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Some particular example clarification on algorithm of Hahn Hellinger Theorem
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Consider the self adjoint matrix $T$
beginbmatrix
1 & 0 & 0 \
0 & 1 & 0 \
0 &0 &2
endbmatrix
The question is the following: I want to understand the algorithm of Hahn-Hellinger (Spectral Theorem) so far I tried is that since 1 is eigenvalue of multiplicity of 2 so it does not have cyclic vector, so $(1,0,1)$ and $(0,1,0)$ are cyclic vectors breaking the space into two parts. My simple question how to find the spectral representation $pi: L^infty(X,mu) mapsto B(L^2(X,mu))$ and unitary such that $pi_1: C(X) mapsto B(mathcalH)$ are equivalent where $X=sigma (T)$, $mu$ is spectral measure
operator-algebras c-star-algebras von-neumann-algebras
edited Jul 23 at 15:51
Martin Argerami
115k1071164
asked Jul 22 at 17:16
mathlover
10518
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up vote
1
down vote
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Consider the self adjoint matrix $T$
beginbmatrix
1 & 0 & 0 \
0 & 1 & 0 \
0 &0 &2
endbmatrix
The question is the following: I want to understand the algorithm of Hahn-Hellinger (Spectral Theorem) so far I tried is that since 1 is eigenvalue of multiplicity of 2 so it does not have cyclic vector, so $(1,0,1)$ and $(0,1,0)$ are cyclic vectors breaking the space into two parts. My simple question how to find the spectral representation $pi: L^infty(X,mu) mapsto B(L^2(X,mu))$ and unitary such that $pi_1: C(X) mapsto B(mathcalH)$ are equivalent where $X=sigma (T)$, $mu$ is spectral measure
operator-algebras c-star-algebras von-neumann-algebras
edited Jul 23 at 15:51
Martin Argerami
115k1071164
asked Jul 22 at 17:16
mathlover
10518
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the self adjoint matrix $T$
beginbmatrix
1 & 0 & 0 \
0 & 1 & 0 \
0 &0 &2
endbmatrix
The question is the following: I want to understand the algorithm of Hahn-Hellinger (Spectral Theorem) so far I tried is that since 1 is eigenvalue of multiplicity of 2 so it does not have cyclic vector, so $(1,0,1)$ and $(0,1,0)$ are cyclic vectors breaking the space into two parts. My simple question how to find the spectral representation $pi: L^infty(X,mu) mapsto B(L^2(X,mu))$ and unitary such that $pi_1: C(X) mapsto B(mathcalH)$ are equivalent where $X=sigma (T)$, $mu$ is spectral measure
operator-algebras c-star-algebras von-neumann-algebras
edited Jul 23 at 15:51
Martin Argerami
115k1071164
asked Jul 22 at 17:16
mathlover
10518
Consider the self adjoint matrix $T$
beginbmatrix
1 & 0 & 0 \
0 & 1 & 0 \
0 &0 &2
endbmatrix
The question is the following: I want to understand the algorithm of Hahn-Hellinger (Spectral Theorem) so far I tried is that since 1 is eigenvalue of multiplicity of 2 so it does not have cyclic vector, so $(1,0,1)$ and $(0,1,0)$ are cyclic vectors breaking the space into two parts. My simple question how to find the spectral representation $pi: L^infty(X,mu) mapsto B(L^2(X,mu))$ and unitary such that $pi_1: C(X) mapsto B(mathcalH)$ are equivalent where $X=sigma (T)$, $mu$ is spectral measure
operator-algebras c-star-algebras von-neumann-algebras
edited Jul 23 at 15:51
Martin Argerami
115k1071164
asked Jul 22 at 17:16
mathlover
10518
edited Jul 23 at 15:51
Martin Argerami
115k1071164
edited Jul 23 at 15:51
Martin Argerami
115k1071164
edited Jul 23 at 15:51
Martin Argerami
115k1071164
115k1071164
asked Jul 22 at 17:16
mathlover
10518
asked Jul 22 at 17:16
mathlover
10518
asked Jul 22 at 17:16
mathlover
10518
10518
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
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up vote
1
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I'm not entirely sure what you are looking for. Here $sigma(T)=1,2$. The C$^*$-algebra generated by $T$ is
$$
leftbeginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrix: a,binmathbb Crightsimeqmathbb C^2,
$$
while
$$
C(sigma(T))=C(1,2)simeq mathbb C^2.
