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Does a bounded linear operator on a Hilbert space conjugate to its adjoint?
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$begingroup$
Let $H$ be a Hilbert space over $mathbb R$ or $mathbb C$ and $fin B(H)$. I wonder if there is a $gin B(H)$ such that $fg=gf^*$, where $f^*$ is the adjoint of $f$.
We know a matrix over a field is always conjugate to its transpose. Is there a generalization of this fact to Hilbert spaces?
functional-analysis operator-theory
asked Mar 18 at 1:54
No OneNo One
2,1051519
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert space over $mathbb R$ or $mathbb C$ and $fin B(H)$. I wonder if there is a $gin B(H)$ such that $fg=gf^*$, where $f^*$ is the adjoint of $f$.
We know a matrix over a field is always conjugate to its transpose. Is there a generalization of this fact to Hilbert spaces?
functional-analysis operator-theory
asked Mar 18 at 1:54
No OneNo One
2,1051519
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert space over $mathbb R$ or $mathbb C$ and $fin B(H)$. I wonder if there is a $gin B(H)$ such that $fg=gf^*$, where $f^*$ is the adjoint of $f$.
We know a matrix over a field is always conjugate to its transpose. Is there a generalization of this fact to Hilbert spaces?
functional-analysis operator-theory
asked Mar 18 at 1:54
No OneNo One
2,1051519
$endgroup$
Let $H$ be a Hilbert space over $mathbb R$ or $mathbb C$ and $fin B(H)$. I wonder if there is a $gin B(H)$ such that $fg=gf^*$, where $f^*$ is the adjoint of $f$.
We know a matrix over a field is always conjugate to its transpose. Is there a generalization of this fact to Hilbert spaces?
functional-analysis operator-theory
functional-analysis operator-theory
asked Mar 18 at 1:54
No OneNo One
2,1051519
asked Mar 18 at 1:54
No OneNo One
2,1051519
asked Mar 18 at 1:54
No OneNo One
2,1051519
asked Mar 18 at 1:54
No OneNo One
2,1051519
asked Mar 18 at 1:54
No OneNo One
2,1051519
2,1051519
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Presumably you mean $g$ to be invertible, otherwise the statement is trivially true (take $g=0$).
The answer is no. For example, if $H$ is infinite-dimensional take $f in B(H)$ to be an isomorphism from $H$ to an infinite-dimensional closed subspace of $H$. $f$ is one-to-one, but $f^*$ is not, so $f$ and $f^*$ are not conjugate.
For an explicit example of this, on $ell^2$ take $f((x_1,x_2,x_3,ldots)) = (x_1, 0, x_2, 0, x_3, ldots)$.
answered Mar 18 at 2:00
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
A matrix is conjugate to its transpose, not its adjoint (in general).
For example:
$$A = begin{bmatrix} i & 0 \ 0 & 1end{bmatrix}, quad A^* = begin{bmatrix} -i & 0 \ 0 & 1end{bmatrix}$$
are clearly not conjugated since $sigma(A) = {i,1}$ and $sigma(A^*) = {-i,1}$.
answered Mar 18 at 9:13
mechanodroidmechanodroid
28.8k62648
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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votes
$begingroup$
Presumably you mean $g$ to be invertible, otherwise the statement is trivially true (take $g=0$).
The answer is no. For example, if $H$ is infinite-dimensional take $f in B(H)$ to be an isomorphism from $H$ to an infinite-dimensional closed subspace of $H$. $f$ is one-to-one, but $f^*$ is not, so $f$ and $f^*$ are not conjugate.
For an explicit example of this, on $ell^2$ take $f((x_1,x_2,x_3,ldots)) = (x_1, 0, x_2, 0, x_3, ldots)$.
answered Mar 18 at 2:00
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
Presumably you mean $g$ to be invertible, otherwise the statement is trivially true (take $g=0$).
The answer is no. For example, if $H$ is infinite-dimensional take $f in B(H)$ to be an isomorphism from $H$ to an infinite-dimensional closed subspace of $H$. $f$ is one-to-one, but $f^*$ is not, so $f$ and $f^*$ are not conjugate.
