Mixing
What does an entire function whose images are normal matrices look like?
$begingroup$
I'd like to ask about the properties of a matrix-valued function $A:mathbb Cto M_n(mathbb C)$ where each entry of $A$ is an entire function and $A(z)$ is invertible and normal for every $zinmathbb C$.
I have difficultly to even come up with an interesting example of such a matrix-valued function. If $A$ is always Hermitian or always unitary, $A$ must be constant. Every non-constant example that I could find turned out to be uninteresting: it could be written in the form of
$$
A(z)=UD(z)U^ast,tag{1}
$$
where $U$ is some constant unitary matrix and $D(z)$ is some diagonal matrix function. This includes the cases where $A=f(z)M$ or $A=e^{f(z)K}$, in which $f$ is an entire scalar function, $M$ is a constant normal matrix and $K$ is a constant skew-Hermitian matrix.
Must $A$ be in the form of $(1)$? If not, is there any complete characterisation of $A$? Answers with the invertibility requirement removed are also welcomed.
linear-algebra matrices complex-analysis
edited Jan 8 at 21:23
amWhy
1
asked Jan 8 at 20:53
user1551user1551
72.5k566127
$endgroup$
add a comment |
$begingroup$
I'd like to ask about the properties of a matrix-valued function $A:mathbb Cto M_n(mathbb C)$ where each entry of $A$ is an entire function and $A(z)$ is invertible and normal for every $zinmathbb C$.
I have difficultly to even come up with an interesting example of such a matrix-valued function. If $A$ is always Hermitian or always unitary, $A$ must be constant. Every non-constant example that I could find turned out to be uninteresting: it could be written in the form of
$$
A(z)=UD(z)U^ast,tag{1}
$$
where $U$ is some constant unitary matrix and $D(z)$ is some diagonal matrix function. This includes the cases where $A=f(z)M$ or $A=e^{f(z)K}$, in which $f$ is an entire scalar function, $M$ is a constant normal matrix and $K$ is a constant skew-Hermitian matrix.
Must $A$ be in the form of $(1)$? If not, is there any complete characterisation of $A$? Answers with the invertibility requirement removed are also welcomed.
linear-algebra matrices complex-analysis
edited Jan 8 at 21:23
amWhy
1
asked Jan 8 at 20:53
user1551user1551
72.5k566127
$endgroup$
1
$begingroup$
Can you link to the deleted question? Users with $geq 20$K in rep can then review the original question to which you refer.
$endgroup$
– amWhy
Jan 8 at 20:59
1
$begingroup$
@amWhy Sorry, no. A user who has decided to delete his/her question may want the question not to be seen again, and I want to respect that.
$endgroup$
– user1551
Jan 8 at 21:02
1
$begingroup$
Okay, then please delete your reference to some non-existent previous post you refuse to link to, because it serves no purpose in your question, and certainly fails to provide context. If you make a claim in a post ("this question was previously deleted..." or "I tried method A, but got no results"), the expectation for any asker is to support such claims, with a link to a previous quesiton, or the inclusion of one's work after trying trying method A, or applying Theory X, whether wrong or right.
$endgroup$
– amWhy
Jan 8 at 21:23
1
$begingroup$
Entire functions are at least a little bit rigid; do you have any examples that you find interesting, where we let the elements share definition on, say, the unit disc?
$endgroup$
– Will Jagy
Jan 8 at 21:29
1
$begingroup$
@WillJagy No. This is more difficult than I thought. Owing to Cauchy-Riemann equations, Hermitian examples and unitary diagonalisation are dead ends even if $A$ is only required to be holomorphic in a neighbourhood.
$endgroup$
– user1551
Jan 8 at 21:38
add a comment |
$begingroup$
I'd like to ask about the properties of a matrix-valued function $A:mathbb Cto M_n(mathbb C)$ where each entry of $A$ is an entire function and $A(z)$ is invertible and normal for every $zinmathbb C$.
I have difficultly to even come up with an interesting example of such a matrix-valued function. If $A$ is always Hermitian or always unitary, $A$ must be constant. Every non-constant example that I could find turned out to be uninteresting: it could be written in the form of
$$
A(z)=UD(z)U^ast,tag{1}
$$
where $U$ is some constant unitary matrix and $D(z)$ is some diagonal matrix function. This includes the cases where $A=f(z)M$ or $A=e^{f(z)K}$, in which $f$ is an entire scalar function, $M$ is a constant normal matrix and $K$ is a constant skew-Hermitian matrix.
