I read the following version of the spectral theorem in Banach Algebra Techniques in Operator Theory by Douglas:

enter image description here

I'm trying to understand why this is a generalization of the following version, which is most common in an undergraduate functional analysis course:

Theorem 1. Suppose $A$ is a compact self-adjoint operator on a Hilbert space $V$. There is an orthonormal basis of $V$ consisting of eigenvectors of $A$. Each eigenvalue is real.

or the version for normal matrix in undergraduate linear algebra:

Theorem 2 An $n\times n$ complex matrix $A$ is normal if and only if there exists a unitary matrix $U$ such that $$ A=U D U^*, $$ where $D$ is a diagonal matrix.

It is pain in the neck to read the abstract version of the theorem without knowing how it generalize the simpler versions. Could anybody explain how is Theorem 4.30 a generalization? Especially,

  • Where is the Gelfand transform in Theorem 1 and Theorem 2?
  • How does the "Gelfand map is a *-isometric isomorphism of $\mathfrak{C}_T$ onto $C(\sigma(T))$" correspond to the eigenvector-eigenvalue statement in Theorem 1 and Theorem 2?

This is not the spectral theorem for normal operators on an arbitrary Hilbert space. It is merely a special case of the Gelfand representation. It is however this special case which gives rise to the full spectral theorem, in about 10 more pages of additional theory.

My answer consists of two parts:

  1. I will prove the spectral theorem from linear algebra using the version you found (Theorem 4.30). The version for compact operators can be deduced in a similar way, but not nearly as easily.
  2. I will try to give an overview of the full spectral theorem and its proof (with references).

Part 1: deducing the finite-dimensional spectral theorem from the Gelfand transform.

Spectral theorem (finite-dimensional version). Let $N$ be an $n\times n$ normal matrix over $\mathbb{C}$. Then there is an orthonormal basis $\mathcal{B} = \{b_1,\ldots,b_n\}$ of $\mathbb{C}^n$ consisting of eigenvectors of $N$.

Proof. Recall that the spectrum $\sigma(N)$ is precisely the set of eigenvalues in the finite-dimensional case. (This is of course no longer true in the infinite-dimensional case.) Thus, $\sigma(N)$ is a non-empty set containing at most $n$ points. Let $\phi : \mathfrak C_N \to C(\sigma(N))$ denote the Gelfand representation from Theorem 4.30. Now the composition $$ C(\sigma(N)) \stackrel{\phi^{-1}}{\longrightarrow} \mathfrak C_N \hookrightarrow B(\mathbb C^n) $$ gives us a unital, isometric $*$-homomorphism $\varphi : C(\sigma(N)) \to B(\mathbb C^n)$ with image $\mathfrak C_N$. This homomorphism is usually called the functional calculus at $N$. It has the property $$ \varphi(\iota) = N, $$ where $\iota : \sigma(N) \to \mathbb{C}$ denotes the inclusion $x\mapsto x$. (In the literature, $\iota$ is usually called $z$.)

Note that $\sigma(N)$ is equipped here with the subspace topology from $\mathbb C$, which is just the discrete topology since it is a finite Hausdorff space. Thus, every function $\sigma(N) \to \mathbb C$ is automatically continuous.

For $\lambda \in \sigma(N)$ let $\chi_\lambda \in C(\sigma(N))$ denote the indicator function of the singleton $\{\lambda\}$, that is: $$ \chi_\lambda(\mu) = \begin{cases} 1,&\quad\text{if $\lambda = \mu$};\\[1ex] 0,&\quad\text{if $\lambda \neq \mu$}. \end{cases} $$ Furthermore, define $P_\lambda := \varphi(\chi_\lambda)$. We prove some properties of the $P_\lambda$. Let $\lambda_1,\ldots,\lambda_m \in \sigma(N)$ denote all the different eigenvalues of $N$.

