Prove that $sigma_T(T_1,T_2)subsetsigma(T_1)timessigma(T_2)$
$begingroup$
Let $(T_1,T_2)$ be a pair of commuting operators acting on $mathcal{H}$. $(lambda_1,lambda_2)in sigma_T(T_1,T_2)$ if and only if the Koszul-Complex of $(T_1-lambda_1,T_2-lambda_2)$ is not exact. This is equivalent that at least one of the following properties fails:
$(a)$ The equation system
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}=0;; wedge ;; (T_2-lambda_2)begin{pmatrix} x_1 \ x_2 end{pmatrix}=0$$
has the unique solution $(x_1,x_2)=(0,0)$.$(b)$ The only solutions for the equation
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}+(T_2-lambda_2)begin{pmatrix} y_1 \ y_2 end{pmatrix}=0$$
are of the form
$$begin{pmatrix} x_1 \ x_2 end{pmatrix}=-(T_2-lambda_2)begin{pmatrix} s_1 \ s_2 end{pmatrix};;text{and};;begin{pmatrix} y_1 \ y_2 end{pmatrix}=(T_1-lambda_1)begin{pmatrix} s_1 \ s_2 end{pmatrix},$$
for some $(s_1,s_2)in mathbb{C}^2$.$(c)$ The equation
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}+(T_2-lambda_2)begin{pmatrix} y_1 \ y_2 end{pmatrix}=begin{pmatrix} b_1 \ b_2 end{pmatrix},$$
has a solution for every $(b_1,b_2)in mathbb{C}^2$.
I want to prove that
$$sigma_T(T_1,T_2)subsetsigma(T_1)timessigma(T_2)$$
For more informations about the Taylor spectrum see:
J. L. Taylor, A joint spectrum for several commuting operators, J. Functional Anal. 6(1970), 172-191.
we have
functional-analysis operator-theory hilbert-spaces
asked Dec 4 '18 at 10:44
StudentStudent
2,4362524
$endgroup$
add a comment |
$begingroup$
Let $(T_1,T_2)$ be a pair of commuting operators acting on $mathcal{H}$. $(lambda_1,lambda_2)in sigma_T(T_1,T_2)$ if and only if the Koszul-Complex of $(T_1-lambda_1,T_2-lambda_2)$ is not exact. This is equivalent that at least one of the following properties fails:
$(a)$ The equation system
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}=0;; wedge ;; (T_2-lambda_2)begin{pmatrix} x_1 \ x_2 end{pmatrix}=0$$
has the unique solution $(x_1,x_2)=(0,0)$.$(b)$ The only solutions for the equation
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}+(T_2-lambda_2)begin{pmatrix} y_1 \ y_2 end{pmatrix}=0$$
are of the form
$$begin{pmatrix} x_1 \ x_2 end{pmatrix}=-(T_2-lambda_2)begin{pmatrix} s_1 \ s_2 end{pmatrix};;text{and};;begin{pmatrix} y_1 \ y_2 end{pmatrix}=(T_1-lambda_1)begin{pmatrix} s_1 \ s_2 end{pmatrix},$$
for some $(s_1,s_2)in mathbb{C}^2$.$(c)$ The equation
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}+(T_2-lambda_2)begin{pmatrix} y_1 \ y_2 end{pmatrix}=begin{pmatrix} b_1 \ b_2 end{pmatrix},$$
has a solution for every $(b_1,b_2)in mathbb{C}^2$.
I want to prove that
$$sigma_T(T_1,T_2)subsetsigma(T_1)timessigma(T_2)$$
For more informations about the Taylor spectrum see:
