Mixing
Tensor product of two positive operators
$begingroup$
Let $E$, $F$ be two complex Hilbert spaces and $mathcal{L}(E)$ (resp. $mathcal{L}(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).
The algebraic tensor product of $E$ and $F$ is given by
$$E otimes F:=left{xi=sum_{i=1}^dv_iotimes w_i:;din mathbb{N}^*,;;v_iin E,;;w_iin F
right}.$$
In $E otimes F$, we define
$$
langle xi,etarangle=sum_{i=1}^nsum_{j=1}^m langle x_i,z_jrangle_1langle y_i ,t_jrangle_2,
$$
for $xi=displaystylesum_{i=1}^nx_iotimes y_iin E otimes F$ and $eta=displaystylesum_{j=1}^mz_jotimes w_jin E otimes F$.
The above sesquilinear form is an inner product in $E otimes F$.
It is well known that $(E otimes F,langlecdot,cdotrangle)$ is not a complete space. Let $E widehat{otimes} F$ be the completion of $E otimes F$ under the inner product $langlecdot,cdotrangle$.
If $Tin mathcal{L}(E)$ and $Sin mathcal{L}(F)$, then the tensor product of $T$ and $S$ is denoted $Totimes S$ and defined as
$$big(Totimes Sbig)bigg(sum_{k=1}^d x_kotimes y_kbigg)=sum_{k=1}^dTx_k otimes Sy_k,;;forall,sum_{k=1}^d x_kotimes y_kin E otimes F,$$
which lies in $mathcal{L}(E otimes F)$. The extension of $Totimes S$ over the Hilbert space $E widehat{otimes} F$, denoted by $T widehat{otimes} S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $mathcal{L}(Ewidehat{otimes}F)$.
An operator $Ainmathcal{L}(E)$ is said to be positive if $langle Axmid xrangle geq 0$ for any $xin E$.
If $T$ and $S$ are positive operators, I want to prove $Totimes S$ is positive on $E otimes F$.
Let $X=sum_{i=1}^nx_iotimes y_iin Eotimes F$. Then
begin{align*}
langle (Totimes S)X,Xrangle
& =sum_{i=1}^nsum_{j=1}^n langle Tx_imid x_jrangle_1langle Sy_imid y_jrangle_2.
end{align*}
My goal is to prove that
$$langle (Totimes S)X,Xrangle geq 0,$$
for any $X in Eotimes F$.
functional-analysis operator-theory
asked Jan 31 at 16:00
StudentStudent
2,4732524
$endgroup$
add a comment |
$begingroup$
Let $E$, $F$ be two complex Hilbert spaces and $mathcal{L}(E)$ (resp. $mathcal{L}(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).
The algebraic tensor product of $E$ and $F$ is given by
$$E otimes F:=left{xi=sum_{i=1}^dv_iotimes w_i:;din mathbb{N}^*,;;v_iin E,;;w_iin F
right}.$$
In $E otimes F$, we define
$$
langle xi,etarangle=sum_{i=1}^nsum_{j=1}^m langle x_i,z_jrangle_1langle y_i ,t_jrangle_2,
$$
for $xi=displaystylesum_{i=1}^nx_iotimes y_iin E otimes F$ and $eta=displaystylesum_{j=1}^mz_jotimes w_jin E otimes F$.
The above sesquilinear form is an inner product in $E otimes F$.
It is well known that $(E otimes F,langlecdot,cdotrangle)$ is not a complete space. Let $E widehat{otimes} F$ be the completion of $E otimes F$ under the inner product $langlecdot,cdotrangle$.
If $Tin mathcal{L}(E)$ and $Sin mathcal{L}(F)$, then the tensor product of $T$ and $S$ is denoted $Totimes S$ and defined as
$$big(Totimes Sbig)bigg(sum_{k=1}^d x_kotimes y_kbigg)=sum_{k=1}^dTx_k otimes Sy_k,;;forall,sum_{k=1}^d x_kotimes y_kin E otimes F,$$
which lies in $mathcal{L}(E otimes F)$. The extension of $Totimes S$ over the Hilbert space $E widehat{otimes} F$, denoted by $T widehat{otimes} S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $mathcal{L}(Ewidehat{otimes}F)$.
