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Hi Boromir, and welcome to MHB!Let A be a communitative unital subalgebra of B(H) (the set of bounded operators on a hilbert space). Let T be in A. Prove that for all x in the spectrum of T, there is a homomorphism h on A such that h(T)=x. Give the homomorphism explicity.

Can you give us some idea of what you know about such algebras? Is the algebra A supposed to be selfadjoint? If so, then the spectral theorem will apply. That might give you a handle on the problem.

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Thanks for the welcome OpalgHi Boromir, and welcome to MHB!

Can you give us some idea of what you know about such algebras? Is the algebra A supposed to be selfadjoint? If so, then the spectral theorem will apply. That might give you a handle on the problem.

I'm afraid I'm in the process of trying to prove the spectral theorem. I will give you more context. I'm trying to prove that for T in A, the spectrum of T is {h(T):h is a non-zero homomorphism A->C}. I have proved that h(T) is in the spectrum so now I need the reverse inclusion. That is, that every member of the spectrum is equal to h(T) for some (non-zero) homomorphism h. I don't want to go down the road of ideals as the proofs on the internet do as I'm specifically focusing on the space B(H).

Nearly forgot, the algebra is adjoint preserving

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Do you think I am trying to 'square the circle' as in there needs to be an assumption of commutativity which is not present in B(H)?Hi Boromir, and welcome to MHB!

Can you give us some idea of what you know about such algebras? Is the algebra A supposed to be selfadjoint? If so, then the spectral theorem will apply. That might give you a handle on the problem.

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The key to this is the commutative Gelfand–Naimark theorem.

The problem is unchanged if we replace $A$ by its closure (in the operator norm topology). So we may as well assume that $A$ is closed. Now that you have added the crucial information that $A$ is selfadjoint (i.e. closed under the adjoint operation), it follows that $A$ is a commutative unital C*-algebra. The G–N theorem then tells you that there is an isometric isomorphism $\phi:A\to C(X)$ from $A$ to the algebra of continuous functions on a compact Hausdorff space $X$ (the*structure space* or *spectrum* of $A$). It follows fairly easily that the spectrum of $T$ is equal to the range of the function $\phi(T)$. Also, the homomorphisms from $A$ to the scalars are exactly the point evaluation functions on $X$, in other words functions of the form $S\mapsto (\phi(S))(x)$ for $x\in X$ and $S\in A$. That is all you need to deduce the result.

The problem is unchanged if we replace $A$ by its closure (in the operator norm topology). So we may as well assume that $A$ is closed. Now that you have added the crucial information that $A$ is selfadjoint (i.e. closed under the adjoint operation), it follows that $A$ is a commutative unital C*-algebra. The G–N theorem then tells you that there is an isometric isomorphism $\phi:A\to C(X)$ from $A$ to the algebra of continuous functions on a compact Hausdorff space $X$ (the

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On a seperate but related note, why is A closed under inverses i.e. if T has an inverse, then the inverse is in AThe key to this is the commutative Gelfand–Naimark theorem.

The problem is unchanged if we replace $A$ by its closure (in the operator norm topology). So we may as well assume that $A$ is closed. Now that you have added the crucial information that $A$ is selfadjoint (i.e. closed under the adjoint operation), it follows that $A$ is a commutative unital C*-algebra. The G–N theorem then tells you that there is an isometric isomorphism $\phi:A\to C(X)$ from $A$ to the algebra of continuous functions on a compact Hausdorff space $X$ (thestructure spaceorspectrumof $A$). It follows fairly easily that the spectrum of $T$ is equal to the range of the function $\phi(T)$. Also, the homomorphisms from $A$ to the scalars are exactly the point evaluation functions on $X$, in other words functions of the form $S\mapsto (\phi(S))(x)$ for $x\in X$ and $S\in A$. That is all you need to deduce the result.

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That also comes from the G–N theorem. If $T$ has an inverse then its Gelfand transform $\phi(T)$ never takes the value $0$ and therefore has an inverse, namely the function $f(x) = 1/((\phi(T))(x))$. The inverse Gelfand transform $\phi^{-1}(f)$ belongs to $A$ and is equal to $T^{-1}$.On a seperate but related note, why is A closed under inverses i.e. if T has an inverse, then the inverse is in A

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Taking as a concrete example the set of polynomials generated by T and it's adjoint, you are saying that the map $p(T,T^*)$->$p(x,x*)$ where x is any fixed member of the spectrum is a homomorphism? Does it preserve the 'multiplication? Well, given polynomials p,g, we have $p(g(T,T^*))$->$p(g(x,x*))$ but that does not equal $p(x,x*)$$g(x,x*)$ which is the multiplication in C. Or am I missing something?

The problem is unchanged if we replace $A$ by its closure (in the operator norm topology). So we may as well assume that $A$ is closed. Now that you have added the crucial information that $A$ is selfadjoint (i.e. closed under the adjoint operation), it follows that $A$ is a commutative unital C*-algebra. The G–N theorem then tells you that there is an isometric isomorphism $\phi:A\to C(X)$ from $A$ to the algebra of continuous functions on a compact Hausdorff space $X$ (thestructure spaceorspectrumof $A$). It follows fairly easily that the spectrum of $T$ is equal to the range of the function $\phi(T)$. Also, the homomorphisms from $A$ to the scalars are exactly the point evaluation functions on $X$, in other words functions of the form $S\mapsto (\phi(S))(x)$ for $x\in X$ and $S\in A$. That is all you need to deduce the result.

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The product of the polynomials $p(T,T^*)$ and $g(T,T^*)$ is $p(T,T^*)g(T,T^*)$, not $p(g(T,T^*))$.Taking as a concrete example the set of polynomials generated by T and it's adjoint, you are saying that the map $p(T,T^*)$->$p(x,x*)$ where x is any fixed member of the spectrum is a homomorphism? Does it preserve the 'multiplication? Well, given polynomials p,g, we have $p(g(T,T^*))$->$p(g(x,x*))$ but that does not equal $p(x,x*)$$g(x,x*)$ which is the multiplication in C. Or am I missing something?

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Yes of course the operation is not composition of functions (polynomials) but composition of operators. Case closed.The product of the polynomials $p(T,T^*)$ and $g(T,T^*)$ is $p(T,T^*)g(T,T^*)$, not $p(g(T,T^*))$.