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0

$begingroup$

Let $T$ be a self-adjoint operator in $mathcalH$ with spectrum $sigma(T)$, Let $P$ be a projection in the commutant of $textvNT$, the von Neumann algebra generated by $T$, question what is the spectrum of $PTP$? What are the relation betweeen $sigma(PTP)$ and $sigma(T)$?

operator-theory operator-algebras spectral-theory von-neumann-algebras

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edited Mar 22 at 9:55

mathlover

asked Mar 22 at 9:49

mathlovermathlover

168110

$endgroup$

add a comment | 

0

$begingroup$

Let $T$ be a self-adjoint operator in $mathcalH$ with spectrum $sigma(T)$, Let $P$ be a projection in the commutant of $textvNT$, the von Neumann algebra generated by $T$, question what is the spectrum of $PTP$? What are the relation betweeen $sigma(PTP)$ and $sigma(T)$?

operator-theory operator-algebras spectral-theory von-neumann-algebras

share|cite|improve this question

edited Mar 22 at 9:55

mathlover

asked Mar 22 at 9:49

mathlovermathlover

168110

$endgroup$

add a comment | 

$begingroup$

Let $T$ be a self-adjoint operator in $mathcalH$ with spectrum $sigma(T)$, Let $P$ be a projection in the commutant of $textvNT$, the von Neumann algebra generated by $T$, question what is the spectrum of $PTP$? What are the relation betweeen $sigma(PTP)$ and $sigma(T)$?

operator-theory operator-algebras spectral-theory von-neumann-algebras

share|cite|improve this question

edited Mar 22 at 9:55

mathlover

asked Mar 22 at 9:49

mathlovermathlover

168110

$endgroup$

share|cite|improve this question

edited Mar 22 at 9:55

mathlover

asked Mar 22 at 9:49

mathlovermathlover

168110

share|cite|improve this question

edited Mar 22 at 9:55

mathlover

asked Mar 22 at 9:49

mathlovermathlover

168110

asked Mar 22 at 9:49

mathlovermathlover

168110

asked Mar 22 at 9:49

mathlovermathlover

168110

1 Answer
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1

$begingroup$

There is no relation. This already shows in $M_2(mathbb C)$. Take any $tin[0,1]$,
$$
T=beginbmatrix t &sqrtt-t^2\ sqrtt-t^2&1-tendbmatrix, P=beginbmatrix 1&0\0&0endbmatrix .
$$
Then $sigma(T)=0,1$, $sigma(PTP)=0,t$.

And this idea can be used even more brutally. Let $K$ be any compact subset of $[0,1]$. Construct a selfadjoint operator $T_0$ with spectrum $K$. Now, on $Hoplus H$, form
$$
T=beginbmatrix T_0 & (T_0-T_0^2)^1/2 \ (T_0-T_0^2)^1/2 & I-T_0endbmatrix, P=beginbmatrix I&0\0&0endbmatrix.
$$
Then $T$ is a projection, so $sigma(T)=0,1$, while $sigma(PTP)=0cupsigma(T_0)=0cup K$.

share|cite|improve this answer

answered Mar 22 at 19:20

Martin ArgeramiMartin Argerami

129k1184185

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1

$begingroup$

There is no relation. This already shows in $M_2(mathbb C)$. Take any $tin[0,1]$,
$$
T=beginbmatrix t &sqrtt-t^2\ sqrtt-t^2&1-tendbmatrix, P=beginbmatrix 1&0\0&0endbmatrix .
$$
Then $sigma(T)=0,1$, $sigma(PTP)=0,t$.

And this idea can be used even more brutally. Let $K$ be any compact subset of $[0,1]$. Construct a selfadjoint operator $T_0$ with spectrum $K$. Now, on $Hoplus H$, form
$$
T=beginbmatrix T_0 & (T_0-T_0^2)^1/2 \ (T_0-T_0^2)^1/2 & I-T_0endbmatrix, P=beginbmatrix I&0\0&0endbmatrix.
$$
Then $T$ is a projection, so $sigma(T)=0,1$, while $sigma(PTP)=0cupsigma(T_0)=0cup K$.

share|cite|improve this answer

answered Mar 22 at 19:20

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

add a comment | 

1

$begingroup$

There is no relation. This already shows in $M_2(mathbb C)$. Take any $tin[0,1]$,
$$
T=beginbmatrix t &sqrtt-t^2\ sqrtt-t^2&1-tendbmatrix, P=beginbmatrix 1&0\0&0endbmatrix .
$$
Then $sigma(T)=0,1$, $sigma(PTP)=0,t$.

And this idea can be used even more brutally. Let $K$ be any compact subset of $[0,1]$. Construct a selfadjoint operator $T_0$ with spectrum $K$. Now, on $Hoplus H$, form
$$
T=beginbmatrix T_0 & (T_0-T_0^2)^1/2 \ (T_0-T_0^2)^1/2 & I-T_0endbmatrix, P=beginbmatrix I&0\0&0endbmatrix.
$$
Then $T$ is a projection, so $sigma(T)=0,1$, while $sigma(PTP)=0cupsigma(T_0)=0cup K$.

share|cite|improve this answer

answered Mar 22 at 19:20

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

add a comment | 

$begingroup$

There is no relation. This already shows in $M_2(mathbb C)$. Take any $tin[0,1]$,
$$
T=beginbmatrix t &sqrtt-t^2\ sqrtt-t^2&1-tendbmatrix, P=beginbmatrix 1&0\0&0endbmatrix .
$$
Then $sigma(T)=0,1$, $sigma(PTP)=0,t$.

And this idea can be used even more brutally. Let $K$ be any compact subset of $[0,1]$. Construct a selfadjoint operator $T_0$ with spectrum $K$. Now, on $Hoplus H$, form
$$
T=beginbmatrix T_0 & (T_0-T_0^2)^1/2 \ (T_0-T_0^2)^1/2 & I-T_0endbmatrix, P=beginbmatrix I&0\0&0endbmatrix.
$$
Then $T$ is a projection, so $sigma(T)=0,1$, while $sigma(PTP)=0cupsigma(T_0)=0cup K$.

share|cite|improve this answer

answered Mar 22 at 19:20

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

share|cite|improve this answer

answered Mar 22 at 19:20

Martin ArgeramiMartin Argerami

129k1184185

answered Mar 22 at 19:20

Martin ArgeramiMartin Argerami

129k1184185

answered Mar 22 at 19:20

Martin ArgeramiMartin Argerami

129k1184185