Must $A$ be in the form of $(1)$? If not, is there any complete characterisation of $A$? Answers with the invertibility requirement removed are also welcomed.
linear-algebra matrices complex-analysis
edited Jan 8 at 21:23
amWhy
1
asked Jan 8 at 20:53
user1551user1551
72.5k566127
$endgroup$
I'd like to ask about the properties of a matrix-valued function $A:mathbb Cto M_n(mathbb C)$ where each entry of $A$ is an entire function and $A(z)$ is invertible and normal for every $zinmathbb C$.
I have difficultly to even come up with an interesting example of such a matrix-valued function. If $A$ is always Hermitian or always unitary, $A$ must be constant. Every non-constant example that I could find turned out to be uninteresting: it could be written in the form of
$$
A(z)=UD(z)U^ast,tag{1}
$$
where $U$ is some constant unitary matrix and $D(z)$ is some diagonal matrix function. This includes the cases where $A=f(z)M$ or $A=e^{f(z)K}$, in which $f$ is an entire scalar function, $M$ is a constant normal matrix and $K$ is a constant skew-Hermitian matrix.
Must $A$ be in the form of $(1)$? If not, is there any complete characterisation of $A$? Answers with the invertibility requirement removed are also welcomed.
linear-algebra matrices complex-analysis
linear-algebra matrices complex-analysis
edited Jan 8 at 21:23
amWhy
1
asked Jan 8 at 20:53
user1551user1551
72.5k566127
edited Jan 8 at 21:23
amWhy
1
asked Jan 8 at 20:53
user1551user1551
72.5k566127
edited Jan 8 at 21:23
amWhy
1
edited Jan 8 at 21:23
amWhy
1
edited Jan 8 at 21:23
amWhy
1
1
asked Jan 8 at 20:53
user1551user1551
72.5k566127
asked Jan 8 at 20:53
user1551user1551
72.5k566127
asked Jan 8 at 20:53
user1551user1551
72.5k566127
72.5k566127
1
$begingroup$
Can you link to the deleted question? Users with $geq 20$K in rep can then review the original question to which you refer.
$endgroup$
– amWhy
Jan 8 at 20:59
1
$begingroup$
@amWhy Sorry, no. A user who has decided to delete his/her question may want the question not to be seen again, and I want to respect that.
$endgroup$
– user1551
Jan 8 at 21:02
1
$begingroup$
Okay, then please delete your reference to some non-existent previous post you refuse to link to, because it serves no purpose in your question, and certainly fails to provide context. If you make a claim in a post ("this question was previously deleted..." or "I tried method A, but got no results"), the expectation for any asker is to support such claims, with a link to a previous quesiton, or the inclusion of one's work after trying trying method A, or applying Theory X, whether wrong or right.
$endgroup$
– amWhy
Jan 8 at 21:23
1
$begingroup$
Entire functions are at least a little bit rigid; do you have any examples that you find interesting, where we let the elements share definition on, say, the unit disc?
$endgroup$
– Will Jagy
Jan 8 at 21:29
1
$begingroup$
@WillJagy No. This is more difficult than I thought. Owing to Cauchy-Riemann equations, Hermitian examples and unitary diagonalisation are dead ends even if $A$ is only required to be holomorphic in a neighbourhood.
$endgroup$
– user1551
Jan 8 at 21:38
add a comment |
1
$begingroup$
Can you link to the deleted question? Users with $geq 20$K in rep can then review the original question to which you refer.
$endgroup$
– amWhy
Jan 8 at 20:59
1
$begingroup$
@amWhy Sorry, no. A user who has decided to delete his/her question may want the question not to be seen again, and I want to respect that.
$endgroup$
– user1551
Jan 8 at 21:02
1
$begingroup$
Okay, then please delete your reference to some non-existent previous post you refuse to link to, because it serves no purpose in your question, and certainly fails to provide context. If you make a claim in a post ("this question was previously deleted..." or "I tried method A, but got no results"), the expectation for any asker is to support such claims, with a link to a previous quesiton, or the inclusion of one's work after trying trying method A, or applying Theory X, whether wrong or right.
$endgroup$
– amWhy
Jan 8 at 21:23
1
$begingroup$
Entire functions are at least a little bit rigid; do you have any examples that you find interesting, where we let the elements share definition on, say, the unit disc?
