Similarity of two self-adjoint operators
$begingroup$
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
edited Jan 8 at 20:43
Aweygan
14.3k21441
asked Jan 8 at 19:56
MathMath
945
$endgroup$
add a comment |
$begingroup$
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
edited Jan 8 at 20:43
Aweygan
14.3k21441
asked Jan 8 at 19:56
MathMath
945
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 8 at 20:43
Aweygan
14.3k21441
asked Jan 8 at 19:56
MathMath
945
$endgroup$
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
edited Jan 8 at 20:43
Aweygan
14.3k21441
asked Jan 8 at 19:56
MathMath
945
edited Jan 8 at 20:43
Aweygan
14.3k21441
asked Jan 8 at 19:56
MathMath
945
edited Jan 8 at 20:43
Aweygan
14.3k21441
edited Jan 8 at 20:43
Aweygan
14.3k21441
edited Jan 8 at 20:43
Aweygan
14.3k21441
14.3k21441
asked Jan 8 at 19:56
MathMath
945
asked Jan 8 at 19:56
MathMath
945
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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).
answered Jan 8 at 20:27
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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).
answered Jan 8 at 20:27
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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).
answered Jan 8 at 20:27
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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).
answered Jan 8 at 20:27
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
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).
answered Jan 8 at 20:27
Martin ArgeramiMartin Argerami
127k1182183
edited Jan 8 at 20:40
answered Jan 8 at 20:27
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 8 at 20:27
Martin ArgeramiMartin Argerami
127k1182183
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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