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Let $H$ be a complex Hilbert space and $\mathcal A$ be a commutative unital $C^{\ast}$-subalgebra of $B (H).$ Then the Gelfand transform gives a isometric-$\ast$-isomorphism from $\mathcal A$ onto $C \left (\Sigma \right ),$ where $\Sigma$ is the set of all non-zero multiplicative linear functionals on $\mathcal A.$ Let $T_f \in \mathcal A$ be the image of $f \in C \left (\Sigma \right )$ under the inverse Gelfand transform. Then $f \mapsto \left \langle T_f u, v \right \rangle$ is a bounded linear functional on $C \left (\Sigma \right )$ for any $u, v \in H.$ So by Riesz-Markov-Kakutani's Theorem we can get hold of a unique complex regular Borel measure $\mu_{u,v}$ (depending upon $u, v \in H$) on $\Sigma$ such that for all $f \in C \left (\Sigma \right )$ $$\left \langle T_f u, v \right \rangle = \int_{\Sigma} f\ d\mu_{u,v}.$$ Now for any $f \in B \left (\Sigma \right )$ (bounded Borel measurable function on $\Sigma$) we consider the map $\psi_f : H \times H \longrightarrow \mathbb C$ defined by $$\psi_f (u, v) = \int_{\Sigma} f\ d\mu_{u, v}.$$ Then it can be easily shown that $\psi_f$ is a bounded sesquilinear form on $H.$ So by Riesz Representation Theorem we can get hold of a unique $T_f \in B (H)$ such that $$\left \langle T_f u, v \right \rangle = \int_{\Sigma} f\ d\mu_{u,v}.$$ Consider the map $f \mapsto T_f$ from $B \left (\Sigma \right)$ into $B(H).$ Then it can be shown that this map is a $\ast$-algebra homomorphism. This is known as $L^{\infty}$-Functional Calculus.

Let $\mathscr B \left (\Sigma \right )$ be the Borel $\sigma$-algebra of subsets of $\Sigma.$ Now define a map $P : \mathscr B \left (\Sigma \right ) \longrightarrow B(H)$ by $P (E) = T_{\chi_{E}}.$ Then for any $E, F \in \mathscr B \left (\Sigma \right )$ the following holds $:$

(a) $P(E) = P(E)^2 = P(E)^{\ast}$ i.e. $P(E)$ is an orthogonal projection.

(b) $P(\varnothing) = 0$ and $P \left (\Sigma \right ) = I.$

(c) $P(E \cap F) = P(E) P(F).$

(d) If $E_1, E_2, \cdots$ are disjoint Borel subsets of $\Sigma,$ then $$P \left (\bigcup\limits_{j = 1}^{\infty} E_j \right ) = \sum\limits_{j = 1}^{\infty} P(E_j),$$ where the infinite sum of projections on the right converges in the strong operator topology.

Such a map $P$ is called a projection valued measure.

Now I would like to prove spectral theorem for normal matrices using the well-known Gelfand's theory that I summarize above. For a normal matrix $A \in M_n \left (\mathbb C \right ),$ we can consider the $C^{\ast}$-subalgebra $\mathcal A$ of $M_n (\mathbb C)$ generated by $A$ and $I.$ Since $A$ is normal, it follows that $\mathcal A$ is a commutaive unital $C^{\ast}$-algebra. Then the map $\varphi \mapsto \varphi (A)$ defines a homeomorphism from $\Sigma$ (the set of non-zero multiplicative linear functionals on $\mathcal A$) onto $\sigma (A)$ (the spectrum of $A$). Here $\sigma (A)$ consists of eigenvalues of $A.$ Since $\sigma (A)$ is finite, the subspace topology on $\Sigma$ would be discrete and hence any function on $\Sigma$ would be continuous. Now for distinct eigenvalues $\lambda, \mu \in \sigma (A),$ the third property of the projection valued measure yields $$P(\{\lambda\}) P(\{\mu \}) = P(\{\mu \}) P(\{\lambda \}) = 0.$$ This shows that $P(\{\lambda\})$ and $P(\{\mu \})$ are mutually orthogonal projections. This reminds me of the spectral theorem for normal matrices, namely, the eigenspaces of a normal matrix corresponding to distinct eigenvalues are orthogonal to each other. But the problem is that I cannot see why $P(\{\lambda \})$ is the orthogonal projection onto the eigenspace $E_{\lambda}$ of $\lambda.$ For that we need to show that $\text {Ran} \left (P \left (\{\lambda \} \right ) \right ) = E_{\lambda}$ for any $\lambda \in \sigma (A).$ It is clear that $P(\{\lambda\}) \in \mathcal A$ (since $\chi_{\{\lambda\}}$ is continuous on $\Sigma$) and hence it commutes with both $A$ and $A^{\ast}.$ Hence $E_{\lambda}$ is an invariant subspace of $P(\{\lambda\}).$ So $E_{\lambda} \subseteq \text {Ran} \left (P \left (\{\lambda \} \right ) \right ).$

