Dilation of an operator
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Let $Tin B(mathcal{H})$. Based on https://en.wikipedia.org/wiki/Dilation_(operator_theory)
dilation of the operator $T$ is an operator on a larger Hilbert space $K$, whose restriction to $mathcal{H}$ composed with the orthogonal projection $P$ onto $mathcal{H}$ is $T$.
I have several questions about this definition. Is there another version of this definition? Is the operator $P$ always projection? I mean could one find a larger Hilbert space $K$ and operators $V, U$ on $K$, when $U$ is not an orthogonal projection, such that $UV|_{mathcal{H}}= T$?
functional-analysis operator-theory
asked Nov 28 at 22:20
saeed
17510
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up vote
2
down vote
favorite
Let $Tin B(mathcal{H})$. Based on https://en.wikipedia.org/wiki/Dilation_(operator_theory)
dilation of the operator $T$ is an operator on a larger Hilbert space $K$, whose restriction to $mathcal{H}$ composed with the orthogonal projection $P$ onto $mathcal{H}$ is $T$.
I have several questions about this definition. Is there another version of this definition? Is the operator $P$ always projection? I mean could one find a larger Hilbert space $K$ and operators $V, U$ on $K$, when $U$ is not an orthogonal projection, such that $UV|_{mathcal{H}}= T$?
functional-analysis operator-theory
asked Nov 28 at 22:20
saeed
17510
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $Tin B(mathcal{H})$. Based on https://en.wikipedia.org/wiki/Dilation_(operator_theory)
dilation of the operator $T$ is an operator on a larger Hilbert space $K$, whose restriction to $mathcal{H}$ composed with the orthogonal projection $P$ onto $mathcal{H}$ is $T$.
I have several questions about this definition. Is there another version of this definition? Is the operator $P$ always projection? I mean could one find a larger Hilbert space $K$ and operators $V, U$ on $K$, when $U$ is not an orthogonal projection, such that $UV|_{mathcal{H}}= T$?
functional-analysis operator-theory
asked Nov 28 at 22:20
saeed
17510
Let $Tin B(mathcal{H})$. Based on https://en.wikipedia.org/wiki/Dilation_(operator_theory)
dilation of the operator $T$ is an operator on a larger Hilbert space $K$, whose restriction to $mathcal{H}$ composed with the orthogonal projection $P$ onto $mathcal{H}$ is $T$.
I have several questions about this definition. Is there another version of this definition? Is the operator $P$ always projection? I mean could one find a larger Hilbert space $K$ and operators $V, U$ on $K$, when $U$ is not an orthogonal projection, such that $UV|_{mathcal{H}}= T$?
functional-analysis operator-theory
functional-analysis operator-theory
asked Nov 28 at 22:20
saeed
17510
asked Nov 28 at 22:20
saeed
17510
asked Nov 28 at 22:20
saeed
17510
asked Nov 28 at 22:20
saeed
17510
asked Nov 28 at 22:20
saeed
17510
17510
add a comment |
add a comment |
2 Answers
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oldest
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up vote
2
down vote
accepted
Your candidate for a definition is exactly the same definition of dilation you quoted. If $UV|_H=T$, then since $PT=T$ you have $PUV|_H=T$, and so $UV$ is a dilation of $T$ in the sense of your first paragraph.
A "dilation" has to be a "dilation". The spirit is that you have $T$ (typically, $T$ is a contraction), and you get a dilation $$tilde T=begin{bmatrix} T&A\ B& Cend{bmatrix}$$ that is better behaved, say $tilde T$ is a unitary, or a projection.
answered Nov 28 at 23:32
Martin Argerami
123k1176174
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up vote
1
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Another version:
Theorem: Let $A$ be a bounded linear operator on a Hilbert space $mathcal{H}$ with $|A|le 1$. There there exists a Hilbert space $mathcal{K}$, an isometry $V : mathcal{H}rightarrowmathcal{K}$, and a unitary operator $U : mathcal{K}rightarrowmathcal{K}$ such that
$$
A^n = V^*U^n V mbox{ for all } n ge 0.
$$
answered Nov 29 at 5:22
DisintegratingByParts
58.3k42579
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your candidate for a definition is exactly the same definition of dilation you quoted. If $UV|_H=T$, then since $PT=T$ you have $PUV|_H=T$, and so $UV$ is a dilation of $T$ in the sense of your first paragraph.
