Theorem 12.23 Rudin's functional analysis - Specialization to a single operator and example of spectral decomposition.

Question

In Rudin's functional analysis the Spectral Theorem theorem is worked out by first proving the following theorem.

Theorem 12.22 If $A$ is a closed normal subalgebra of $\mathcal{B}(H)$ which contains the identity operator $I$ and if $\Delta$ is the maximal ideal space of $A$, then the following assertions are true:
a) There exist a unique resolution $E$ of the identity on the Borel subsets of $\Delta$ which satisfies $$ T = \int_{\Delta} \hat{T}dE \;\; (1) $$ for every $T \in A$, where $\hat{T}$ is the Gelfand transform of $T$.
(b) The inverse Gelfand transform (i.e. the map that takes $\hat{T}$ back to $T$) extends to an isometric *-isomorphism $\Phi$ of the algebra $L^{\infty}(E)$ onto a closed subalgebra $B$ of $\mathcal{B}(H)$, $A \subset B$, given by $$ \Phi f = \int_{\Delta} f dE \;\; (f \in L^{\infty}(E)) \;\; (2) $$ Explicitly, $\Phi$ is a lionear and multiplicative and satisfies $$ \Phi \overline{f} = (\Phi f)^* , \left\lVert\ \Phi f \right\rVert = \left\lVert f \right\rVert_{\infty} \;\; (3) $$ c) $B$ is in the closure [in the norm topology of $\mathcal{B}(H)$] of the set of all finite linear combinations of the projections $E(\omega)$.
d) If $\omega \subset \Delta$ is open and non empty, then $E(\omega) \neq 0$
e) An operator $S \in \mathcal{B}(H)$ commutes with every $T \in A$ if and only if $S$ commutes with every projection $E(\omega)$.

After the proof of theorem 12.22 we have the statement

We now specilize this theorem to a single operator.

And we have the proof of

Theorem 12.23. If $T \in \mathcal{B}(H)$ and $T$ is normal, then there exists a unique resolution of the identity $E$ of the Borel subsets of $\sigma(T)$ which satisfies $$ T = \int_{\sigma(T)} \lambda dE(\lambda) $$ Furthermore, every projection $E(\omega)$ commutes with every $S \in \mathcal{B}(H)$ which commutes with $T$.

I have two questions:

1. This might be silly but why does the author say "specialize to a single operator". I am trying to compare statement (a) of theorem 12.22 against the statement of theorem 12.23 but I cannot spot the reason why it is claimed 12.23 is "specialized to a single operator".

2. Suppose we consider $T$ as orthonormal matrix in $\mathbb{R}^n$ then the matrix can be diagonalized, and I believe the spectral decompositon in this case can be written as

$$ T = \int_{\sigma(T)} \lambda dE(\lambda) = \sum_{i = 1}^n \lambda_i \left\langle u_i, \cdot \right\rangle \left\langle \cdot, u_i \right\rangle $$

where $u_1,\ldots, u_n$ are the eigenvectors of $T$. I am not sure I am going to phrase this correctly but I assume that from the equality

$$ \int_{\sigma(T)} \lambda dE(\lambda) = \sum_{i = 1}^n \lambda_i \left\langle u_i, \cdot \right\rangle \left\langle \cdot, u_i \right\rangle $$

I should be able to derive an expression for the spectral measure $E$ or some form of relationship, which I cannot work out, can anyone elaborate on that?

(If the question isn't clear please point which bit isn't and I'll try to phrase it better).

Answer 1

For 1., look at the first line of the proof of Theorem 12.23: "Let A be the smallest closed subalgebra of that contains...". Theorem 12.22 is formulated in forms of subalgebras (that contain more than a single operator).

For 2., I don't really get what your question is, but in your case of an orthonormal matrix $T$, we have $$ T = \sum_{i = 1}^{n}{\lambda_i \langle u_i, \cdot \rangle u_i}. $$ So the corresponding projection-valued measure is $$ E(A) = \sum_{i = 1}^{n}{\delta_{\lambda_i}(A) \langle u_i, \cdot \rangle u_i } $$ for $A \in \mathcal{B}(\mathbb{R})$.