Is the reverse inclusion also trivial in some sense from whatever I know? Any suggestion in this regard would be highly appreciated.

Thanks in advance.

Anacardium
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1 Answers1

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Let $x: \sigma(A) \to \sigma(A)$ denote the identity map and recall that $A = T_x$. Let the spectrum of $A$ be $\{\lambda_1, \cdots, \lambda_k\}$, where $\lambda = \lambda_1$. We first note that, on $\sigma(A)$,

$$\chi_{\{\lambda\}}(x) = \frac{\prod_{i=2}^k (x - \lambda_i)}{\prod_{i=2}^k (\lambda - \lambda_i)}$$

So if $v$ is s.t. $Av = \lambda v$, then,

$$P(\{\lambda\})v = T_{\chi_{\{\lambda\}}}v = \frac{\prod_{i=2}^k (A - \lambda_i)}{\prod_{i=2}^k (\lambda - \lambda_i)}v = \frac{\prod_{i=2}^k (\lambda - \lambda_i)}{\prod_{i=2}^k (\lambda - \lambda_i)}v = v$$

This shows $E_\lambda \subset \text{range}(P(\{\lambda\}))$. Conversely, since $x \cdot \chi_{\{\lambda\}} = \lambda \chi_{\{\lambda\}}$, we have, for any $v \in \text{range}(P(\{\lambda\}))$,

$$Av = AP(\{\lambda\})v = T_{x \cdot \chi_{\{\lambda\}}}v = T_{\lambda \chi_{\{\lambda\}}}v = \lambda P(\{\lambda\})v = \lambda v$$

So $v \in E_\lambda$, i.e., $\text{range}(P(\{\lambda\})) \subset E_\lambda$ as well.

Let me remark that the second half of the argument works for any normal operator on any complex Hilbert space, while the first half assumes the spectrum is finite. However, the result that $E_\lambda = \text{range}(P(\{\lambda\}))$ is actually true for all normal operators, so I’ll provide an alternative argument that applies to the general case here:

Let $v$ be s.t. $Av = \lambda v$. Again, as $A = T_x$, for any polynomial $f$, it is straightforward to check that $T_fv = f(\lambda)v$. As polynomials are dense in $C(\sigma(A))$ by Stone-Weierstrass, we have $T_fv = f(\lambda)v$ for any $f \in C(\sigma(A))$. But then, for any $w \in H$ and $f \in C(\sigma(A))$,

$$\langle T_fv, w \rangle = \langle v, w \rangle f(\lambda)$$

So $\mu_{v, w} = \langle v, w \rangle \delta_\lambda$, where $\delta_\lambda$ is the Dirac delta at $\lambda$. Hence,

$$\langle P(\{\lambda\})v, w \rangle = \langle v, w \rangle \chi_{\{\lambda\}}(\lambda) = \langle v, w \rangle$$

As this holds for all $w \in H$, we conclude that $P(\{\lambda\})v = v$, i.e., $v \in \text{range}(P(\{\lambda\}))$.

David Gao
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