Spectrum of a pair of commuting operators
Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
$(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
$$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$
Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
$$sigma_H(I,A)={(1,1);(1,-1)}.$$
functional-analysis operator-theory hilbert-spaces
asked Nov 28 at 12:57
Student
2,4762524
add a comment |
Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
$(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
$$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$
Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
$$sigma_H(I,A)={(1,1);(1,-1)}.$$
functional-analysis operator-theory hilbert-spaces
asked Nov 28 at 12:57
Student
2,4762524
add a comment |
Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
$(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
$$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$
Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
$$sigma_H(I,A)={(1,1);(1,-1)}.$$
functional-analysis operator-theory hilbert-spaces
asked Nov 28 at 12:57
Student
2,4762524
Definition: Let ${bf A}=(A_1,A_2,cdots,A_d)in mathcal{B}(mathcal{H})^d$ be a $d$-tuple of commuting operators on a complex Hilbert space $mathcal{H}.$ Let $sigma_H({bf A})$ denotes the Harte spectrum of ${bf A}$.
$(lambda_1,lambda_2,cdots,lambda_d)notin sigma({bf A})$ if there exist operators $U_1,cdots,U_d,V_1,cdots,V_d in mathcal{B}(mathcal{H}))$ such that
$$sum_{1leq k leq d}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq d}(A_k-lambda_k I)V_k =I.$$
Let $A= begin{pmatrix}0&1\1&0end{pmatrix}$ and $I= begin{pmatrix}1&0\0&1end{pmatrix}$. I don't understant how to prove that
$$sigma_H(I,A)={(1,1);(1,-1)}.$$
functional-analysis operator-theory hilbert-spaces
functional-analysis operator-theory hilbert-spaces
asked Nov 28 at 12:57
Student
2,4762524
asked Nov 28 at 12:57
Student
2,4762524
edited Nov 28 at 15:11
asked Nov 28 at 12:57
Student
2,4762524
asked Nov 28 at 12:57
Student
2,4762524
asked Nov 28 at 12:57
Student
2,4762524
2,4762524
add a comment |
add a comment |
1 Answer
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First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 at 16:59
s.harp
8,39312049
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 at 11:12
add a comment |
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1 Answer
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1 Answer
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First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 at 16:59
s.harp
8,39312049
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 at 11:12
add a comment |
First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 at 16:59
s.harp
8,39312049
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 at 11:12
add a comment |
First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 at 16:59
s.harp
8,39312049
First a remark:
If $(lambda_1,...,lambda_n)in sigma_H(A_1,..,A_n)$ then $lambda_iinsigma(A_i)$ for all $i$. If this is not the case, ie there is a $lambda_i$ with $A_i-lambda_i I$ invertible, then $U_i=V_i=(A_i-lambda_i I)^{-1}$ and the other $V_j=U_j=0$ shows that $(lambda_1,..,lambda_n)notinsigma_H(A_1,...,A_n)$.
This means $sigma_H(A_1,..,.A_n)subset sigma(A_1)times...times sigma(A_n)$. In our case we've got then that $sigma_H(I,A)subset {1}times {1,-1}$.
So the only thing you've got to do is prove that both those points lie in the spectrum, then you're done. This is easy, as $I-1cdot I$ is always $0$ and the question reduces to showing that $Apm I$ is not invertible.
answered Nov 28 at 16:59
s.harp
8,39312049
answered Nov 28 at 16:59
s.harp
8,39312049
answered Nov 28 at 16:59
s.harp
8,39312049
answered Nov 28 at 16:59
s.harp
8,39312049
8,39312049
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 at 11:12
add a comment |
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 at 11:12
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 at 11:12
Please see my question related to the Taylor spectrum: math.stackexchange.com/questions/3019776/…
– Student
Dec 2 at 11:12
add a comment |
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