns out to hold, then I am done.
$endgroup$
– Math
Jan 8 at 20:11




2




2




$begingroup$
Hint: Any two diagonal matrices commute with each other.
$endgroup$
– Daniel Schepler
Jan 8 at 20:12




$begingroup$
Hint: Any two diagonal matrices commute with each other.
$endgroup$
– Daniel Schepler
Jan 8 at 20:12












$begingroup$
Daniel, are you saying that the result is not true?
$endgroup$
– Math
Jan 8 at 20:25




$begingroup$
Daniel, are you saying that the result is not true?
$endgroup$
– Math
Jan 8 at 20:25










1 Answer
1






active

oldest

votes


















4












$begingroup$

Even if $sigma(A)=sigma(B)$, similarity sees multiplicity. For instance let
$$
A=begin{bmatrix} 1&0&0\ 0&1&0\ 0&0&2end{bmatrix}, B=begin{bmatrix} 1&0&0\ 0&2&0\0&0&2end{bmatrix}.
$$
Then $AB=BA$, $sigma(A)=sigma(B)={1,2}$, but they are not similar (for instance, because they don't have the same trace).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great Martin! I also wanted two invertible $A$ and $B$ but this is perhaps easy to cure. Moreover, I forgot to mention that $sigma(A)=sigma(B)$ is also an assumption on my hand. Could you provide a proof now? Thanks!
    $endgroup$
    – Math
    Jan 8 at 20:32










  • $begingroup$
    Changed the answer to fit your requirements.
    $endgroup$
    – Martin Argerami
    Jan 8 at 20:40










  • $begingroup$
    Yes! and Thanks! Martin you've been of great help! Math
    $endgroup$
    – Math
    Jan 8 at 20:42











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Even if $sigma(A)=sigma(B)$, similarity sees multiplicity. For instance let
$$
A=begin{bmatrix} 1&0&0\ 0&1&0\ 0&0&2end{bmatrix}, B=begin{bmatrix} 1&0&0\ 0&2&0\0&0&2end{bmatrix}.
$$
Then $AB=BA$, $sigma(A)=sigma(B)={1,2}$, but they are not similar (for instance, because they don't have the same trace).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great Martin! I also wanted two invertible $A$ and $B$ but this is perhaps easy to cure. Moreover, I forgot to mention that $sigma(A)=sigma(B)$ is also an assumption on my hand. Could you provide a proof now? Thanks!
    $endgroup$
    – Math
    Jan 8 at 20:32










  • $begingroup$
    Changed the answer to fit your requirements.
    $endgroup$
    – Martin Argerami
    Jan 8 at 20:40










  • $begingroup$
    Yes! and Thanks! Martin you've been of great help! Math
    $endgroup$
    – Math
    Jan 8 at 20:42
















4












$begingroup$

Even if $sigma(A)=sigma(B)$, similarity sees multiplicity. For instance let
$$
A=begin{bmatrix} 1&0&0\ 0&1&0\ 0&0&2end{bmatrix}, B=begin{bmatrix} 1&0&0\ 0&2&0\0&0&2end{bmatrix}.
$$
Then $AB=BA$, $sigma(A)=sigma(B)={1,2}$, but they are not similar (for instance, because they don't have the same trace).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great Martin! I also wanted two invertible $A$ and $B$ but this is perhaps easy to cure. Moreover, I forgot to mention that $sigma(A)=sigma(B)$ is also an assumption on my hand. Could you provide a proof now? Thanks!
    $endgroup$
    – Math
    Jan 8 at 20:32










  • $begingroup$
    Changed the answer to fit your requirements.
    $endgroup$
    – Martin Argerami
    Jan 8 at 20:40










  • $begingroup$
    Yes! and Thanks! Martin you've been of great help! Math
    $endgroup$
    – Math
    Jan 8 at 20:42














4












4








4





$begingroup$

Even if $sigma(A)=sigma(B)$, similarity sees multiplicity. For instance let
$$
A=begin{bmatrix} 1&0&0\ 0&1&0\ 0&0&2end{bmatrix}, B=begin{bmatrix} 1&0&0\ 0&2&0\0&0&2end{bmatrix}.
$$
Then $AB=BA$, $sigma(A)=sigma(B)={1,2}$, but they are not similar (for instance, because they don't have the same trace).






share|cite|improve this answer











$endgroup$



Even if $sigma(A)=sigma(B)$, similarity sees multiplicity. For instance let
$$
A=begin{bmatrix} 1&0&0\ 0&1&0\ 0&0&2end{bmatrix}, B=begin{bmatrix} 1&0&0\ 0&2&0\0&0&2end{bmatrix}.
$$
Then $AB=BA$, $sigma(A)=sigma(B)={1,2}$, but they are not similar (for instance, because they don't have the same trace).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 20:40

























answered Jan 8 at 20:27









Martin ArgeramiMartin Argerami

127k1182183




127k1182183












  • $begingroup$
    Great Martin! I also wanted two invertible $A$ and $B$ but this is perhaps easy to cure. Moreover, I forgot to mention that $sigma(A)=sigma(B)$ is also an assumption on my hand. Could you provide a proof now? Thanks!
    $endgroup$
    – Math
    Jan 8 at 20:32










  • $begingroup$
    Changed the answer to fit your requirements.
    $endgroup$
    – Martin Argerami
    Jan 8 at 20:40










  • $begingroup$
    Yes! and Thanks! Martin you've been of great help! Math
    $endgroup$
    – Math
    Jan 8 at 20:42


















  • $begingroup$
    Great Martin! I also wanted two invertible $A$ and $B$ but this is perhaps easy to cure. Moreover, I forgot to mention that $sigma(A)=sigma(B)$ is also an assumption on my hand. Could you provide a proof now? Thanks!
    $endgroup$
    – Math
    Jan 8 at 20:32










  • $begingroup$
    Changed the answer to fit your requirements.
    $endgroup$
    – Martin Argerami
    Jan 8 at 20:40










  • $begingroup$
    Yes! and Thanks! Martin you've been of great help! Math
    $endgroup$
    – Math
    Jan 8 at 20:42
















$begingroup$
Great Martin! I also wanted two invertible $A$ and $B$ but this is perhaps easy to cure. Moreover, I forgot to mention that $sigma(A)=sigma(B)$ is also an assumption on my hand. Could you provide a proof now? Thanks!
$endgroup$
– Math
Jan 8 at 20:32




$begingroup$
Great Martin! I also wanted two invertible $A$ and $B$ but this is perhaps easy to cure. Moreover, I forgot to mention that $sigma(A)=sigma(B)$ is also an assumption on my hand. Could you provide a proof now? Thanks!
$endgroup$
– Math
Jan 8 at 20:32












$begingroup$
Changed the answer to fit your requirements.
$endgroup$
– Martin Argerami
Jan 8 at 20:40




$begingroup$
Changed the answer to fit your requirements.
$endgroup$
– Martin Argerami
Jan 8 at 20:40












$begingroup$
Yes! and Thanks! Martin you've been of great help! Math
$endgroup$
– Math
Jan 8 at 20:42




$begingroup$
Yes! and Thanks! Martin you've been of great help! Math
$endgroup$
– Math
Jan 8 at 20:42


















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