Criteria to find a common non orthonormal basis for two linear operators
I can't find any criteria to determine, in finite dimension, if two operators has a non orthogonal common basis, for example, given two operators A and B if I check
- $A=A^+$
- $B=B^+$
- $[A, B] =0$
In this situation I can affirm that A and B have a common orthonormal basis of eigenvector, or if the 1 and 2 properties are not true for A and B but the third one is, I can say that A and B have only a common eigenvector.
But what properties should I check to know if two operators have a common non orthonormal basis? Can I ask this question to myself or it is incorrect itself?
Often on old tests I find "determine if this two operator have a common eigenvector basis. What kind of basis is this and why?" but I only find on my book criteria to find the orthonormal one.
Thank you and sorry for bad English.
eigenvalues-eigenvectors operator-theory orthonormal normal-operator
add a comment |
I can't find any criteria to determine, in finite dimension, if two operators has a non orthogonal common basis, for example, given two operators A and B if I check
- $A=A^+$
- $B=B^+$
- $[A, B] =0$
In this situation I can affirm that A and B have a common orthonormal basis of eigenvector, or if the 1 and 2 properties are not true for A and B but the third one is, I can say that A and B have only a common eigenvector.
But what properties should I check to know if two operators have a common non orthonormal basis? Can I ask this question to myself or it is incorrect itself?
Often on old tests I find "determine if this two operator have a common eigenvector basis. What kind of basis is this and why?" but I only find on my book criteria to find the orthonormal one.
Thank you and sorry for bad English.
eigenvalues-eigenvectors operator-theory orthonormal normal-operator
If they have a common basis of eigenvectors, then they satisfy 3. (it is enough to check this on the basis of common eigenvectors and there it is easy).
– Severin Schraven
Jan 4 at 16:55
For the other direction you might want to check this out mathoverflow.net/questions/124779/…
– Severin Schraven
Jan 4 at 16:58
add a comment |
I can't find any criteria to determine, in finite dimension, if two operators has a non orthogonal common basis, for example, given two operators A and B if I check
- $A=A^+$
- $B=B^+$
- $[A, B] =0$
In this situation I can affirm that A and B have a common orthonormal basis of eigenvector, or if the 1 and 2 properties are not true for A and B but the third one is, I can say that A and B have only a common eigenvector.
But what properties should I check to know if two operators have a common non orthonormal basis? Can I ask this question to myself or it is incorrect itself?
Often on old tests I find "determine if this two operator have a common eigenvector basis. What kind of basis is this and why?" but I only find on my book criteria to find the orthonormal one.
Thank you and sorry for bad English.
eigenvalues-eigenvectors operator-theory orthonormal normal-operator
I can't find any criteria to determine, in finite dimension, if two operators has a non orthogonal common basis, for example, given two operators A and B if I check
- $A=A^+$
- $B=B^+$
- $[A, B] =0$
In this situation I can affirm that A and B have a common orthonormal basis of eigenvector, or if the 1 and 2 properties are not true for A and B but the third one is, I can say that A and B have only a common eigenvector.
But what properties should I check to know if two operators have a common non orthonormal basis? Can I ask this question to myself or it is incorrect itself?
Often on old tests I find "determine if this two operator have a common eigenvector basis. What kind of basis is this and why?" but I only find on my book criteria to find the orthonormal one.
Thank you and sorry for bad English.
eigenvalues-eigenvectors operator-theory orthonormal normal-operator
eigenvalues-eigenvectors operator-theory orthonormal normal-operator
edited Jan 4 at 17:09
pter26
asked Jan 4 at 16:40
pter26pter26
317111
317111
If they have a common basis of eigenvectors, then they satisfy 3. (it is enough to check this on the basis of common eigenvectors and there it is easy).
– Severin Schraven
Jan 4 at 16:55
For the other direction you might want to check this out mathoverflow.net/questions/124779/…
– Severin Schraven
Jan 4 at 16:58
add a comment |
If they have a common basis of eigenvectors, then they satisfy 3. (it is enough to check this on the basis of common eigenvectors and there it is easy).
– Severin Schraven
Jan 4 at 16:55
For the other direction you might want to check this out mathoverflow.net/questions/124779/…
– Severin Schraven
Jan 4 at 16:58
If they have a common basis of eigenvectors, then they satisfy 3. (it is enough to check this on the basis of common eigenvectors and there it is easy).
– Severin Schraven
Jan 4 at 16:55
If they have a common basis of eigenvectors, then they satisfy 3. (it is enough to check this on the basis of common eigenvectors and there it is easy).
– Severin Schraven
Jan 4 at 16:55
For the other direction you might want to check this out mathoverflow.net/questions/124779/…
– Severin Schraven
Jan 4 at 16:58
For the other direction you might want to check this out mathoverflow.net/questions/124779/…
– Severin Schraven
Jan 4 at 16:58
add a comment |
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If they have a common basis of eigenvectors, then they satisfy 3. (it is enough to check this on the basis of common eigenvectors and there it is easy).
– Severin Schraven
Jan 4 at 16:55
For the other direction you might want to check this out mathoverflow.net/questions/124779/…
– Severin Schraven
Jan 4 at 16:58