An operator $Ainmathcal{L}(E)$ is said to be positive if $langle Axmid xrangle geq 0$ for any $xin E$.
If $T$ and $S$ are positive operators, I want to prove $Totimes S$ is positive on $E otimes F$.
Let $X=sum_{i=1}^nx_iotimes y_iin Eotimes F$. Then
begin{align*}
langle (Totimes S)X,Xrangle
& =sum_{i=1}^nsum_{j=1}^n langle Tx_imid x_jrangle_1langle Sy_imid y_jrangle_2.
end{align*}
My goal is to prove that
$$langle (Totimes S)X,Xrangle geq 0,$$
for any $X in Eotimes F$.
functional-analysis operator-theory
asked Jan 31 at 16:00
StudentStudent
2,4732524
$endgroup$
add a comment |
$begingroup$
Let $E$, $F$ be two complex Hilbert spaces and $mathcal{L}(E)$ (resp. $mathcal{L}(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).
The algebraic tensor product of $E$ and $F$ is given by
$$E otimes F:=left{xi=sum_{i=1}^dv_iotimes w_i:;din mathbb{N}^*,;;v_iin E,;;w_iin F
right}.$$
In $E otimes F$, we define
$$
langle xi,etarangle=sum_{i=1}^nsum_{j=1}^m langle x_i,z_jrangle_1langle y_i ,t_jrangle_2,
$$
for $xi=displaystylesum_{i=1}^nx_iotimes y_iin E otimes F$ and $eta=displaystylesum_{j=1}^mz_jotimes w_jin E otimes F$.
The above sesquilinear form is an inner product in $E otimes F$.
It is well known that $(E otimes F,langlecdot,cdotrangle)$ is not a complete space. Let $E widehat{otimes} F$ be the completion of $E otimes F$ under the inner product $langlecdot,cdotrangle$.
If $Tin mathcal{L}(E)$ and $Sin mathcal{L}(F)$, then the tensor product of $T$ and $S$ is denoted $Totimes S$ and defined as
$$big(Totimes Sbig)bigg(sum_{k=1}^d x_kotimes y_kbigg)=sum_{k=1}^dTx_k otimes Sy_k,;;forall,sum_{k=1}^d x_kotimes y_kin E otimes F,$$
which lies in $mathcal{L}(E otimes F)$. The extension of $Totimes S$ over the Hilbert space $E widehat{otimes} F$, denoted by $T widehat{otimes} S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $mathcal{L}(Ewidehat{otimes}F)$.
An operator $Ainmathcal{L}(E)$ is said to be positive if $langle Axmid xrangle geq 0$ for any $xin E$.
If $T$ and $S$ are positive operators, I want to prove $Totimes S$ is positive on $E otimes F$.
Let $X=sum_{i=1}^nx_iotimes y_iin Eotimes F$. Then
begin{align*}
langle (Totimes S)X,Xrangle
& =sum_{i=1}^nsum_{j=1}^n langle Tx_imid x_jrangle_1langle Sy_imid y_jrangle_2.
end{align*}
My goal is to prove that
$$langle (Totimes S)X,Xrangle geq 0,$$
for any $X in Eotimes F$.
functional-analysis operator-theory
asked Jan 31 at 16:00
StudentStudent
2,4732524
$endgroup$
Let $E$, $F$ be two complex Hilbert spaces and $mathcal{L}(E)$ (resp. $mathcal{L}(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).
The algebraic tensor product of $E$ and $F$ is given by
$$E otimes F:=left{xi=sum_{i=1}^dv_iotimes w_i:;din mathbb{N}^*,;;v_iin E,;;w_iin F
right}.$$
In $E otimes F$, we define
$$
langle xi,etarangle=sum_{i=1}^nsum_{j=1}^m langle x_i,z_jrangle_1langle y_i ,t_jrangle_2,
$$
for $xi=displaystylesum_{i=1}^nx_iotimes y_iin E otimes F$ and $eta=displaystylesum_{j=1}^mz_jotimes w_jin E otimes F$.
The above sesquilinear form is an inner product in $E otimes F$.
It is well known that $(E otimes F,langlecdot,cdotrangle)$ is not a complete space. Let $E widehat{otimes} F$ be the completion of $E otimes F$ under the inner product $langlecdot,cdotrangle$.