$endgroup$
– Will Jagy
Jan 8 at 21:29
1
$begingroup$
@WillJagy No. This is more difficult than I thought. Owing to Cauchy-Riemann equations, Hermitian examples and unitary diagonalisation are dead ends even if $A$ is only required to be holomorphic in a neighbourhood.
$endgroup$
– user1551
Jan 8 at 21:38
1
1
$begingroup$
Can you link to the deleted question? Users with $geq 20$K in rep can then review the original question to which you refer.
$endgroup$
– amWhy
Jan 8 at 20:59
$begingroup$
Can you link to the deleted question? Users with $geq 20$K in rep can then review the original question to which you refer.
$endgroup$
– amWhy
Jan 8 at 20:59
1
1
$begingroup$
@amWhy Sorry, no. A user who has decided to delete his/her question may want the question not to be seen again, and I want to respect that.
$endgroup$
– user1551
Jan 8 at 21:02
$begingroup$
@amWhy Sorry, no. A user who has decided to delete his/her question may want the question not to be seen again, and I want to respect that.
$endgroup$
– user1551
Jan 8 at 21:02
1
1
$begingroup$
Okay, then please delete your reference to some non-existent previous post you refuse to link to, because it serves no purpose in your question, and certainly fails to provide context. If you make a claim in a post ("this question was previously deleted..." or "I tried method A, but got no results"), the expectation for any asker is to support such claims, with a link to a previous quesiton, or the inclusion of one's work after trying trying method A, or applying Theory X, whether wrong or right.
$endgroup$
– amWhy
Jan 8 at 21:23
$begingroup$
Okay, then please delete your reference to some non-existent previous post you refuse to link to, because it serves no purpose in your question, and certainly fails to provide context. If you make a claim in a post ("this question was previously deleted..." or "I tried method A, but got no results"), the expectation for any asker is to support such claims, with a link to a previous quesiton, or the inclusion of one's work after trying trying method A, or applying Theory X, whether wrong or right.
$endgroup$
– amWhy
Jan 8 at 21:23
1
1
$begingroup$
Entire functions are at least a little bit rigid; do you have any examples that you find interesting, where we let the elements share definition on, say, the unit disc?
$endgroup$
– Will Jagy
Jan 8 at 21:29
$begingroup$
Entire functions are at least a little bit rigid; do you have any examples that you find interesting, where we let the elements share definition on, say, the unit disc?
$endgroup$
– Will Jagy
Jan 8 at 21:29
1
1
$begingroup$
@WillJagy No. This is more difficult than I thought. Owing to Cauchy-Riemann equations, Hermitian examples and unitary diagonalisation are dead ends even if $A$ is only required to be holomorphic in a neighbourhood.
$endgroup$
– user1551
Jan 8 at 21:38
$begingroup$
@WillJagy No. This is more difficult than I thought. Owing to Cauchy-Riemann equations, Hermitian examples and unitary diagonalisation are dead ends even if $A$ is only required to be holomorphic in a neighbourhood.
$endgroup$
– user1551
Jan 8 at 21:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When $tinmathbb{R}rightarrow A(t)$ is real analytic and $A(t)$ is normal, then, LOCALLY, there are analytic parametrizations of the eigenvalues and the eigenvectors of $A(t)$.
cf. https://www.mat.univie.ac.at/~michor/DC-perturb.pdf
I think that a similar result stands (but we have to show it!) for $zinmathbb{C}rightarrow A(z)$ complex holomorphic and $A(z)$ complex normal (it's not the same situation as for the application $(u,v)inmathbb{R}^2rightarrow f(u,v)+ig(u,v)$, where $f,g$ are analytic real, because, here, there are relations between the derivatives of $f,g$).
If it's true, locally, one has $A(z)=U(z)D(z)U^*(z)$ where $U(z)$ is holomorphic unitary and $D(z)=diag(f_i(z))$ where $f_i$ is holomorphic; since the columns vectors of $U$ have a constant length, these vectors are constant on a connected open subset. Finally $A(z)=UD(z)U^*$ where $U$ is constant unitary.
answered Jan 9 at 16:25
loup blancloup blanc
23k21850
$endgroup$
$begingroup$
Thanks, but isn't "we have to show it"... the hard part? ;-D The trouble with a complex holomorphic $A(z)$ is that it seems hard to make inferences about the eigenvectors. E.g. in the paper you linked to, property (O) suggests that the eigenvalues are at best Lipchitz. Nothing about the eigenvectors are mentioned. Even if the eigenvectors can be chosen to be Lipschitz locally, it seems a great leap to infer that $A(z)=UD(z)U^ast$ from $A(z)=U(z)D(z)U^ast(z)$.