A "dilation" has to be a "dilation". The spirit is that you have $T$ (typically, $T$ is a contraction), and you get a dilation $$tilde T=begin{bmatrix} T&A\ B& Cend{bmatrix}$$ that is better behaved, say $tilde T$ is a unitary, or a projection.
answered Nov 28 at 23:32
Martin Argerami
123k1176174
add a comment |
up vote
2
down vote
accepted
Your candidate for a definition is exactly the same definition of dilation you quoted. If $UV|_H=T$, then since $PT=T$ you have $PUV|_H=T$, and so $UV$ is a dilation of $T$ in the sense of your first paragraph.
A "dilation" has to be a "dilation". The spirit is that you have $T$ (typically, $T$ is a contraction), and you get a dilation $$tilde T=begin{bmatrix} T&A\ B& Cend{bmatrix}$$ that is better behaved, say $tilde T$ is a unitary, or a projection.
answered Nov 28 at 23:32
Martin Argerami
123k1176174
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your candidate for a definition is exactly the same definition of dilation you quoted. If $UV|_H=T$, then since $PT=T$ you have $PUV|_H=T$, and so $UV$ is a dilation of $T$ in the sense of your first paragraph.
A "dilation" has to be a "dilation". The spirit is that you have $T$ (typically, $T$ is a contraction), and you get a dilation $$tilde T=begin{bmatrix} T&A\ B& Cend{bmatrix}$$ that is better behaved, say $tilde T$ is a unitary, or a projection.
answered Nov 28 at 23:32
Martin Argerami
123k1176174
Your candidate for a definition is exactly the same definition of dilation you quoted. If $UV|_H=T$, then since $PT=T$ you have $PUV|_H=T$, and so $UV$ is a dilation of $T$ in the sense of your first paragraph.
A "dilation" has to be a "dilation". The spirit is that you have $T$ (typically, $T$ is a contraction), and you get a dilation $$tilde T=begin{bmatrix} T&A\ B& Cend{bmatrix}$$ that is better behaved, say $tilde T$ is a unitary, or a projection.
answered Nov 28 at 23:32
Martin Argerami
123k1176174
answered Nov 28 at 23:32
Martin Argerami
123k1176174
answered Nov 28 at 23:32
Martin Argerami
123k1176174
answered Nov 28 at 23:32
Martin Argerami
123k1176174
123k1176174
add a comment |
add a comment |
up vote
1
down vote
Another version:
Theorem: Let $A$ be a bounded linear operator on a Hilbert space $mathcal{H}$ with $|A|le 1$. There there exists a Hilbert space $mathcal{K}$, an isometry $V : mathcal{H}rightarrowmathcal{K}$, and a unitary operator $U : mathcal{K}rightarrowmathcal{K}$ such that
$$
A^n = V^*U^n V mbox{ for all } n ge 0.
$$
answered Nov 29 at 5:22
DisintegratingByParts
58.3k42579
add a comment |
up vote
1
down vote
Another version:
Theorem: Let $A$ be a bounded linear operator on a Hilbert space $mathcal{H}$ with $|A|le 1$. There there exists a Hilbert space $mathcal{K}$, an isometry $V : mathcal{H}rightarrowmathcal{K}$, and a unitary operator $U : mathcal{K}rightarrowmathcal{K}$ such that
$$
A^n = V^*U^n V mbox{ for all } n ge 0.
$$
answered Nov 29 at 5:22
DisintegratingByParts
58.3k42579
add a comment |
up vote
1
down vote
up vote
1
down vote
Another version:
Theorem: Let $A$ be a bounded linear operator on a Hilbert space $mathcal{H}$ with $|A|le 1$. There there exists a Hilbert space $mathcal{K}$, an isometry $V : mathcal{H}rightarrowmathcal{K}$, and a unitary operator $U : mathcal{K}rightarrowmathcal{K}$ such that
$$
A^n = V^*U^n V mbox{ for all } n ge 0.
$$
answered Nov 29 at 5:22
DisintegratingByParts
58.3k42579
Another version:
Theorem: Let $A$ be a bounded linear operator on a Hilbert space $mathcal{H}$ with $|A|le 1$. There there exists a Hilbert space $mathcal{K}$, an isometry $V : mathcal{H}rightarrowmathcal{K}$, and a unitary operator $U : mathcal{K}rightarrowmathcal{K}$ such that
$$
A^n = V^*U^n V mbox{ for all } n ge 0.
$$
answered Nov 29 at 5:22
DisintegratingByParts
58.3k42579
answered Nov 29 at 5:22
DisintegratingByParts
58.3k42579
answered Nov 29 at 5:22
DisintegratingByParts
58.3k42579
answered Nov 29 at 5:22
DisintegratingByParts
58.3k42579
58.3k42579
add a comment |
add a comment |
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