If $Tin mathcal{L}(E)$ and $Sin mathcal{L}(F)$, then the tensor product of $T$ and $S$ is denoted $Totimes S$ and defined as
$$big(Totimes Sbig)bigg(sum_{k=1}^d x_kotimes y_kbigg)=sum_{k=1}^dTx_k otimes Sy_k,;;forall,sum_{k=1}^d x_kotimes y_kin E otimes F,$$
which lies in $mathcal{L}(E otimes F)$. The extension of $Totimes S$ over the Hilbert space $E widehat{otimes} F$, denoted by $T widehat{otimes} S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $mathcal{L}(Ewidehat{otimes}F)$.
An operator $Ainmathcal{L}(E)$ is said to be positive if $langle Axmid xrangle geq 0$ for any $xin E$.
If $T$ and $S$ are positive operators, I want to prove $Totimes S$ is positive on $E otimes F$.
Let $X=sum_{i=1}^nx_iotimes y_iin Eotimes F$. Then
begin{align*}
langle (Totimes S)X,Xrangle
& =sum_{i=1}^nsum_{j=1}^n langle Tx_imid x_jrangle_1langle Sy_imid y_jrangle_2.
end{align*}
My goal is to prove that
$$langle (Totimes S)X,Xrangle geq 0,$$
for any $X in Eotimes F$.
functional-analysis operator-theory
functional-analysis operator-theory
asked Jan 31 at 16:00
StudentStudent
2,4732524
asked Jan 31 at 16:00
StudentStudent
2,4732524
asked Jan 31 at 16:00
StudentStudent
2,4732524
asked Jan 31 at 16:00
StudentStudent
2,4732524
asked Jan 31 at 16:00
StudentStudent
2,4732524
2,4732524
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm not sure that it can be done directly as you want. But if you know that positive operators admit positive square roots, you know that $T=T_0^*T_0$, $S=S_0^*S_0$. Then
$$
Totimes S=T_0^*T_0otimes S_0^*S_0=(T_0otimes S_0)^*(T_0otimes S_0)geq0.
$$
edited Jan 31 at 18:56
Student
2,4732524
answered Jan 31 at 17:09
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Thanks a lot for your answer. However, I think you have shown that $T widehat{otimes} S$ is a positive operator in $Ewidehat{otimes}F$ because $(T_0otimes S_0)^*$ does not make sense in $Eotimes F$ because it is not a Hilbert space.
$endgroup$
– Student
Jan 31 at 18:57
$begingroup$
Yes, but the restriction to $Eotimes F$ is $Totimes S$.
$endgroup$
– Martin Argerami
Jan 31 at 19:00
$begingroup$
Yes thank you professor.
$endgroup$
– Student
Jan 31 at 19:04
add a comment |
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1 Answer
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$begingroup$
I'm not sure that it can be done directly as you want. But if you know that positive operators admit positive square roots, you know that $T=T_0^*T_0$, $S=S_0^*S_0$. Then
$$
Totimes S=T_0^*T_0otimes S_0^*S_0=(T_0otimes S_0)^*(T_0otimes S_0)geq0.
$$
edited Jan 31 at 18:56
Student
2,4732524
answered Jan 31 at 17:09
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Thanks a lot for your answer. However, I think you have shown that $T widehat{otimes} S$ is a positive operator in $Ewidehat{otimes}F$ because $(T_0otimes S_0)^*$ does not make sense in $Eotimes F$ because it is not a Hilbert space.
$endgroup$
– Student
Jan 31 at 18:57
$begingroup$
Yes, but the restriction to $Eotimes F$ is $Totimes S$.
$endgroup$
– Martin Argerami
Jan 31 at 19:00
$begingroup$
Yes thank you professor.
$endgroup$
– Student
Jan 31 at 19:04
add a comment |
$begingroup$
I'm not sure that it can be done directly as you want. But if you know that positive operators admit positive square roots, you know that $T=T_0^*T_0$, $S=S_0^*S_0$. Then
$$
Totimes S=T_0^*T_0otimes S_0^*S_0=(T_0otimes S_0)^*(T_0otimes S_0)geq0.
$$
edited Jan 31 at 18:56
Student
2,4732524
answered Jan 31 at 17:09
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Thanks a lot for your answer. However, I think you have shown that $T widehat{otimes} S$ is a positive operator in $Ewidehat{otimes}F$ because $(T_0otimes S_0)^*$ does not make sense in $Eotimes F$ because it is not a Hilbert space.