J. L. Taylor, A joint spectrum for several commuting operators, J. Functional Anal. 6(1970), 172-191.
we have
functional-analysis operator-theory hilbert-spaces
asked Dec 4 '18 at 10:44
StudentStudent
2,4362524
$endgroup$
add a comment |
$begingroup$
Let $(T_1,T_2)$ be a pair of commuting operators acting on $mathcal{H}$. $(lambda_1,lambda_2)in sigma_T(T_1,T_2)$ if and only if the Koszul-Complex of $(T_1-lambda_1,T_2-lambda_2)$ is not exact. This is equivalent that at least one of the following properties fails:
$(a)$ The equation system
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}=0;; wedge ;; (T_2-lambda_2)begin{pmatrix} x_1 \ x_2 end{pmatrix}=0$$
has the unique solution $(x_1,x_2)=(0,0)$.$(b)$ The only solutions for the equation
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}+(T_2-lambda_2)begin{pmatrix} y_1 \ y_2 end{pmatrix}=0$$
are of the form
$$begin{pmatrix} x_1 \ x_2 end{pmatrix}=-(T_2-lambda_2)begin{pmatrix} s_1 \ s_2 end{pmatrix};;text{and};;begin{pmatrix} y_1 \ y_2 end{pmatrix}=(T_1-lambda_1)begin{pmatrix} s_1 \ s_2 end{pmatrix},$$
for some $(s_1,s_2)in mathbb{C}^2$.$(c)$ The equation
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}+(T_2-lambda_2)begin{pmatrix} y_1 \ y_2 end{pmatrix}=begin{pmatrix} b_1 \ b_2 end{pmatrix},$$
has a solution for every $(b_1,b_2)in mathbb{C}^2$.
I want to prove that
$$sigma_T(T_1,T_2)subsetsigma(T_1)timessigma(T_2)$$
For more informations about the Taylor spectrum see:
J. L. Taylor, A joint spectrum for several commuting operators, J. Functional Anal. 6(1970), 172-191.
we have
functional-analysis operator-theory hilbert-spaces
asked Dec 4 '18 at 10:44
StudentStudent
2,4362524
$endgroup$
Let $(T_1,T_2)$ be a pair of commuting operators acting on $mathcal{H}$. $(lambda_1,lambda_2)in sigma_T(T_1,T_2)$ if and only if the Koszul-Complex of $(T_1-lambda_1,T_2-lambda_2)$ is not exact. This is equivalent that at least one of the following properties fails:
$(a)$ The equation system
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}=0;; wedge ;; (T_2-lambda_2)begin{pmatrix} x_1 \ x_2 end{pmatrix}=0$$
has the unique solution $(x_1,x_2)=(0,0)$.$(b)$ The only solutions for the equation
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}+(T_2-lambda_2)begin{pmatrix} y_1 \ y_2 end{pmatrix}=0$$
are of the form
$$begin{pmatrix} x_1 \ x_2 end{pmatrix}=-(T_2-lambda_2)begin{pmatrix} s_1 \ s_2 end{pmatrix};;text{and};;begin{pmatrix} y_1 \ y_2 end{pmatrix}=(T_1-lambda_1)begin{pmatrix} s_1 \ s_2 end{pmatrix},$$
for some $(s_1,s_2)in mathbb{C}^2$.$(c)$ The equation
$$(T_1-lambda_1)begin{pmatrix} x_1 \ x_2 end{pmatrix}+(T_2-lambda_2)begin{pmatrix} y_1 \ y_2 end{pmatrix}=begin{pmatrix} b_1 \ b_2 end{pmatrix},$$
has a solution for every $(b_1,b_2)in mathbb{C}^2$.
I want to prove that
$$sigma_T(T_1,T_2)subsetsigma(T_1)timessigma(T_2)$$
For more informations about the Taylor spectrum see:
J. L. Taylor, A joint spectrum for several commuting operators, J. Functional Anal. 6(1970), 172-191.
we have
functional-analysis operator-theory hilbert-spaces
functional-analysis operator-theory hilbert-spaces
asked Dec 4 '18 at 10:44
StudentStudent
2,4362524
asked Dec 4 '18 at 10:44
StudentStudent
2,4362524
edited Dec 4 '18 at 16:58
Student
asked Dec 4 '18 at 10:44
StudentStudent
2,4362524
asked Dec 4 '18 at 10:44
StudentStudent
2,4362524
asked Dec 4 '18 at 10:44
StudentStudent
2,4362524
2,4362524
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You seem to be assuming that $mathcal H=mathbb C^2$. There is no need for that.
Take $(lambda,mu)$ such that at least $lambdanotinsigma(T_1)$ or $munotinsigma(T_2)$. Suppose first that $lambdanotinsigma(T_1)$ (the other case is similar). So $T_1-lambda I$ is invertible. For typing simplicity, from now on I will write $T_1$ instead of $T_1-lambda I$, and $T_2$ instead of $T_2-mu I$.
As $T_1$ is invertible, $ker T_1={0}$, so
$$tag1
ker T_1cap ker T_2={0}.
$$
For the second condition, if $T_1x+T_2y=0$, then $x=-T_1^{-1}T_2y=T_2T_1^{-1}y$. Let $z=T_1^{-1}y$. Then
$$tag2
x=-T_2z, y=T_1z,
$$
so the second condition is satisfied.
