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In my functional analysis lecture, we have covered two spectral theorems so far. The first one being:

$\underline{\textbf{Theorem 1:}}$ Suppose $\mathfrak{H}$ is a Hilbert space and $A\in\mathcal{K}(\mathfrak{H})$ is a compact symmetric (i.e., self-adjoint) operator. Then there exists a possibly finite sequence of real nonzero eigenvalues $(\alpha_j)_{j=1}^N$ listed with multiplicity which converges to $0$ if $N=\infty$. One can choose corresponding orthonormal eigenvectors $u_j$ such that every $f\in\mathfrak{H}$ can be written as $$ f = \sum_{j=1}^N \langle u_j,f\rangle u_j + h, $$ where $h$ is in the kernel of $A$, that is, $Ah=0$. Moreover, $u_j\in\operatorname{Ker}(A)^\perp$.

And the second one being:

$\underline{\textbf{Theorem 2:}}$ If $X$ is a $C^\ast$ algebra and $x\in X$ is self-adjoint, then there is an isometric isomorphism $$\Phi:C(\sigma(x))\to C^\ast(x) $$ such that $f(t)=t$ maps to $\Phi(t)=x$ and $f(t)=1$ maps to $\Phi(1)=e$. Moreover, for every $f\in C(\sigma(x))$ we have $$ \sigma(f(x)) = f(\sigma(x)), $$ where $f(x):=\Phi(f)$.

Since the lecture notes do not provide any mention of a relation between these two, I'm writing this post. Any intuition or help is highly appreciated!

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    $\begingroup$ $$\underline{\textbf{Please don't type like this!}}$$ $\endgroup$ Commented 13 hours ago

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I have never seen Theorem 2 mentioned as "Spectral Theorem". Rather, it's usually called Gelfand Representation or simply Continuous Functional Calculus.

The connection between the two results is precisely that they allow you to define functional calculus. Theorem 1 tells you that a compact selfadjoint operator $A$ is of the form $$ A=\sum_k\alpha_kP_k, $$ where $\{P_k\}$ are pairwise orthogonal rank-one projections. Concretely $P_kh=\langle h,u_k\rangle \,u_k$. This allows you to define, for a function $g:\sigma(A)\cup\{0\}\to\mathbb C$ continuous at $0$, $$ g(A)=\sum_kg(\alpha_k)P_k $$ (the continuity requirement is so that $g(A)$ is compact, but not necessary otherwise). The map $g\longmapsto g(A)$ is a $*$-homomorphism, so in particular for a polynomial $p$ the operator $p(A)$ is precisely what you would expect it to be; and the obvious manipulations work, like $(fg)(A)=f(A)g(A)$. There is a version of this theorem for arbitrary selfadjoint (or, more generally, normal) operators and Borel functions, where your series is replaced by an operator-valued integral in terms of an operator-valued measure.

Theorem 2 is an extremely powerful and everyday tool in the theory of C$^*$-algebras.

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  • $\begingroup$ First of all, thank you! If you are interested, it's Theorem 5.10 (p. 173-174) in these lecture notes. $\endgroup$ Commented 13 hours ago
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In the case of $X$ being the unital $C^*$-algebra generated by a compact operator $A$ (as a subalgebra of $B(H)$), it's easy to show that Theorem 1 implies Theorem 2 directly. But Theorem 2 is more general than 1 in the sense that it's about all commutative $C^*$-algebras, not just those generated by a single element (we need to generalize theorem 1 to commuting families of self-adjoint or normal operators to match 2 in this perspective).

But Theorem 2 to Theorem 1 is not obvious at all. Theorem 2 is about the intrinsic characterization of a commutative $C^*$-algebra, while Theorem 1 is a theorem about representation (of the algebra generated by $A$). If we compare them, the information about multiplicity of each eigenspace is lost when passing from the spectral decomposition of a self-adjoint operator to the Gelfand spectrum. And when $A$ is not compact, multiplicities are difficult to define (instead of direct sum of eigenspaces, we need a direct integral against a projection valued measure), and that's why the spectral theorem for compact operators are easier than general self-adjoint or normal operators.

To make an analogy with algebraic geometry, theorem 2 is about the spectrum of a ring $R$, while theorem 1 is about modules over $R$. In the special case of (finite dimensional) linear algebra, we can say theorem 1 is about diagonalization of a (self-adjoint) matrix $A$, while theorem 2 is saying essentially that if the minimal polynomial $p(x)$ has no repeated root over $\mathbb C$, then $$\mathbb C[x]/\langle p(x) \rangle =\mathbb C[x]/\langle (x-c_1)(x-c_2)\cdots (x-c_n)\rangle \simeq \prod_{i=1}^n \mathbb C[x]/(x-c_i)\simeq\prod_{i=1}^n \mathbb C$$

But we all know that the minimal polynomial doesn't capture all needed to know about a matrix (the multiplicities of eigenspaces are lost).

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