For an explicit example of this, on $ell^2$ take $f((x_1,x_2,x_3,ldots)) = (x_1, 0, x_2, 0, x_3, ldots)$.
answered Mar 18 at 2:00
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
Presumably you mean $g$ to be invertible, otherwise the statement is trivially true (take $g=0$).
The answer is no. For example, if $H$ is infinite-dimensional take $f in B(H)$ to be an isomorphism from $H$ to an infinite-dimensional closed subspace of $H$. $f$ is one-to-one, but $f^*$ is not, so $f$ and $f^*$ are not conjugate.
For an explicit example of this, on $ell^2$ take $f((x_1,x_2,x_3,ldots)) = (x_1, 0, x_2, 0, x_3, ldots)$.
answered Mar 18 at 2:00
Robert IsraelRobert Israel
330k23219473
$endgroup$
Presumably you mean $g$ to be invertible, otherwise the statement is trivially true (take $g=0$).
The answer is no. For example, if $H$ is infinite-dimensional take $f in B(H)$ to be an isomorphism from $H$ to an infinite-dimensional closed subspace of $H$. $f$ is one-to-one, but $f^*$ is not, so $f$ and $f^*$ are not conjugate.
For an explicit example of this, on $ell^2$ take $f((x_1,x_2,x_3,ldots)) = (x_1, 0, x_2, 0, x_3, ldots)$.
answered Mar 18 at 2:00
Robert IsraelRobert Israel
330k23219473
answered Mar 18 at 2:00
Robert IsraelRobert Israel
330k23219473
answered Mar 18 at 2:00
Robert IsraelRobert Israel
330k23219473
answered Mar 18 at 2:00
Robert IsraelRobert Israel
330k23219473
330k23219473
add a comment |
add a comment |
$begingroup$
A matrix is conjugate to its transpose, not its adjoint (in general).
For example:
$$A = begin{bmatrix} i & 0 \ 0 & 1end{bmatrix}, quad A^* = begin{bmatrix} -i & 0 \ 0 & 1end{bmatrix}$$
are clearly not conjugated since $sigma(A) = {i,1}$ and $sigma(A^*) = {-i,1}$.
answered Mar 18 at 9:13
mechanodroidmechanodroid
28.8k62648
$endgroup$
add a comment |
$begingroup$
A matrix is conjugate to its transpose, not its adjoint (in general).
For example:
$$A = begin{bmatrix} i & 0 \ 0 & 1end{bmatrix}, quad A^* = begin{bmatrix} -i & 0 \ 0 & 1end{bmatrix}$$
are clearly not conjugated since $sigma(A) = {i,1}$ and $sigma(A^*) = {-i,1}$.
answered Mar 18 at 9:13
mechanodroidmechanodroid
28.8k62648
$endgroup$
add a comment |
$begingroup$
A matrix is conjugate to its transpose, not its adjoint (in general).
For example:
$$A = begin{bmatrix} i & 0 \ 0 & 1end{bmatrix}, quad A^* = begin{bmatrix} -i & 0 \ 0 & 1end{bmatrix}$$
are clearly not conjugated since $sigma(A) = {i,1}$ and $sigma(A^*) = {-i,1}$.
answered Mar 18 at 9:13
mechanodroidmechanodroid
28.8k62648
$endgroup$
A matrix is conjugate to its transpose, not its adjoint (in general).
For example:
$$A = begin{bmatrix} i & 0 \ 0 & 1end{bmatrix}, quad A^* = begin{bmatrix} -i & 0 \ 0 & 1end{bmatrix}$$
are clearly not conjugated since $sigma(A) = {i,1}$ and $sigma(A^*) = {-i,1}$.
answered Mar 18 at 9:13
mechanodroidmechanodroid
28.8k62648
answered Mar 18 at 9:13
mechanodroidmechanodroid
28.8k62648
answered Mar 18 at 9:13
mechanodroidmechanodroid
28.8k62648
answered Mar 18 at 9:13
mechanodroidmechanodroid
28.8k62648
28.8k62648
add a comment |
add a comment |
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