$endgroup$
– user1551
Jan 9 at 22:58
$begingroup$
It's better to look at (A) and (L). For $1$ real variable, an analytic parametrization exists; when there is more than one real variable, there is a finite analytic covering associated to the eigenvectors. Clearly, we have to show that we can avoid the second case for an holomorphic function. Anyway, the implicit function theorem exists for holomorphic functions. When $n=2$, it isn't difficult to see that the required solutions are in the form $f(z)I_2+g(z)N$ where $N$ is a fixed normal matrix.
$endgroup$
– loup blanc
Jan 10 at 9:28
add a comment |
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$begingroup$
When $tinmathbb{R}rightarrow A(t)$ is real analytic and $A(t)$ is normal, then, LOCALLY, there are analytic parametrizations of the eigenvalues and the eigenvectors of $A(t)$.
cf. https://www.mat.univie.ac.at/~michor/DC-perturb.pdf
I think that a similar result stands (but we have to show it!) for $zinmathbb{C}rightarrow A(z)$ complex holomorphic and $A(z)$ complex normal (it's not the same situation as for the application $(u,v)inmathbb{R}^2rightarrow f(u,v)+ig(u,v)$, where $f,g$ are analytic real, because, here, there are relations between the derivatives of $f,g$).
If it's true, locally, one has $A(z)=U(z)D(z)U^*(z)$ where $U(z)$ is holomorphic unitary and $D(z)=diag(f_i(z))$ where $f_i$ is holomorphic; since the columns vectors of $U$ have a constant length, these vectors are constant on a connected open subset. Finally $A(z)=UD(z)U^*$ where $U$ is constant unitary.
answered Jan 9 at 16:25
loup blancloup blanc
23k21850
$endgroup$
$begingroup$
Thanks, but isn't "we have to show it"... the hard part? ;-D The trouble with a complex holomorphic $A(z)$ is that it seems hard to make inferences about the eigenvectors. E.g. in the paper you linked to, property (O) suggests that the eigenvalues are at best Lipchitz. Nothing about the eigenvectors are mentioned. Even if the eigenvectors can be chosen to be Lipschitz locally, it seems a great leap to infer that $A(z)=UD(z)U^ast$ from $A(z)=U(z)D(z)U^ast(z)$.
$endgroup$
– user1551
Jan 9 at 22:58
$begingroup$
It's better to look at (A) and (L). For $1$ real variable, an analytic parametrization exists; when there is more than one real variable, there is a finite analytic covering associated to the eigenvectors. Clearly, we have to show that we can avoid the second case for an holomorphic function. Anyway, the implicit function theorem exists for holomorphic functions. When $n=2$, it isn't difficult to see that the required solutions are in the form $f(z)I_2+g(z)N$ where $N$ is a fixed normal matrix.
$endgroup$
– loup blanc
Jan 10 at 9:28
add a comment |
$begingroup$
When $tinmathbb{R}rightarrow A(t)$ is real analytic and $A(t)$ is normal, then, LOCALLY, there are analytic parametrizations of the eigenvalues and the eigenvectors of $A(t)$.
cf. https://www.mat.univie.ac.at/~michor/DC-perturb.pdf
I think that a similar result stands (but we have to show it!) for $zinmathbb{C}rightarrow A(z)$ complex holomorphic and $A(z)$ complex normal (it's not the same situation as for the application $(u,v)inmathbb{R}^2rightarrow f(u,v)+ig(u,v)$, where $f,g$ are analytic real, because, here, there are relations between the derivatives of $f,g$).
If it's true, locally, one has $A(z)=U(z)D(z)U^*(z)$ where $U(z)$ is holomorphic unitary and $D(z)=diag(f_i(z))$ where $f_i$ is holomorphic; since the columns vectors of $U$ have a constant length, these vectors are constant on a connected open subset. Finally $A(z)=UD(z)U^*$ where $U$ is constant unitary.
answered Jan 9 at 16:25
loup blancloup blanc
23k21850
$endgroup$
$begingroup$
Thanks, but isn't "we have to show it"... the hard part? ;-D The trouble with a complex holomorphic $A(z)$ is that it seems hard to make inferences about the eigenvectors. E.g. in the paper you linked to, property (O) suggests that the eigenvalues are at best Lipchitz. Nothing about the eigenvectors are mentioned. Even if the eigenvectors can be chosen to be Lipschitz locally, it seems a great leap to infer that $A(z)=UD(z)U^ast$ from $A(z)=U(z)D(z)U^ast(z)$.