Finally, since $T_1$ is onto, the sum of the images of $T_1$ and $T_2$ is $mathcal H$. Together with $(1)$ and $(2)$, this shows that if $(lambda,mu)notinsigma(T_1)timessigma(T_2)$, then $(lambda,mu)notinsigma_T(T_1,T_2)$. Thus
$$
sigma_T(T_1,T_2)subsetsigma(T_1)times sigma(T_2).
$$
answered Dec 4 '18 at 18:07
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
If $mathcal H$ is not $mathbb{C}^2$ the inclusion is not in general true? Also I don't see where exactely you use the fact that $mathcal H =mathbb{C}^2$? Thank you for your help.
$endgroup$
– Student
Dec 4 '18 at 18:49
$begingroup$
In the paper of Taylor $X$ is a Banach space.
$endgroup$
– Student
Dec 4 '18 at 18:53
$begingroup$
Did you read my answer? In your question, you used $mathcal H=mathbb C^2$; in my answer, I'm saying that such a thing is not necessary.
$endgroup$
– Martin Argerami
Dec 4 '18 at 18:53
$begingroup$
Now I understand. Yes in my question I use $mathbb{C}^2$ in order to understand with an example.
$endgroup$
– Student
Dec 4 '18 at 18:58
$begingroup$
I don't really know what $sigma_T(mathbf T)$ is. I would still expect the same argument to work, though.
$endgroup$
– Martin Argerami
Dec 4 '18 at 20:44
|
show 2 more comments
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1 Answer
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1 Answer
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$begingroup$
You seem to be assuming that $mathcal H=mathbb C^2$. There is no need for that.
Take $(lambda,mu)$ such that at least $lambdanotinsigma(T_1)$ or $munotinsigma(T_2)$. Suppose first that $lambdanotinsigma(T_1)$ (the other case is similar). So $T_1-lambda I$ is invertible. For typing simplicity, from now on I will write $T_1$ instead of $T_1-lambda I$, and $T_2$ instead of $T_2-mu I$.
As $T_1$ is invertible, $ker T_1={0}$, so
$$tag1
ker T_1cap ker T_2={0}.
$$
For the second condition, if $T_1x+T_2y=0$, then $x=-T_1^{-1}T_2y=T_2T_1^{-1}y$. Let $z=T_1^{-1}y$. Then
$$tag2
x=-T_2z, y=T_1z,
$$
so the second condition is satisfied.
Finally, since $T_1$ is onto, the sum of the images of $T_1$ and $T_2$ is $mathcal H$. Together with $(1)$ and $(2)$, this shows that if $(lambda,mu)notinsigma(T_1)timessigma(T_2)$, then $(lambda,mu)notinsigma_T(T_1,T_2)$. Thus
$$
sigma_T(T_1,T_2)subsetsigma(T_1)times sigma(T_2).
$$
answered Dec 4 '18 at 18:07
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
If $mathcal H$ is not $mathbb{C}^2$ the inclusion is not in general true? Also I don't see where exactely you use the fact that $mathcal H =mathbb{C}^2$? Thank you for your help.
$endgroup$
– Student
Dec 4 '18 at 18:49
$begingroup$
In the paper of Taylor $X$ is a Banach space.
$endgroup$
– Student
Dec 4 '18 at 18:53
$begingroup$
Did you read my answer? In your question, you used $mathcal H=mathbb C^2$; in my answer, I'm saying that such a thing is not necessary.
$endgroup$
– Martin Argerami
Dec 4 '18 at 18:53
$begingroup$
Now I understand. Yes in my question I use $mathbb{C}^2$ in order to understand with an example.
$endgroup$
– Student
Dec 4 '18 at 18:58
$begingroup$
I don't really know what $sigma_T(mathbf T)$ is. I would still expect the same argument to work, though.
$endgroup$
– Martin Argerami
Dec 4 '18 at 20:44
|
show 2 more comments
$begingroup$
You seem to be assuming that $mathcal H=mathbb C^2$. There is no need for that.
Take $(lambda,mu)$ such that at least $lambdanotinsigma(T_1)$ or $munotinsigma(T_2)$. Suppose first that $lambdanotinsigma(T_1)$ (the other case is similar). So $T_1-lambda I$ is invertible. For typing simplicity, from now on I will write $T_1$ instead of $T_1-lambda I$, and $T_2$ instead of $T_2-mu I$.