$$
The Gelfand transform here is the map $Gamma: C^*(T)to C(sigma(T))$, which in this case is explicitly given by
$$
Gammaleft(beginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrixright)=beginbmatrix a\ bendbmatrix.
$$
The natural embedding of $C(sigma(T))$ as multiplication operators on $B(L^2(sigma(T))$ is given by
$$
pi_1(a,b)=beginbmatrix a&0\0&bendbmatrixin M_2(mathbb C)=B(mathbb C^2).
$$
answered Jul 23 at 15:50
Martin Argerami
115k1071164
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I'm not entirely sure what you are looking for. Here $sigma(T)=1,2$. The C$^*$-algebra generated by $T$ is
$$
leftbeginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrix: a,binmathbb Crightsimeqmathbb C^2,
$$
while
$$
C(sigma(T))=C(1,2)simeq mathbb C^2.
$$
The Gelfand transform here is the map $Gamma: C^*(T)to C(sigma(T))$, which in this case is explicitly given by
$$
Gammaleft(beginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrixright)=beginbmatrix a\ bendbmatrix.
$$
The natural embedding of $C(sigma(T))$ as multiplication operators on $B(L^2(sigma(T))$ is given by
$$
pi_1(a,b)=beginbmatrix a&0\0&bendbmatrixin M_2(mathbb C)=B(mathbb C^2).
$$
answered Jul 23 at 15:50
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
I'm not entirely sure what you are looking for. Here $sigma(T)=1,2$. The C$^*$-algebra generated by $T$ is
$$
leftbeginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrix: a,binmathbb Crightsimeqmathbb C^2,
$$
while
$$
C(sigma(T))=C(1,2)simeq mathbb C^2.
$$
The Gelfand transform here is the map $Gamma: C^*(T)to C(sigma(T))$, which in this case is explicitly given by
$$
Gammaleft(beginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrixright)=beginbmatrix a\ bendbmatrix.
$$
The natural embedding of $C(sigma(T))$ as multiplication operators on $B(L^2(sigma(T))$ is given by
$$
pi_1(a,b)=beginbmatrix a&0\0&bendbmatrixin M_2(mathbb C)=B(mathbb C^2).
$$
answered Jul 23 at 15:50
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I'm not entirely sure what you are looking for. Here $sigma(T)=1,2$. The C$^*$-algebra generated by $T$ is
$$
leftbeginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrix: a,binmathbb Crightsimeqmathbb C^2,
$$
while
$$
C(sigma(T))=C(1,2)simeq mathbb C^2.
$$
The Gelfand transform here is the map $Gamma: C^*(T)to C(sigma(T))$, which in this case is explicitly given by
$$
Gammaleft(beginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrixright)=beginbmatrix a\ bendbmatrix.
$$
The natural embedding of $C(sigma(T))$ as multiplication operators on $B(L^2(sigma(T))$ is given by
$$
pi_1(a,b)=beginbmatrix a&0\0&bendbmatrixin M_2(mathbb C)=B(mathbb C^2).
$$
answered Jul 23 at 15:50
Martin Argerami
115k1071164
I'm not entirely sure what you are looking for. Here $sigma(T)=1,2$. The C$^*$-algebra generated by $T$ is
$$
leftbeginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrix: a,binmathbb Crightsimeqmathbb C^2,
$$
while
$$
C(sigma(T))=C(1,2)simeq mathbb C^2.
$$
The Gelfand transform here is the map $Gamma: C^*(T)to C(sigma(T))$, which in this case is explicitly given by
$$
Gammaleft(beginbmatrix a&0&0\ 0&a&0\ 0&0&bendbmatrixright)=beginbmatrix a\ bendbmatrix.
$$
The natural embedding of $C(sigma(T))$ as multiplication operators on $B(L^2(sigma(T))$ is given by
$$
pi_1(a,b)=beginbmatrix a&0\0&bendbmatrixin M_2(mathbb C)=B(mathbb C^2).
$$
answered Jul 23 at 15:50
Martin Argerami
115k1071164
answered Jul 23 at 15:50
Martin Argerami
115k1071164
answered Jul 23 at 15:50
Martin Argerami
115k1071164
answered Jul 23 at 15:50
Martin Argerami
115k1071164
115k1071164
add a comment |Â
add a comment |Â
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