$endgroup$
– user1551
Jan 9 at 22:58
$begingroup$
It's better to look at (A) and (L). For $1$ real variable, an analytic parametrization exists; when there is more than one real variable, there is a finite analytic covering associated to the eigenvectors. Clearly, we have to show that we can avoid the second case for an holomorphic function. Anyway, the implicit function theorem exists for holomorphic functions. When $n=2$, it isn't difficult to see that the required solutions are in the form $f(z)I_2+g(z)N$ where $N$ is a fixed normal matrix.
$endgroup$
– loup blanc
Jan 10 at 9:28
add a comment |
$begingroup$
When $tinmathbb{R}rightarrow A(t)$ is real analytic and $A(t)$ is normal, then, LOCALLY, there are analytic parametrizations of the eigenvalues and the eigenvectors of $A(t)$.
cf. https://www.mat.univie.ac.at/~michor/DC-perturb.pdf
I think that a similar result stands (but we have to show it!) for $zinmathbb{C}rightarrow A(z)$ complex holomorphic and $A(z)$ complex normal (it's not the same situation as for the application $(u,v)inmathbb{R}^2rightarrow f(u,v)+ig(u,v)$, where $f,g$ are analytic real, because, here, there are relations between the derivatives of $f,g$).
If it's true, locally, one has $A(z)=U(z)D(z)U^*(z)$ where $U(z)$ is holomorphic unitary and $D(z)=diag(f_i(z))$ where $f_i$ is holomorphic; since the columns vectors of $U$ have a constant length, these vectors are constant on a connected open subset. Finally $A(z)=UD(z)U^*$ where $U$ is constant unitary.
answered Jan 9 at 16:25
loup blancloup blanc
23k21850
$endgroup$
When $tinmathbb{R}rightarrow A(t)$ is real analytic and $A(t)$ is normal, then, LOCALLY, there are analytic parametrizations of the eigenvalues and the eigenvectors of $A(t)$.
cf. https://www.mat.univie.ac.at/~michor/DC-perturb.pdf
I think that a similar result stands (but we have to show it!) for $zinmathbb{C}rightarrow A(z)$ complex holomorphic and $A(z)$ complex normal (it's not the same situation as for the application $(u,v)inmathbb{R}^2rightarrow f(u,v)+ig(u,v)$, where $f,g$ are analytic real, because, here, there are relations between the derivatives of $f,g$).
If it's true, locally, one has $A(z)=U(z)D(z)U^*(z)$ where $U(z)$ is holomorphic unitary and $D(z)=diag(f_i(z))$ where $f_i$ is holomorphic; since the columns vectors of $U$ have a constant length, these vectors are constant on a connected open subset. Finally $A(z)=UD(z)U^*$ where $U$ is constant unitary.
answered Jan 9 at 16:25
loup blancloup blanc
23k21850
answered Jan 9 at 16:25
loup blancloup blanc
23k21850
answered Jan 9 at 16:25
loup blancloup blanc
23k21850
answered Jan 9 at 16:25
loup blancloup blanc
23k21850
23k21850
$begingroup$
Thanks, but isn't "we have to show it"... the hard part? ;-D The trouble with a complex holomorphic $A(z)$ is that it seems hard to make inferences about the eigenvectors. E.g. in the paper you linked to, property (O) suggests that the eigenvalues are at best Lipchitz. Nothing about the eigenvectors are mentioned. Even if the eigenvectors can be chosen to be Lipschitz locally, it seems a great leap to infer that $A(z)=UD(z)U^ast$ from $A(z)=U(z)D(z)U^ast(z)$.
$endgroup$
– user1551
Jan 9 at 22:58
$begingroup$
It's better to look at (A) and (L). For $1$ real variable, an analytic parametrization exists; when there is more than one real variable, there is a finite analytic covering associated to the eigenvectors. Clearly, we have to show that we can avoid the second case for an holomorphic function. Anyway, the implicit function theorem exists for holomorphic functions. When $n=2$, it isn't difficult to see that the required solutions are in the form $f(z)I_2+g(z)N$ where $N$ is a fixed normal matrix.