  1. Note that $P_\lambda$ is an orthogonal projection, since we have \begin{equation} P_\lambda^2 = \varphi(\chi_\lambda)^2 = \varphi(\chi_\lambda^2) = \varphi(\chi_\lambda) = P_\lambda,\\[1ex] P_\lambda^* = \varphi(\chi_\lambda)^* = \varphi(\chi_\lambda^*) = \varphi(\chi_\lambda) = P_\lambda. \end{equation}
  2. Secondly, for $\lambda,\mu\in\sigma(N)$ with $\lambda \neq \mu$ we have $$ P_\lambda P_\mu \: = \: \varphi(\chi_\lambda)\varphi(\chi_\mu) \: = \: \varphi(\chi_\lambda\cdot \chi_\mu) \: = \: \varphi(0) \: = \: 0. $$ It follows from this that $\text{im}(P_\lambda)$ and $\text{im}(P_\mu)$ are orthogonal subspaces of $\mathbb{C}^n$ (exercise).
  3. Thirdly, we have $\mathbb{1} = \chi_{\lambda_1} + \cdots + \chi_{\lambda_m}$ in $C(\sigma(N))$, where $\mathbb{1} : \sigma(N) \to \mathbb C$ denotes the constant one function $x \mapsto 1$. Thus, since $\varphi$ is unital, we have $$ I \: = \: \varphi(\mathbb{1}) \: = \: \varphi\big(\chi_{\lambda_1} + \cdots + \chi_{\lambda_m}\big) \: = \: P_{\lambda_1} + \cdots + P_{\lambda_m}. $$
  4. Analogously, $\iota$ can be written as $\iota \: = \: \lambda_1 \cdot \chi_{\lambda_1} + \cdots + \lambda_m \cdot \chi_{\lambda_m}$, so we have $$ N \: = \: \varphi(\iota) \: = \: \varphi\big(\lambda_1 \cdot \chi_{\lambda_1} + \cdots + \lambda_m \cdot \chi_{\lambda_m}\big) \: = \: \lambda_1P_{\lambda_1} + \cdots + \lambda_m P_{\lambda_m}. $$

It follows from 1–3 that we have a decomposition of $\mathbb{C}^n$ into pairwise orthogonal subspaces: $$ \mathbb C^n \: = \: \text{im}(P_{\lambda_1}) \oplus \cdots \oplus \text{im}(P_{\lambda_m}). $$ Thus, we can choose an orthonormal basis $\mathcal B = \{b_1,\ldots,b_n\}$ of $\mathbb C^n$ such that every $b_j$ is a member of some $\text{im}(P_{\lambda_k})$. We prove that every member of $\mathcal B$ is an eigenvector of $N$. Let $b_j \in \text{im}(P_{\lambda_k})$ be given, then we have $b_j = P_{\lambda_k}b_j$, hence $$ Nb_j \: = \: NP_{\lambda_k}b_j \: = \: \big(\lambda_1 P_{\lambda_1} + \cdots + \lambda_m P_{\lambda_m}\big)P_{\lambda_k}b_j \: = \: \lambda_k P_{\lambda_k}^2b_j \: = \: \lambda_k\cdot b_j. $$ Indeed this shows that $b_j$ is an eigenvector, completing the proof.$\hspace{3mm}\blacksquare$

The version for compact self-adjoint operators can be deduced in a similar way, but this requires several theorems about compact operators. (You have to prove that the spectrum has no limit points except possibly $0$, that every non-zero $\lambda \in \sigma(N)$ is an eigenvalue and that every non-zero eigenvalue has finite multiplicity.)

Part 2: the spectral theorem for normal operators on an arbitrary Hilbert space.

The general spectral theorem is stronger yet than the one from Theorem 4.30. For a normal operator $N \in B(\mathscr H)$, the continuous functional calculus at $N$ is the unique unital $*$-homomorphism $\varphi : C(\sigma(N)) \to B(\mathscr H)$ such that $\varphi(\iota) = N$ holds, where $\iota : \sigma(N) \to \mathbb C$ once again denotes the inclusion $x \mapsto x$. The continuous functional calculus is automatically an isometry. It can be obtained through the Gelfand representation and it exists even for arbitrary unital $C^*$-algebras.

For a compact Hausdorff space $X$ we let $B^\infty(X)$ denote the space of bounded Borel measurable functions $X\to \mathbb C$. This becomes a $C^*$-algebra when equipped with the supremum norm.

Claim. For the specific $C^*$-algebra $B(\mathscr H)$ we can extend the continuous functional calculus $\varphi : C(\sigma(N)) \to B(\mathscr H)$ to a unital $*$-homomorphism $\tilde\varphi : B^\infty(\sigma(N)) \to B(\mathscr H)$.

Sketch of proof. Fix $x,y \in \mathscr H$, then we may consider the linear functional $\psi_{x,y} : C(\sigma(N)) \to \mathbb C$ defined by $$ \psi_{x,y}(f) = \langle \varphi(f)x,y\rangle. $$ This is a bounded linear functional with norm at most $||x||\cdot ||y||$. By the Riesz representation theorem there exists some measure $\mu_{x,y}$ on $\sigma(N)$ such that $$ \psi_{x,y}(f) = \int f\: d\mu_{x,y},\qquad\text{for all $f\in C(\sigma(N))$.} $$ One easily shows that the map $(x,y) \mapsto \mu_{x,y}$ is sesquilinear. Also, for fixed $f \in B^\infty(\sigma(N))$ we can show that the map $(x,y) \mapsto \int f\: d\mu_{x,y}$ is a bounded sesquilinear form. Hence there exists a unique $A_f \in B(\mathscr H)$ such that $\langle A_fx,y\rangle = \int f\: d\mu_{x,y}$ holds for all $x,y\in\mathscr H$. A large technical proof now shows that the map $f \mapsto A_f$ defines a unital $*$-homomorphism $B^\infty(\sigma(N)) \to B(\mathscr H)$ extending $\varphi$. $\hspace{3mm}\blacksquare$