As $T_1$ is invertible, $ker T_1={0}$, so
$$tag1
ker T_1cap ker T_2={0}.
$$
For the second condition, if $T_1x+T_2y=0$, then $x=-T_1^{-1}T_2y=T_2T_1^{-1}y$. Let $z=T_1^{-1}y$. Then
$$tag2
x=-T_2z, y=T_1z,
$$
so the second condition is satisfied.
Finally, since $T_1$ is onto, the sum of the images of $T_1$ and $T_2$ is $mathcal H$. Together with $(1)$ and $(2)$, this shows that if $(lambda,mu)notinsigma(T_1)timessigma(T_2)$, then $(lambda,mu)notinsigma_T(T_1,T_2)$. Thus
$$
sigma_T(T_1,T_2)subsetsigma(T_1)times sigma(T_2).
$$
answered Dec 4 '18 at 18:07
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
If $mathcal H$ is not $mathbb{C}^2$ the inclusion is not in general true? Also I don't see where exactely you use the fact that $mathcal H =mathbb{C}^2$? Thank you for your help.
$endgroup$
– Student
Dec 4 '18 at 18:49
$begingroup$
In the paper of Taylor $X$ is a Banach space.
$endgroup$
– Student
Dec 4 '18 at 18:53
$begingroup$
Did you read my answer? In your question, you used $mathcal H=mathbb C^2$; in my answer, I'm saying that such a thing is not necessary.
$endgroup$
– Martin Argerami
Dec 4 '18 at 18:53
$begingroup$
Now I understand. Yes in my question I use $mathbb{C}^2$ in order to understand with an example.
$endgroup$
– Student
Dec 4 '18 at 18:58
$begingroup$
I don't really know what $sigma_T(mathbf T)$ is. I would still expect the same argument to work, though.
$endgroup$
– Martin Argerami
Dec 4 '18 at 20:44
|
show 2 more comments
$begingroup$
You seem to be assuming that $mathcal H=mathbb C^2$. There is no need for that.
Take $(lambda,mu)$ such that at least $lambdanotinsigma(T_1)$ or $munotinsigma(T_2)$. Suppose first that $lambdanotinsigma(T_1)$ (the other case is similar). So $T_1-lambda I$ is invertible. For typing simplicity, from now on I will write $T_1$ instead of $T_1-lambda I$, and $T_2$ instead of $T_2-mu I$.
As $T_1$ is invertible, $ker T_1={0}$, so
$$tag1
ker T_1cap ker T_2={0}.
$$
For the second condition, if $T_1x+T_2y=0$, then $x=-T_1^{-1}T_2y=T_2T_1^{-1}y$. Let $z=T_1^{-1}y$. Then
$$tag2
x=-T_2z, y=T_1z,
$$
so the second condition is satisfied.
Finally, since $T_1$ is onto, the sum of the images of $T_1$ and $T_2$ is $mathcal H$. Together with $(1)$ and $(2)$, this shows that if $(lambda,mu)notinsigma(T_1)timessigma(T_2)$, then $(lambda,mu)notinsigma_T(T_1,T_2)$. Thus
$$
sigma_T(T_1,T_2)subsetsigma(T_1)times sigma(T_2).
$$
answered Dec 4 '18 at 18:07
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
You seem to be assuming that $mathcal H=mathbb C^2$. There is no need for that.
Take $(lambda,mu)$ such that at least $lambdanotinsigma(T_1)$ or $munotinsigma(T_2)$. Suppose first that $lambdanotinsigma(T_1)$ (the other case is similar). So $T_1-lambda I$ is invertible. For typing simplicity, from now on I will write $T_1$ instead of $T_1-lambda I$, and $T_2$ instead of $T_2-mu I$.
As $T_1$ is invertible, $ker T_1={0}$, so
$$tag1
ker T_1cap ker T_2={0}.
$$
For the second condition, if $T_1x+T_2y=0$, then $x=-T_1^{-1}T_2y=T_2T_1^{-1}y$. Let $z=T_1^{-1}y$. Then
$$tag2
x=-T_2z, y=T_1z,
$$
so the second condition is satisfied.