$endgroup$
– loup blanc
Jan 10 at 9:28
add a comment |
$begingroup$
Thanks, but isn't "we have to show it"... the hard part? ;-D The trouble with a complex holomorphic $A(z)$ is that it seems hard to make inferences about the eigenvectors. E.g. in the paper you linked to, property (O) suggests that the eigenvalues are at best Lipchitz. Nothing about the eigenvectors are mentioned. Even if the eigenvectors can be chosen to be Lipschitz locally, it seems a great leap to infer that $A(z)=UD(z)U^ast$ from $A(z)=U(z)D(z)U^ast(z)$.
$endgroup$
– user1551
Jan 9 at 22:58
$begingroup$
It's better to look at (A) and (L). For $1$ real variable, an analytic parametrization exists; when there is more than one real variable, there is a finite analytic covering associated to the eigenvectors. Clearly, we have to show that we can avoid the second case for an holomorphic function. Anyway, the implicit function theorem exists for holomorphic functions. When $n=2$, it isn't difficult to see that the required solutions are in the form $f(z)I_2+g(z)N$ where $N$ is a fixed normal matrix.
$endgroup$
– loup blanc
Jan 10 at 9:28
$begingroup$
Thanks, but isn't "we have to show it"... the hard part? ;-D The trouble with a complex holomorphic $A(z)$ is that it seems hard to make inferences about the eigenvectors. E.g. in the paper you linked to, property (O) suggests that the eigenvalues are at best Lipchitz. Nothing about the eigenvectors are mentioned. Even if the eigenvectors can be chosen to be Lipschitz locally, it seems a great leap to infer that $A(z)=UD(z)U^ast$ from $A(z)=U(z)D(z)U^ast(z)$.
$endgroup$
– user1551
Jan 9 at 22:58
$begingroup$
Thanks, but isn't "we have to show it"... the hard part? ;-D The trouble with a complex holomorphic $A(z)$ is that it seems hard to make inferences about the eigenvectors. E.g. in the paper you linked to, property (O) suggests that the eigenvalues are at best Lipchitz. Nothing about the eigenvectors are mentioned. Even if the eigenvectors can be chosen to be Lipschitz locally, it seems a great leap to infer that $A(z)=UD(z)U^ast$ from $A(z)=U(z)D(z)U^ast(z)$.
$endgroup$
– user1551
Jan 9 at 22:58
$begingroup$
It's better to look at (A) and (L). For $1$ real variable, an analytic parametrization exists; when there is more than one real variable, there is a finite analytic covering associated to the eigenvectors. Clearly, we have to show that we can avoid the second case for an holomorphic function. Anyway, the implicit function theorem exists for holomorphic functions. When $n=2$, it isn't difficult to see that the required solutions are in the form $f(z)I_2+g(z)N$ where $N$ is a fixed normal matrix.
$endgroup$
– loup blanc
Jan 10 at 9:28
$begingroup$
It's better to look at (A) and (L). For $1$ real variable, an analytic parametrization exists; when there is more than one real variable, there is a finite analytic covering associated to the eigenvectors. Clearly, we have to show that we can avoid the second case for an holomorphic function. Anyway, the implicit function theorem exists for holomorphic functions. When $n=2$, it isn't difficult to see that the required solutions are in the form $f(z)I_2+g(z)N$ where $N$ is a fixed normal matrix.
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– loup blanc
Jan 10 at 9:28
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1
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Can you link to the deleted question? Users with $geq 20$K in rep can then review the original question to which you refer.
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– amWhy
Jan 8 at 20:59
1
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@amWhy Sorry, no. A user who has decided to delete his/her question may want the question not to be seen again, and I want to respect that.
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– user1551
Jan 8 at 21:02
1
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Okay, then please delete your reference to some non-existent previous post you refuse to link to, because it serves no purpose in your question, and certainly fails to provide context. If you make a claim in a post ("this question was previously deleted..." or "I tried method A, but got no results"), the expectation for any asker is to support such claims, with a link to a previous quesiton, or the inclusion of one's work after trying trying method A, or applying Theory X, whether wrong or right.
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– amWhy
Jan 8 at 21:23
1
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Entire functions are at least a little bit rigid; do you have any examples that you find interesting, where we let the elements share definition on, say, the unit disc?
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– Will Jagy
Jan 8 at 21:29
1
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@WillJagy No. This is more difficult than I thought. Owing to Cauchy-Riemann equations, Hermitian examples and unitary diagonalisation are dead ends even if $A$ is only required to be holomorphic in a neighbourhood.
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– user1551
Jan 8 at 21:38