# A Sledgehammer Proof of the Spectral Theorem for Normal Matrices

My research is about the structure of homomorphisms between certain C*-algebras. To illustrate what I mean and its possible importance, consider an elementary case given below.

Proposition. Let $X$ be a compact, Hausdorff space, $N$ be a positive integer, and $\phi\colon C(X) \to M_N$ be a homomorphism. There exist mutually orthogonal rank one projections $p_n$  and points $\xi_n\in X$ for $n=1,2,\dotsc,N$ such that

$\phi(f) = \sum_{n=1}^{N} f(\xi_n) p_n$

for all $f \in C(X)$.

Proof: Since $C(X)/\ker\,\phi$ is a commutative C*-algebra, by Gelfand’s representation theorem, there exists a compact Hausdorff space $Y$ such that $C(X)/\ker\,\phi \cong C(Y)$. Furthermore, since there is a surjective homomorphism from $C(X)$ onto $C(Y)$, there is a continuous embedding from $Y$ into $X$. We assume, without loss of generality, that $Y\subseteq X$ and that the canonical projection homomorphism is given by restriction.

Let $\phi'\colon C(Y) \to M_N$ be the injective homomorphism induced by $\phi$. Since $\phi'$ is an injective linear map and $M_N$ is finite dimensional, $Y$ is finite (i.e. consists of finitely many points). Label these points $\xi_n$ and let $\chi_n \in C(Y)$ denote the indicator functions for $\{\xi_n\}$. Notice that $\phi'(\chi_n)$ are mutually orthogonal projections. Each projection $\phi'(\chi_n)$ can be decomposed into the sum of mutually orthogonal rank one projections $p_n$. By relabeling and repeating $\xi_n$ as necessary, we assume that $p_n$ is associated with the corresponding point $\xi_n$.

Let $\pi_n\colon M_N \to p_n M_N p_n$ be the usual positive linear map. Since $p_n$ has rank one, $p_n M_N p_n \cong \mathbb{C}$. So it is easy to see that $\pi_n \circ \phi'$ is a state of $C(Y)$ and so $\pi_n \circ \phi'(f) = f(\xi_n)$. Therefore,

$\phi(f) = \sum_{n=1}^{N}\pi_n\circ \phi(f) = \sum_{n=1}^{N} f(\xi_n) p_n$.    $\Box$

We can rephrase this proposition to resemble linear algebra more. Since rank one projections correspond to one-dimensional subspaces, we can choose an orthonormal basis $(\delta_n)$, where $\delta_n$ is a unit vector in the range of $p_n$. So there exists a unitary $u \in M_N$ such that $u\delta_n = \epsilon_n$, where $\epsilon_n$ is the standard basis for $\mathbb{C}^n$. So our proposition above can be restated as:

Proposition. Let $X$ be a compact, Hausdorff space, $N$ be a positive integer, and $\phi\colon C(X) \to M_N$ be a homomorphism. There exist points $\xi_n\in X$ for $n=1,2,\dotsc,N$ and a unitary matrix $u\in M_N$ such that

$u\phi(f)u^{*} = \left(\begin{array}{c c c c}f(\xi_1) & 0 & \cdots & 0 \\ 0 & f(\xi_2) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & f(\xi_N) \end{array}\right)$

for all $f \in C(X)$.

To apply this theorem, let $a \in M_N$ be a normal matrix. Then by continuous functional calculus, there is a homomorphism from $C(X)$ to $M_N$ denoted $f \mapsto f(a)$, where $X = \mathrm{spec}(a)$. In particular, the inclusion function $z \mapsto z$ is sent to the matrix $a$. By applying the proposition to this homomorphism and plugging in the inclusion function, we obtain:

Theorem (Spectral Theorem for Normal Matrices). Let $N$ be a positive integer. For any normal matrix $a \in M_N$, there exist $\xi_n\in \mathbb{C}$ for $n=1,2,\dotsc,N$ and a unitary matrix $u\in M_N$ such that

$u a u^{*} = \left(\begin{array}{c c c c}\xi_1 & 0 & \cdots & 0 \\ 0 & \xi_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \xi_N \end{array}\right)$

As stated, this is the rather significant finite dimensional spectral theorem using rather excessively powerful tools for the proof. But we do obtain a stronger version for “free”. Let $S\subseteq M_N$ be a set of commuting normal matrices. Using the spectral theorem above, $C^{*}(S)$ is a commutative C*-subalgebra of $M_N$ and so by Gelfand’s representation theorem, $C^{*}(S) \cong C(X)$ for some compact Hausdorff space $X$. So we obtain a homomorphism from $C(X)$ to $M_N$ defined by composing the Gelfand transform with inclusion. Since each element of $S$, corresponds to some continuous function of $X$, by applying our proposition and plugging those functions, we obtain a simultaneous diagonalization of the matrices in $S$:

Theorem. Let $N$ be a positive integer. For any set $S \subseteq M_N$ of commuting normal matrices, there exists a unitary $u \in M_N$, such that for any $a\in S$, there exist $\xi_n\in \mathbb{C}$ for $n=1,2,\dotsc,N$ such that

$u a u^{*} = \left(\begin{array}{c c c c}\xi_1 & 0 & \cdots & 0 \\ 0 & \xi_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \xi_N \end{array}\right)$