Finally, since $T_1$ is onto, the sum of the images of $T_1$ and $T_2$ is $mathcal H$. Together with $(1)$ and $(2)$, this shows that if $(lambda,mu)notinsigma(T_1)timessigma(T_2)$, then $(lambda,mu)notinsigma_T(T_1,T_2)$. Thus
$$
sigma_T(T_1,T_2)subsetsigma(T_1)times sigma(T_2).
$$
answered Dec 4 '18 at 18:07
Martin ArgeramiMartin Argerami
126k1182181
edited Dec 5 '18 at 14:44
answered Dec 4 '18 at 18:07
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 4 '18 at 18:07
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 4 '18 at 18:07
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
$begingroup$
If $mathcal H$ is not $mathbb{C}^2$ the inclusion is not in general true? Also I don't see where exactely you use the fact that $mathcal H =mathbb{C}^2$? Thank you for your help.
$endgroup$
– Student
Dec 4 '18 at 18:49
$begingroup$
In the paper of Taylor $X$ is a Banach space.
$endgroup$
– Student
Dec 4 '18 at 18:53
$begingroup$
Did you read my answer? In your question, you used $mathcal H=mathbb C^2$; in my answer, I'm saying that such a thing is not necessary.
$endgroup$
– Martin Argerami
Dec 4 '18 at 18:53
$begingroup$
Now I understand. Yes in my question I use $mathbb{C}^2$ in order to understand with an example.
$endgroup$
– Student
Dec 4 '18 at 18:58
$begingroup$
I don't really know what $sigma_T(mathbf T)$ is. I would still expect the same argument to work, though.
$endgroup$
– Martin Argerami
Dec 4 '18 at 20:44
|
show 2 more comments
$begingroup$
If $mathcal H$ is not $mathbb{C}^2$ the inclusion is not in general true? Also I don't see where exactely you use the fact that $mathcal H =mathbb{C}^2$? Thank you for your help.
$endgroup$
– Student
Dec 4 '18 at 18:49
$begingroup$
In the paper of Taylor $X$ is a Banach space.
$endgroup$
– Student
Dec 4 '18 at 18:53
$begingroup$
Did you read my answer? In your question, you used $mathcal H=mathbb C^2$; in my answer, I'm saying that such a thing is not necessary.
$endgroup$
– Martin Argerami
Dec 4 '18 at 18:53
$begingroup$
Now I understand. Yes in my question I use $mathbb{C}^2$ in order to understand with an example.
$endgroup$
– Student
Dec 4 '18 at 18:58
$begingroup$
I don't really know what $sigma_T(mathbf T)$ is. I would still expect the same argument to work, though.
$endgroup$
– Martin Argerami
Dec 4 '18 at 20:44
$begingroup$
If $mathcal H$ is not $mathbb{C}^2$ the inclusion is not in general true? Also I don't see where exactely you use the fact that $mathcal H =mathbb{C}^2$? Thank you for your help.
$endgroup$
– Student
Dec 4 '18 at 18:49
$begingroup$
If $mathcal H$ is not $mathbb{C}^2$ the inclusion is not in general true? Also I don't see where exactely you use the fact that $mathcal H =mathbb{C}^2$? Thank you for your help.
$endgroup$
– Student
Dec 4 '18 at 18:49
$begingroup$
In the paper of Taylor $X$ is a Banach space.
$endgroup$
– Student
Dec 4 '18 at 18:53
$begingroup$
In the paper of Taylor $X$ is a Banach space.
$endgroup$
– Student
Dec 4 '18 at 18:53
$begingroup$
Did you read my answer? In your question, you used $mathcal H=mathbb C^2$; in my answer, I'm saying that such a thing is not necessary.
$endgroup$
– Martin Argerami
Dec 4 '18 at 18:53
$begingroup$
Did you read my answer? In your question, you used $mathcal H=mathbb C^2$; in my answer, I'm saying that such a thing is not necessary.
$endgroup$
– Martin Argerami
Dec 4 '18 at 18:53
$begingroup$
Now I understand. Yes in my question I use $mathbb{C}^2$ in order to understand with an example.
$endgroup$
– Student
Dec 4 '18 at 18:58
$begingroup$
Now I understand. Yes in my question I use $mathbb{C}^2$ in order to understand with an example.
$endgroup$
– Student
Dec 4 '18 at 18:58
$begingroup$
I don't really know what $sigma_T(mathbf T)$ is. I would still expect the same argument to work, though.
$endgroup$
– Martin Argerami
Dec 4 '18 at 20:44
$begingroup$
I don't really know what $sigma_T(mathbf T)$ is. I would still expect the same argument to work, though.
$endgroup$
– Martin Argerami
Dec 4 '18 at 20:44
|
show 2 more comments
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