Cayley-Transformation Example
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I searching for an example to the Cayley-Transformation. I know that I need an symmetric Operator and the self-adjoint Extension. I try it with the Derivation Operator but I don't think its correct.
It would be nice if someone can give me an example and how they use the Cayley-Transformation, when I have some self-adjoint Extension. My definition of CT is:
$$V=(i-T)(-i-T)^{-1},$$
and $T$ is symmetric Operator.
Mfg Neuling.
complex-analysis functional-analysis
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Neuling is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
favorite
I searching for an example to the Cayley-Transformation. I know that I need an symmetric Operator and the self-adjoint Extension. I try it with the Derivation Operator but I don't think its correct.
It would be nice if someone can give me an example and how they use the Cayley-Transformation, when I have some self-adjoint Extension. My definition of CT is:
$$V=(i-T)(-i-T)^{-1},$$
and $T$ is symmetric Operator.
Mfg Neuling.
complex-analysis functional-analysis
New contributor
Neuling is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 17 at 10:15
If $T$ is symmetric, not selfajdoint, and has a selfadjoint extension, then $(iIpm T)$ have ranges that are not full. So $V$ won't be defined on the full space. And you can write your $V$ as $(-i-T+2i)(-i-T)^{-1}=I+2i(-i-T)^{-1}$, which won't be defined on the full space.
– DisintegratingByParts
Nov 17 at 14:37
add a comment |
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up vote
0
down vote
favorite
I searching for an example to the Cayley-Transformation. I know that I need an symmetric Operator and the self-adjoint Extension. I try it with the Derivation Operator but I don't think its correct.
It would be nice if someone can give me an example and how they use the Cayley-Transformation, when I have some self-adjoint Extension. My definition of CT is:
$$V=(i-T)(-i-T)^{-1},$$
and $T$ is symmetric Operator.
Mfg Neuling.
complex-analysis functional-analysis
New contributor
Neuling is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I searching for an example to the Cayley-Transformation. I know that I need an symmetric Operator and the self-adjoint Extension. I try it with the Derivation Operator but I don't think its correct.
It would be nice if someone can give me an example and how they use the Cayley-Transformation, when I have some self-adjoint Extension. My definition of CT is:
$$V=(i-T)(-i-T)^{-1},$$
and $T$ is symmetric Operator.
Mfg Neuling.
complex-analysis functional-analysis
complex-analysis functional-analysis
New contributor
Neuling is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Neuling is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 17 at 13:40


Nosrati
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26k62252
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asked Nov 17 at 10:10
Neuling
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Neuling is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Neuling is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Neuling is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 17 at 10:15
If $T$ is symmetric, not selfajdoint, and has a selfadjoint extension, then $(iIpm T)$ have ranges that are not full. So $V$ won't be defined on the full space. And you can write your $V$ as $(-i-T+2i)(-i-T)^{-1}=I+2i(-i-T)^{-1}$, which won't be defined on the full space.
– DisintegratingByParts
Nov 17 at 14:37
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 17 at 10:15
If $T$ is symmetric, not selfajdoint, and has a selfadjoint extension, then $(iIpm T)$ have ranges that are not full. So $V$ won't be defined on the full space. And you can write your $V$ as $(-i-T+2i)(-i-T)^{-1}=I+2i(-i-T)^{-1}$, which won't be defined on the full space.
– DisintegratingByParts
Nov 17 at 14:37
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 17 at 10:15
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 17 at 10:15
If $T$ is symmetric, not selfajdoint, and has a selfadjoint extension, then $(iIpm T)$ have ranges that are not full. So $V$ won't be defined on the full space. And you can write your $V$ as $(-i-T+2i)(-i-T)^{-1}=I+2i(-i-T)^{-1}$, which won't be defined on the full space.
– DisintegratingByParts
Nov 17 at 14:37
If $T$ is symmetric, not selfajdoint, and has a selfadjoint extension, then $(iIpm T)$ have ranges that are not full. So $V$ won't be defined on the full space. And you can write your $V$ as $(-i-T+2i)(-i-T)^{-1}=I+2i(-i-T)^{-1}$, which won't be defined on the full space.
– DisintegratingByParts
Nov 17 at 14:37
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 17 at 10:15
If $T$ is symmetric, not selfajdoint, and has a selfadjoint extension, then $(iIpm T)$ have ranges that are not full. So $V$ won't be defined on the full space. And you can write your $V$ as $(-i-T+2i)(-i-T)^{-1}=I+2i(-i-T)^{-1}$, which won't be defined on the full space.
– DisintegratingByParts
Nov 17 at 14:37