The above result is the full spectral theorem for bounded normal operators on an arbitrary Hilbert space, and the extended unital $*$-homomorphism $\tilde\varphi : B^\infty(\sigma(N)) \to B(\mathscr H)$ is called the Borel functional calculus at $N$. The above result makes the theory of Hilbert space operators a lot easier. For instance, try the following exercise with and without the spectral theorem.

Exercise. Let $T \in B(\mathscr H)$ be self-adjoint. Now $T^2 = T^*T$ is a positive operator, so it has a unique positive square root $|T|$. Define $T^+ = \tfrac{1}{2}(|T| + T)$ and $T^- = \tfrac{1}{2}(|T| - T)$.

  1. Show that $T^+$ and $T^-$ are positive.
  2. Prove that $T^+T^- = T^-T^+ = 0$ holds.
  3. Show that $\mathscr H$ has an orthonormal basis $\mathcal B$ such that $|\langle Tb,b\rangle| = \langle |T|b,b\rangle$ holds for all $b\in\mathcal B$.

Further reading:

  • John B. Conway, A Course In Functional Analysis, Sections IX.1 and IX.2. (After presenting the spectral theorem for bounded normal operators on a Hilbert space in about 10 pages, he asks the reader to deduce the spectral theorem for compact, self-adjoint operators as a special case in exercise IX.2.3. Furthermore, section X.4 contains the spectral theorem for unbounded normal operators on a Hilbert space.)
  • Ronald G. Douglas, Banach Algebra Techniques in Operator Theory, remainder of chapter 4 up until theorem 4.71. (I personally feel that the presentation of the material is not as clear as Conway's, which is partially due to the lack of subdivisions in the chapters. Perhaps this also reflects my unfamiliarity with this particular book.)
  • Gerard J. Murphy, $C^*$-algebras and Operator Theory, section 2.5. (Again the presentation is not as clear as Conway's, but the rest of the book is a great resource for the theory of $C^*$-algebras.)
  • $\begingroup$ Thank you very much for your answer. In your proof of the finite dimension version spectral theorem, is $\varphi$ is inverse of the Gelfand transform? is $B(\mathbb{C}^n)$ the $C^*$-algebra generated by the normal matrix $N$? $\endgroup$ – Jack Feb 28 '16 at 23:40
  • $\begingroup$ Ah, no, $B(\mathbb C^n)$ is just the $C^*$-algebra of bounded linear operators $\mathbb C^n \to \mathbb C^n$. Douglas calls this $\mathfrak L(\mathscr H)$. If you compose the inverse Gelfand transform $\phi^{-1} : C(\sigma(N)) \to \mathfrak C_N$ with the inclusion $\mathfrak C_N \hookrightarrow B(\mathbb C^n)$, then you get the homomorphism I call $\varphi$. $\endgroup$ – Josse van Dobben de Bruyn Mar 1 '16 at 17:04

If $A$ is a normal matrix, then $$ A = \lambda_1 P_1 + \lambda_2 P_2 + \cdots +\lambda_k P_k, $$ where the $P_k$ are orthogonal projections (i.e., $P_k^2=P_k = P_k^{\star}$) such that $$ P_1 + P_2 + \cdots + P_k = I \\ P_k P_j = 0,\;\;\; j \ne k \\ P_k P_k = P_k. $$ The algebra of operators is in correspondence with continuous functions on $\sigma(A)=\{\lambda_1,\lambda_2,\cdots,\lambda_k\}$ through the correspondence $$ f(A) = f(\lambda_1)P_1 + f(\lambda_2)P_2 + \cdots + f(\lambda_k)P_k. $$ It's easy to check that $(fg)(A)=f(A)g(A)$ and $1(A)=I$. Let $e_j = \chi_{\{\lambda_j\}}$ be the characteristic function of $\lambda_j$. Then $P_j=e_j(A)$. Every operator $N$ in the $C^{\star}$ algebra generated by $A$ is uniquely represented as $N=f(A)$. The point evaluations $\omega_j$ of $f$ at elements of the spectrum are the complex homomorphisms of the $C^{\star}$ algebra. In this version of the spectral theorem, $\omega_j$ is associated with $\lambda_j$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.