$endgroup$
– Student
Jan 31 at 18:57
$begingroup$
Yes, but the restriction to $Eotimes F$ is $Totimes S$.
$endgroup$
– Martin Argerami
Jan 31 at 19:00
$begingroup$
Yes thank you professor.
$endgroup$
– Student
Jan 31 at 19:04
add a comment |
$begingroup$
I'm not sure that it can be done directly as you want. But if you know that positive operators admit positive square roots, you know that $T=T_0^*T_0$, $S=S_0^*S_0$. Then
$$
Totimes S=T_0^*T_0otimes S_0^*S_0=(T_0otimes S_0)^*(T_0otimes S_0)geq0.
$$
edited Jan 31 at 18:56
Student
2,4732524
answered Jan 31 at 17:09
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
I'm not sure that it can be done directly as you want. But if you know that positive operators admit positive square roots, you know that $T=T_0^*T_0$, $S=S_0^*S_0$. Then
$$
Totimes S=T_0^*T_0otimes S_0^*S_0=(T_0otimes S_0)^*(T_0otimes S_0)geq0.
$$
edited Jan 31 at 18:56
Student
2,4732524
answered Jan 31 at 17:09
Martin ArgeramiMartin Argerami
129k1184185
edited Jan 31 at 18:56
Student
2,4732524
edited Jan 31 at 18:56
Student
2,4732524
edited Jan 31 at 18:56
Student
2,4732524
2,4732524
answered Jan 31 at 17:09
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 31 at 17:09
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 31 at 17:09
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
$begingroup$
Thanks a lot for your answer. However, I think you have shown that $T widehat{otimes} S$ is a positive operator in $Ewidehat{otimes}F$ because $(T_0otimes S_0)^*$ does not make sense in $Eotimes F$ because it is not a Hilbert space.
$endgroup$
– Student
Jan 31 at 18:57
$begingroup$
Yes, but the restriction to $Eotimes F$ is $Totimes S$.
$endgroup$
– Martin Argerami
Jan 31 at 19:00
$begingroup$
Yes thank you professor.
$endgroup$
– Student
Jan 31 at 19:04
add a comment |
$begingroup$
Thanks a lot for your answer. However, I think you have shown that $T widehat{otimes} S$ is a positive operator in $Ewidehat{otimes}F$ because $(T_0otimes S_0)^*$ does not make sense in $Eotimes F$ because it is not a Hilbert space.
$endgroup$
– Student
Jan 31 at 18:57
$begingroup$
Yes, but the restriction to $Eotimes F$ is $Totimes S$.
$endgroup$
– Martin Argerami
Jan 31 at 19:00
$begingroup$
Yes thank you professor.
$endgroup$
– Student
Jan 31 at 19:04
$begingroup$
Thanks a lot for your answer. However, I think you have shown that $T widehat{otimes} S$ is a positive operator in $Ewidehat{otimes}F$ because $(T_0otimes S_0)^*$ does not make sense in $Eotimes F$ because it is not a Hilbert space.
$endgroup$
– Student
Jan 31 at 18:57
$begingroup$
Thanks a lot for your answer. However, I think you have shown that $T widehat{otimes} S$ is a positive operator in $Ewidehat{otimes}F$ because $(T_0otimes S_0)^*$ does not make sense in $Eotimes F$ because it is not a Hilbert space.
$endgroup$
– Student
Jan 31 at 18:57
$begingroup$
Yes, but the restriction to $Eotimes F$ is $Totimes S$.
$endgroup$
– Martin Argerami
Jan 31 at 19:00
$begingroup$
Yes, but the restriction to $Eotimes F$ is $Totimes S$.
$endgroup$
– Martin Argerami
Jan 31 at 19:00
$begingroup$
Yes thank you professor.
$endgroup$
– Student
Jan 31 at 19:04
$begingroup$
Yes thank you professor.
$endgroup$
– Student
Jan 31 at 19:04
add a comment |
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